Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 15 - Tangents And Normals

Download PdfDownload Pdf

Share with your friends

Rd Sharma XII Vol 1 2020 Solutions for Class 12 Science Maths Chapter 15 Tangents And Normals are provided here with simple step-by-step explanations. These solutions for Tangents And Normals are extremely popular among Class 12 Science students for Maths Tangents And Normals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Science Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 15.10:

Question 1:

Find the slopes of the tangent and the normal to the following curves at the indicted points:

(i) $y = \sqrt{x^{3}} at x = 4$
(ii) $y = \sqrt{x} at x = 9$
(iii) y = x³ − x at x = 2
(iv) y = 2x² + 3 sin x at x = 0
(v) x = a (θ − sin θ), y = a(1 − cos θ) at θ = −π/2
(vi) x = a cos³ θ, y = a sin³ θ at θ = π/4
(vii) x = a (θ − sin θ), y = a(1 − cos θ) at θ = π/2
(viii) y = (sin 2x + cot x + 2)² at x = π/2
(ix) x² + 3y + y² = 5 at (1, 1)
(x) xy = 6 at (1, 6)

Answer:

$(i) y = \sqrt{x^{3}} = x^{\frac{3}{2}} \Rightarrow \frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} = \frac{3}{2} \sqrt{x} When x =4, y = \sqrt{x^{3}} = \sqrt{64} = 8 Now, Slope of the tangent= {(\frac{d y}{d x})}_{(4, 8)} = \frac{3}{2} \sqrt{4} = 3 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(4, 8)}} = \frac{- 1}{3}$

$(i i) y = \sqrt{x} = x^{\frac{1}{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{2} x^{\frac{- 1}{2}} = \frac{1}{2 \sqrt{x}} When x =9, y = \sqrt{x} = \sqrt{9} = 3 Now, Slope of the tangent= {(\frac{d y}{d x})}_{(9, 3)} = \frac{1}{2 \sqrt{9}} = \frac{1}{6} Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(9, 3)}} = \frac{- 1}{(\frac{1}{6})} =-6$

$(i i i) y = x^{3} - x \Rightarrow \frac{d y}{d x} = 3 x^{2} - 1 When x =2, y = x^{3} - x = 2^{3} - 2 = 6 Now, Slope of the tangent= {(\frac{d y}{d x})}_{(2, 6)} =3 {(2)}^{2} -1=11 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(2, 6)}} = \frac{- 1}{11}$

$(i v) y = 2 x^{2} + 3 \sin x \Rightarrow \frac{d y}{d x} = 4 x + 3 \cos x When x =0, y = 2 x^{2} + 3 \sin x = 2 {(0)}^{2} + 3 \sin 0 = 0 Now, Slope of the tangent= {(\frac{d y}{d x})}_{(0, 0)} =4 (0) + 3 cos 0=3 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(0, 0)}} = \frac{- 1}{3}$

$(v) x = a (θ - \sin θ) \Rightarrow \frac{d x}{d θ} = a (1 - \cos θ) y = a (1 + \cos θ) \Rightarrow \frac{d y}{d θ} = a (- \sin θ) ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a (- \sin θ)}{a (1 - \cos θ)} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} = - c o t \frac{θ}{2} Now, Slope of the tangent= {(\frac{d y}{d x})}_{θ = \frac{- π}{2}} =-cot (\frac{\frac{- π}{2}}{2}) =-cot (\frac{- π}{4}) =1 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{θ = \frac{- π}{2}}} = \frac{- 1}{1} =-1$

$(v i) x = a \cos^{3} θ \Rightarrow \frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ y = a \sin^{3} θ \Rightarrow \frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 a \sin^{2} θ \cos θ}{- 3 a \cos^{2} θ \sin θ} = - \tan θ Now, Slope of the tangent= {(\frac{d y}{d x})}_{θ = \frac{π}{4}} =-tan \frac{π}{4} =-1 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{θ = \frac{π}{4}}} = \frac{- 1}{- 1} =1$

$(v i i) x = a (θ - \sin θ) \Rightarrow \frac{d x}{d θ} = a (1 - \cos θ) y = a (1 - \cos θ) \Rightarrow \frac{d y}{d θ} = a (\sin θ) ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a (\sin θ)}{a (1 - \cos θ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} = \cot \frac{θ}{2} Now, Slope of the tangent= {(\frac{d y}{d x})}_{θ = \frac{π}{2}} =cot (\frac{\frac{π}{2}}{2}) =cot (\frac{π}{4}) =1 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{θ = \frac{π}{2}}} = \frac{- 1}{1} =-1$

$(v i i i) y = {(\sin 2 x + \cot x + 2)}^{2} \Rightarrow \frac{d y}{d x} = 2 (\sin 2 x + \cot x + 2) (2 \cos 2 x - \cos e c^{2} x) Now, Slope of the tangent= {(\frac{d y}{d x})}_{x = \frac{π}{2}} = 2 [\sin 2 (\frac{π}{2}) + \cot (\frac{π}{2}) + 2] [2 \cos 2 (\frac{π}{2}) - {cosec}^{2} (\frac{π}{2})] = 2 (0 + 0 + 2) (- 2 - 1) = - 12 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{x = \frac{π}{2}}} = \frac{- 1}{- 12} = \frac{1}{12}$

$(i x) x^{2} + 3 y + y^{2} = 5 On differentiating both sides w.r.t. x, we get 2 x + 3 \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (3 + 2 y) = - 2 x \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{3 + 2 y} Now, Slope of the tangent= {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 2 x}{3 + 2 y} = \frac{- 2}{3 + 2} = \frac{- 2}{5} Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(1, 1)}} = \frac{- 1}{(\frac{- 2}{5})} = \frac{5}{2}$

$(x) x y = 6 On differentiating both sides w.r.t. x, we get x \frac{d y}{d x} + y = 0 \Rightarrow x \frac{d y}{d x} = - y \Rightarrow \frac{d y}{d x} = \frac{- y}{x} Now, Slope of the tangent= {(\frac{d y}{d x})}_{(1, 6)} = \frac{- y}{x} = \frac{- 6}{1} =-6 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(1, 6)}} = \frac{- 1}{- 6} = \frac{1}{6}$

Page No 15.10:

Question 2:

Find the values of a and b if the slope of the tangent to the curve xy + ax + by = 2 at (1, 1) is 2.

Answer:

$Given : x y + a x + b y = 2 . . . (1) On differentiating both sides w.r.t. x, we get x \frac{d y}{d x} + y + a + b \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (x + b) = - a - y \Rightarrow \frac{d y}{d x} = \frac{- a - y}{x + b} Now, {(\frac{d y}{d x})}_{(1, 1)} = 2 \Rightarrow \frac{- a - 1}{1 + b} = 2 \Rightarrow - a - 1 = 2 + 2 b \Rightarrow - a = 3 + 2 b \Rightarrow a = - (3 + 2 b) On substituting a = - (3 + 2 b), x =1 and y =1 in eq. (1), we get 1 - (3 + 2 b) + b = 2 \Rightarrow 1 - 3 - 2 b + b = 2 \Rightarrow b = - 4 and a = - (3 + 2 b) = - (3 - 8) = 5 ∴ a = 5 and b = - 4$

Page No 15.10:

Question 3:

If the tangent to the curve y = x³ + ax + b at (1, − 6) is parallel to the line x − y + 5 = 0, find a and b.

Answer:

$Given: x - y + 5 = 0 \Rightarrow y = x + 5 \Rightarrow \frac{d y}{d x} = 1 Now, y = x^{3} + a x + b . . . (1) \Rightarrow \frac{d y}{d x} = 3 x^{2} + a Slope of the tangent at (1, - 6) = Slope of the given line \Rightarrow {(\frac{d y}{d x})}_{(1, - 6)} = 1 \Rightarrow 3 + a = 1 \Rightarrow a = - 2 On substituting a= - 2, x =1 and y =-6 in eq. (1), we get - 6 = 1 - 2 + b \Rightarrow b = - 5 ∴ a = - 2 and b = - 5$

Page No 15.10:

Question 4:

Find a point on the curve y = x³ − 3x where the tangent is parallel to the chord joining (1, −2) and (2, 2).

Answer:

Let (x₁, y₁) be the required point.

$Slope of the chord = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 + 2}{2 - 1} = 4 y = x^{3} - 3 x \Rightarrow \frac{d y}{d x} = 3 x^{2} - 3 . . . (1) Slope of the tangent= {(\frac{d y}{d x})}_{(x_{1}, y_{1})} {=3x}_{1}^{2} -3 It is given that the tangent and the chord are parallel. ∴ Slope of the tangent = Slope of the chord \Rightarrow 3 {x_{1}}^{2} - 3 = 4 \Rightarrow 3 {x_{1}}^{2} = 7 \Rightarrow {x_{1}}^{2} = \frac{7}{3} \Rightarrow x_{1} = \pm \sqrt{\frac{7}{3}} = \sqrt{\frac{7}{3}} or - \sqrt{\frac{7}{3}} Case 1 When x_{1} = \sqrt{\frac{7}{3}} On substituting this in eq. (1), we get y_{1} = {(\sqrt{\frac{7}{3}})}^{3} - 3 (\sqrt{\frac{7}{3}}) = \frac{7}{3} \sqrt{\frac{7}{3}} - 3 \sqrt{\frac{7}{3}} = \frac{- 2}{3} \sqrt{\frac{7}{3}} ∴ (x_{1}, y_{1}) = (\sqrt{\frac{7}{3}}, \frac{- 2}{3} \sqrt{\frac{7}{3}}) Case 2 When x_{1} = - \sqrt{\frac{7}{3}} On substituting this in eq. (1), we get y_{1} = {(- \sqrt{\frac{7}{3}})}^{3} - 3 (- \sqrt{\frac{7}{3}}) = \frac{- 7}{3} \sqrt{\frac{7}{3}} + 3 \sqrt{\frac{7}{3}} = \frac{2}{3} \sqrt{\frac{7}{3}} ∴ (x_{1}, y_{1}) = (- \sqrt{\frac{7}{3}}, \frac{2}{3} \sqrt{\frac{7}{3}})$

Page No 15.10:

Question 5:

Find the points on the curve y = x³ − 2x² − 2x at which the tangent lines are parallel to the line y = 2x − 3.

Answer:

Let (x₁, y₁) be the required point.
Given:

$y = 2 x - 3 ∴ Slope of the line = \frac{d y}{d x} = 2 y = x^{3} - 2 x^{2} - 2 x Since (x_{1} y_{1}) lies on curve, y_{1} = {x_{1}}^{3} - 2 {x_{1}}^{2} - 2 x_{1} . . . (1) \Rightarrow {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 3 {x_{1}}^{2} - 4 x_{1} - 2 It is given that the tangent and the given line are parallel. ∴ Slope of the tangent = Slope of the given line 3 {x_{1}}^{2} - 4 x_{1} - 2 = 2 \Rightarrow 3 {x_{1}}^{2} - 4 x_{1} - 4 = 0 \Rightarrow 3 {x_{1}}^{2} - 6 x_{1} + 2 x_{1} - 4 = 0 \Rightarrow 3 x_{1} (x_{1} - 2) + 2 (x_{1} - 2) = 0 \Rightarrow (x_{1} - 2) (3 x_{1} + 2) = 0 \Rightarrow x_{1} = 2 or x_{1} = \frac{- 2}{3} Case 1 When x_{1} = 2 On s ubstituting the value of x_{1} in eq. (1), we get y_{1} = 8 - 8 - 4 = - 4 ∴ (x_{1}, y_{1}) = (2, - 4) Case 2 When x_{1} = \frac{- 2}{3} On s ubstituting the value of x_{1} in eq. (1), we get y_{1} = \frac{- 8}{27} - \frac{8}{9} + \frac{4}{3} = \frac{- 8 - 24 + 36}{27} = \frac{4}{27} ∴ (x_{1}, y_{1}) = (\frac{- 2}{3}, \frac{4}{27})$

Page No 15.10:

Question 6:

Find the points on the curve y² = 2x³ at which the slope of the tangent is 3.

Answer:

Let (x₁, y₁) be the required point.
Given:
$y^{2} = 2 x^{3} Since (x_{1} y_{1}) lies on a curve, {y_{1}}^{2} = 2 {x_{1}}^{3} . . . . (1) \Rightarrow 2 y \frac{d y}{d x} = 6 x^{2} \Rightarrow \frac{d y}{d x} = \frac{6 x^{2}}{2 y} = \frac{3 x^{2}}{y} Slope of the tangent at (x, y) = \frac{3 {x_{1}}^{2}}{y_{1}} S lope of the tangent=3 [Given] ∴ \frac{3 {x_{1}}^{2}}{y_{1}} = 3 . . . . (2) \Rightarrow y_{1} = {x_{1}}^{2} On substituting the value of y_{1} in eq. (1), we get {x_{1}}^{4} = 2 {x_{1}}^{3} \Rightarrow {x_{1}}^{3} (x_{1} - 2) = 0 \Rightarrow x_{1} = 0, 2 Case 1 When x_{1} = 0, y_{1} = x^{2} = 0 . Thus, we get the point (0, 0) . But, it does not satisfy eq. (2). So, we can ignore (0, 0). Case 2 When x_{1} = 2, y_{1} = {x_{1}}^{2} = 4 . Thus, we get the point (2, 4) .$

Page No 15.10:

Question 7:

Find the points on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45° with the x-axis.

Answer:

Let the required point be (x₁, y₁).
Slope of the tangent at this point = tan 45 $°$ = 1
Given:
$x y + 4 = 0 . . . (1) Since the point satisfies the above equation, x_{1} y_{1} + 4 = 0 . . . (2) On differentiating equation (2) both sides with respect to x, we get x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x, y)} = \frac{- y_{1}}{x_{1}} Slope of the tangent =1 [Given] ∴ \frac{- y_{1}}{x_{1}} = 1 \Rightarrow x_{1} = - y_{1} On substituting the value of x_{1} in eq. (2), we get - {y_{1}}^{2} + 4 = 0 \Rightarrow {y_{1}}^{2} = 4 \Rightarrow y_{1} = \pm 2 Case 1 When y_{1} = 2, x_{1} = - y_{1} = - 2 ∴ (x_{1}, y_{1}) = (-2, 2) Case 2 When y_{1} = - 2, x_{1} = - y_{1} = 2 ∴ (x_{1}, y_{1}) = (2, -2)$

Page No 15.10:

Question 8:

Find the point on the curve y = x² where the slope of the tangent is equal to the x-coordinate of the point.

Answer:

Let the required point be (x₁, y₁).
Given:
$y = x^{2} Point (x_{1}, y_{1}) lies on a curve . ∴ y_{1} = {x_{1}}^{2} . . . (1) Now, y = x^{2} \Rightarrow \frac{d y}{d x} = 2 x Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 2 x_{1} S lope of the tangent = x coordinate of the point [Given] ∴ 2 x_{1} = x_{1} This happens only when x_{1} = 0. On putting x_{1} = 0 in eq . (1), we get y_{1} = {x_{1}}^{2} = 0^{2} = 0 Thus, the required point is (0, 0) .$

Page No 15.10:

Question 9:

At what points on the circle x² + y² − 2x − 4y + 1 = 0, the tangent is parallel to x-axis?

Answer:

Let the required point be (x₁, y₁).
We know that the slope of the x-axis is 0.
Given:

$x^{2} + y^{2} - 2 x - 4 y + 1 = 0 (x_{1}, y_{1}) lies on a curve . ∴ {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 4 y_{1} + 1 = 0 . . . (1) Now, x^{2} + y^{2} - 2 x - 4 y + 1 = 0 \Rightarrow 2 x + 2 y \frac{d y}{d x} - 2 - 4 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (2 y - 4) = 2 - 2 x \Rightarrow \frac{d y}{d x} = \frac{2 - 2 x}{2 y - 4} = \frac{1 - x}{y - 2} Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1} - 2} ...(2) Slope of the tangent = 0 [Given] ∴ \frac{1 - x_{1}}{y_{1} - 2} = 0 \Rightarrow 1 - x_{1} = 0 \Rightarrow x_{1} = 1 On substituting the value of x_{1} in eq. (1), we get 1 + {y_{1}}^{2} - 2 - 4 y_{1} + 1 = 0 \Rightarrow {y_{1}}^{2} - 4 y_{1} = 0 \Rightarrow y_{1} (y_{1} - 4) = 0 \Rightarrow y_{1} = 0, 4 Thus, the required points are (1, 0) and (1, 4).$

Page No 15.10:

Question 10:

At what point of the curve y = x² does the tangent make an angle of 45° with the x-axis?

Answer:

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = x_{1} Now, y^{2} = x \Rightarrow 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1}{2 y_{1}} Given: \frac{1}{2 y_{1}} = 1 \Rightarrow 2 y_{1} = 1 \Rightarrow y_{1} = \frac{1}{2} Now, x_{1} = {y_{1}}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} ∴ (x_{1}, y_{1}) = (\frac{1}{4}, \frac{1}{2})$

Page No 15.10:

Question 11:

Find the points on the curve y = 3x² − 9x + 8 at which the tangents are equally inclined with the axes.

Answer:

Let (x₁, y₁) be the required point.
It is given that the tangent at this point is equally inclined to the axes. It means that the angle made by the tangent with the x-axis is $\pm$ 45 $°$ .
∴ Slope of the tangent = tan ( $\pm$ 45) = $\pm$ 1 ...(1)

$Since, the point lies on the curve . Hence, y_{1} = 3 {x_{1}}^{2} - 9 x_{1} + 8 Now, y = 3 x^{2} - 9 x + 8 \Rightarrow \frac{d y}{d x} = 6 x - 9 Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =6 x_{1} -9 ...(2) From eq. (1) and eq. (2), we get 6 x_{1} - 9 = \pm 1 \Rightarrow 6 x_{1} - 9 = 1 or 6 x_{1} - 9 = - 1 \Rightarrow 6 x_{1} = 10 or 6 x_{1} = 8 \Rightarrow x_{1} = \frac{10}{6} = \frac{5}{3} or x_{1} = \frac{8}{6} = \frac{4}{3} Also, y_{1} = 3 {(\frac{5}{3})}^{2} - 9 (\frac{5}{3}) + 8 or y_{1} = 3 {(\frac{4}{3})}^{2} - 9 (\frac{4}{3}) + 8 \Rightarrow y_{1} = \frac{25}{3} - \frac{45}{3} + 8 or y_{1} = \frac{16}{3} - \frac{36}{3} + 8 \Rightarrow y_{1} = \frac{4}{3} or y_{1} = \frac{4}{3} Thus, the required points are (\frac{5}{3}, \frac{4}{3}) and (\frac{4}{3}, \frac{4}{3}) .$

Page No 15.10:

Question 12:

At what points on the curve y = 2x² − x + 1 is the tangent parallel to the line y = 3x + 4?

Answer:

Let (x₁, y₁) be the required point.
The slope of line y = 3x + 4 is 3.

$Since, the point lies on the curve . Hence, y_{1} = 2 {x_{1}}^{2} - x_{1} + 1 Now, y = 2 x^{2} - x + 1 \frac{d y}{d x} = 4 x - 1 Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =4 x_{1} -1 Slope of the tangent at (x_{1}, y_{1}) = Slope of the given line [Given] ∴ 4 x_{1} - 1 = 3 \Rightarrow 4 x_{1} = 4 \Rightarrow x_{1} = 1 and y_{1} = 2 {x_{1}}^{2} - x_{1} + 1 = 2 - 1 + 1 = 2 Thus, the required point is (1, 2) .$

Page No 15.10:

Question 13:

Find the point on the curve y = 3x² + 4 at which the tangent is perpendicular to the line whose slop is $- \frac{1}{6}$ .

Answer:

Let (x₁, y₁) be the required point.
Slope of the given line = $\frac{- 1}{6}$
∴ Slope of the line perpendicular to it = 6

$Since, the point lies on the curve . Hence, y_{1} = 3 {x_{1}}^{2} + 4 Now, y = 3 x^{2} + 4 ∴ \frac{d y}{d x} = 6 x Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =6 x_{1} Slope of the tangent at (x_{1}, y_{1}) = Slope of the given line [Given] ∴ 6 x_{1} = 6 \Rightarrow x_{1} = 1 and y_{1} = 3 {x_{1}}^{2} + 4 = 3 + 4 = 7 Thus, the required point is (1, 7) .$

Page No 15.11:

Question 14:

Find the points on the curve x² + y² = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.

Answer:

Let (x₁, y₁) represent the required point.
The slope of line 2x + 3y = 7 is $\frac{- 2}{3}$ .

$Since, the point lies on the curve . Hence, {x_{1}}^{2} + {y_{1}}^{2} = 13 . . . (1) Now, x^{2} + y^{2} = 13 On differentiating both sides w.r.t. x, we get 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- x_{1}}{y_{1}} Slope of the tangent at (x_{1}, y_{1}) = Slope of the given line [Given] \Rightarrow \frac{- x_{1}}{y_{1}} = \frac{- 2}{3} \Rightarrow x_{1} = \frac{2 y_{1}}{3} . . . (2) From eq. (1), we get {(\frac{2 y_{1}}{3})}^{2} + {y_{1}}^{2} = 13 \Rightarrow \frac{13 {y_{1}}^{2}}{9} = 13 \Rightarrow {y_{1}}^{2} = 9 \Rightarrow y_{1} = \pm 3 \Rightarrow y_{1} = 3 or y_{1} = - 3 and x_{1} = 2 or x_{1} = - 2 [From eq. (2)] Thus, the required points are (2, 3) and (- 2, - 3) .$

Page No 15.11:

Question 15:

Find the points on the curve 2a²y = x³ − 3ax² where the tangent is parallel to x-axis.

Answer:

Let (x₁, y₁) represent the required points.
The slope of the x-axis is 0.
Here,
$2 a^{2} y = x^{3} - 3 a x^{2} Since, the point lies on the curve . Hence, 2 a^{2} y_{1} = {x_{1}}^{3} - 3 a {x_{1}}^{2} . . . (1) Now, 2 a^{2} y = x^{3} - 3 a x^{2} On differentiating both sides w.r.t. x, we get 2 a^{2} \frac{d y}{d x} = 3 x^{2} - 6 a x \Rightarrow \frac{d y}{d x} = \frac{3 x^{2} - 6 a x}{2 a^{2}} Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{3 {x_{1}}^{2} - 6 a x_{1}}{2 a^{2}} Given: Slope of the tangent at (x_{1}, y_{1}) = Slope of the x -axis \Rightarrow \frac{3 {x_{1}}^{2} - 6 a x_{1}}{2 a^{2}} = 0 \Rightarrow 3 {x_{1}}^{2} - 6 a x_{1} = 0 \Rightarrow x_{1} (3 x_{1} - 6 a) = 0 \Rightarrow x_{1} = 0 or x_{1} = 2 a Also, 2 a^{2} y_{1} = 0 or 2 a^{2} y_{1} = 8 a^{3} - 12 a^{3} [From eq. (1)] \Rightarrow y_{1} = 0 or y_{1} = - 2 a Thus, the required points are (0, 0) and (2 a, - 2 a) .$

Page No 15.11:

Question 16:

At what points on the curve y = x² − 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

Answer:

Let (x₁, y₁) be the required point.
Slope of the given line = $\frac{- 1}{2}$
Slope of the line perpendicular to this line = 2

$Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{2} - 4 x_{1} + 5 . . . (1) Now, y = x^{2} - 4 x + 5 ∴ \frac{d y}{d x} = 2 x - 4 Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =2 x_{1} -4 Slope of the tangent at (x_{1}, y_{1}) =Slope of the given line [Given] ∴ 2 x_{1} - 4 = 2 \Rightarrow 2 x_{1} = 6 \Rightarrow x_{1} = 3 Also, y_{1} = 9 - 12 + 5 = 2 [From eq. (1)] Thus, the required point is (3, 2) .$

Page No 15.11:

Question 17:

Find the points on the curve $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$ at which the tangents are parallel to the (i) x-axis (ii) y-axis.

Answer:

(i) The slope of the x-axis is 0.
Now, let (x₁, y₁) be the required point.
$Since, the point lies on the curve . Hence, \frac{{x_{1}}^{2}}{4} + \frac{{y_{1}}^{2}}{25} = 1 . . . (1) Now, \frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 ∴ \frac{2 x}{4} + \frac{2 y}{25} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{25} \frac{d y}{d x} = \frac{- x}{2} \Rightarrow \frac{d y}{d x} = \frac{- 25 x}{4 y} Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 25 x_{1}}{4 y_{1}} Slope of the tangent at (x_{1}, y_{1}) =Slope of the x -axis [Given] ∴ \frac{- 25 x_{1}}{4 y_{1}} = 0 \Rightarrow x_{1} = 0 Also, 0 + \frac{{y_{1}}^{2}}{25} = 1 [From eq. (1)] \Rightarrow {y_{1}}^{2} = 25 \Rightarrow y_{1} = \pm 5 Thus, the required points are (0, 5) and (0, - 5) .$

(ii) The slope of the y-axis is $\infty$ .
Now, let (x₁, y₁) be the required point.
$Since, the point lies on the curve . Hence, \frac{{x_{1}}^{2}}{4} + \frac{{y_{1}}^{2}}{25} = 1 . . . (1) Now, \frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 ∴ \frac{2 x}{4} + \frac{2 y}{25} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{25} \frac{d y}{d x} = \frac{- x}{2} \Rightarrow \frac{d y}{d x} = \frac{- 25 x}{4 y} Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 25 x_{1}}{4 y_{1}} Slope of the tangent at (x_{1}, y_{1}) =Slope of the y -axis [Given] ∴ \frac{- 25 x_{1}}{4 y_{1}} = \infty \Rightarrow \frac{4 y_{1}}{- 25 x_{1}} = 0 \Rightarrow y_{1} = 0 Also, \frac{{x_{1}}^{2}}{4} = 1 [From eq. (1)] \Rightarrow {x_{1}}^{2} = 4 \Rightarrow x_{1} = \pm 2 Thus, the required points are (2, 0) and (- 2, 0) .$

Page No 15.11:

Question 18:

Find the points on the curve x² + y² − 2x − 3 = 0 at which the tangents are parallel to the x-axis.

Answer:

Let (x₁, y₁) be the required point.

$Since the point lie on the curve . Hence {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 3 = 0 . . . (1) Now, x^{2} + y^{2} - 2 x - 3 = 0 \Rightarrow 2 x + 2 y \frac{d y}{d x} - 2 = 0 ∴ \frac{d y}{d x} = \frac{2 - 2 x}{2 y} = \frac{1 - x}{y} Now, Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1}} S lope of the tangent = 0 (Given) ∴ \frac{1 - x_{1}}{y_{1}} = 0 \Rightarrow 1 - x_{1} = 0 \Rightarrow x_{1} = 1 From (1), we get {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 3 = 0 \Rightarrow 1 + {y_{1}}^{2} - 2 - 3 = 0 \Rightarrow {y_{1}}^{2} - 4 = 0 \Rightarrow y_{1} = \pm 2 Hence, the points are (1, 2) and (1, - 2) .$

Page No 15.11:

Question 19:

Find the points on the curve $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.

Answer:

(i) The slope of the x-axis is 0.
Now, let (x₁, y₁) be the required point.
$Since, the point lies on the curve . Hence, \frac{{x_{1}}^{2}}{9} + \frac{{y_{1}}^{2}}{16} = 1 . . . (1) Now, \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \Rightarrow \frac{2 x}{9} + \frac{2 y}{16} \frac{d y}{d x} = 0 \Rightarrow \frac{y}{16} \frac{d y}{d x} = \frac{- x}{9} \Rightarrow \frac{d y}{d x} = \frac{- 16 x}{9 y} Now, Slope of the tangent at (x, y) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 16 x_{1}}{9 y_{1}} Slope of the tangent at (x, y) = Slope of the x -axis [Given] ∴ \frac{- 16 x_{1}}{9 y_{1}} = 0 \Rightarrow x_{1} = 0 0 + \frac{{y_{1}}^{2}}{16} = 1 [From eq. (1)] Also, {y_{1}}^{2} = 16 y_{1} = \pm 4 Thus, the required points are (0, 4) and (0, - 4) .$

(ii) The slope of the y-axis is $\infty$ .
Let (x₁, y₁) be the required point.
Given:
$Since, the point lies on the curve . Hence, \frac{{x_{1}}^{2}}{9} + \frac{{y_{1}}^{2}}{16} = 1 . . . (1) \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \Rightarrow \frac{2 x}{9} + \frac{2 y}{16} \frac{d y}{d x} = 0 \Rightarrow \frac{y}{16} \frac{d y}{d x} = \frac{- x}{9} \Rightarrow \frac{d y}{d x} = \frac{- 16 x}{9 y} Now, Slope of the tangent at (x, y) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 16 x_{1}}{9 y_{1}} Slope of the tangent at (x_{1}, y_{1}) = Slope of the y -axis [Given] ∴ \frac{- 16 x_{1}}{9 y_{1}} = \infty \Rightarrow \frac{9 y_{1}}{- 16 x_{1}} = 0 \Rightarrow y_{1} = 0 \Rightarrow \frac{{x_{1}}^{2}}{9} + 0 = 1 [From eq. (1)] \Rightarrow {x_{1}}^{2} = 9 \Rightarrow x_{1} = \pm 3 Thus, the required points are (3, 0) and (- 3, 0) .$

Page No 15.11:

Question 20:

Who that the tangents to the curve y = 7x³ + 11 at the points x = 2 and x = −2 are parallel.

Answer:

$Given : y = 7 x^{3} + 11 ∴ \frac{d y}{d x} = 21 x^{2} Now, Slope of the tangent at (x = 2) = {(\frac{d y}{d x})}_{x = 2} = 21 {(2)}^{2} = 84 Slope of the tangent at (x = - 2) = {(\frac{d y}{d x})}_{x = - 2} = 21 {(- 2)}^{2} = 84$

Both slopes are the same. Hence, the tangents at points x = 2 and x = −2 are parallel.

Page No 15.11:

Question 21:

Find the points on the curve y = x³ where the slope of the tangent is equal to the x-coordinate of the point.

Answer:

Let (x₁, y₁) be the required point.
x coordinate of the point is x₁.

$Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{3} . . . (1) Now, y = x^{3} \Rightarrow \frac{d y}{d x} = 3 x^{2} Slope of tangent at (x, y) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 3 {x_{1}}^{2} Given that Slope of tangent at (x_{1}, y_{1}) = x co-ordinate of the point \Rightarrow 3 {x_{1}}^{2} = x_{1} \Rightarrow x_{1} (3 x_{1} - 1) = 0 \Rightarrow x_{1} = 0 or x_{1} = \frac{1}{3} \Rightarrow y_{1} = 0^{3} or y_{1} = {(\frac{1}{3})}^{3} (From (1)) \Rightarrow y_{1} = 0 or y_{1} = \frac{1}{27} So, the points are (x_{1}, y_{1}) = (0, 0), (\frac{1}{3}, \frac{1}{27})$

Page No 15.27:

Question 1:

Find the equation of the tangent to the curve $\sqrt{x} + \sqrt{y} = a$ , at the point $(\frac{a^{2}}{4}, \frac{a^{2}}{4})$ .

Answer:

$\sqrt{x} + \sqrt{y} = a Differentiating both sides w.r.t. x, \Rightarrow \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- \sqrt{y}}{\sqrt{x}} Given (x_{1}, y_{1}) = (\frac{a^{2}}{4}, \frac{a^{2}}{4}) Slope of tangent, m = {(\frac{d y}{d x})}_{(\frac{a^{2}}{4}, \frac{a^{2}}{4})} = \frac{- \sqrt{\frac{a^{2}}{4}}}{\sqrt{\frac{a^{2}}{4}}} =-1 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{a^{2}}{4} = - 1 (x - \frac{a^{2}}{4}) \Rightarrow y - \frac{a^{2}}{4} = - x + \frac{a^{2}}{4} \Rightarrow x + y = \frac{a^{2}}{2}$

Page No 15.27:

Question 2:

Find the equation of the normal to y = 2x³ − x² + 3 at (1, 4).

Answer:

$y = 2 x^{3} - x^{2} + 3 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 6 x^{2} - 2 x Slope of tangent = {(\frac{d y}{d x})}_{(1, 4)} =6 {(1)}^{2} -2 (1) =4 Slope of normal = \frac{- 1}{Slope of tangent} = \frac{- 1}{4} Given (x_{1}, y_{1}) = (1, 4) Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 4 = \frac{- 1}{4} (x - 1) \Rightarrow 4 y - 16 = - x + 1 \Rightarrow x + 4 y = 17$

Page No 15.27:

Question 3:

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) y = x⁴ − bx³ + 13x² − 10x + 5 at (0, 5) [NCERT]
(ii) y = x⁴ − 6x³ + 13x² − 10x + 5 at x = 1 [NCERT, CBSE 2011]
(iii) y = x² at (0, 0) [NCERT]
(iv) y = 2x² − 3x − 1 at (1, −2)
(v) $y^{2} = \frac{x^{3}}{4 - x} at (2, - 2)$
(vi) y = x² + 4x + 1 at x = 3 [CBSE 2004]
(vii) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at (a \cos θ, b \sin θ)$
(viii) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at (a \sec θ, b \tan θ)$
(ix) y² = 4ax at $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(x) $c^{2} (x^{2} + y^{2}) = x^{2} y^{2} at (\frac{c}{\cos θ}, \frac{c}{\sin θ})$
(ix) xy = c² at $(c t, \frac{c}{t})$
(xii) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at (x_{1}, y_{1})$
(xiii) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at (x_{0}, y_{0})$ [NCERT]
(xiv) $x^{\frac{2}{3}} + y^{\frac{2}{3}}$ = 2 at (1, 1) [NCERT]
(xv) x² = 4y at (2, 1)
(xvi) y² = 4x at (1, 2) [NCERT]
(xvii) 4x² + 9y² = 36 at (3cosθ, 2sinθ) [CBSE 2011]
(xviii) y² = 4ax at (x₁, y₁) [CBSE 2012]
(xix) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at (\sqrt{2} a, b)$ [CBSE 2014]

Answer:

(i)
$y = x^{4} - b x^{3} + 13 x^{2} - 10 x + 5 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 4 x^{3} - 3 b x^{2} + 26 x - 10 Slope of tangent, m = {(\frac{d y}{d x})}_{(0, 5)} =-10 Given (x_{1}, y_{1}) = (0, 5) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 5 = - 10 (x - 0) \Rightarrow y - 5 = - 10 x \Rightarrow y + 10 x - 5 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 5 = \frac{1}{10} (x - 0) \Rightarrow 10 y - 50 = x \Rightarrow x - 10 y + 50 = 0$

(ii)
$y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 When x = 1, y = 1 - 6 + 13 - 10 + 5 = 3 So, (x_{1}, y_{1}) = (1, 3) Now, y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10 Slope of tangent, m = {(\frac{d y}{d x})}_{(1, 3)} =4-18+ 26 - 10 = 2 Equation of tangent is, y - y_{1} = 2 (x - x_{1}) \Rightarrow y - 3 = 2 (x - 1) \Rightarrow y - 3 = 2 x - 2 \Rightarrow 2 x - y + 1 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 3 = \frac{- 1}{2} (x - 1) \Rightarrow 3 y - 6 = - x + 1 \Rightarrow x + 3 y - 7 = 0$

(iii)
$y = x^{2} Differentiating both sides w.r.t. x, \frac{d y}{d x} = 2 x Given (x_{1}, y_{1}) = (0, 0) Slope of tangent, m = {(\frac{d y}{d x})}_{(0, 0)} =2 (0) =0 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = 0 (x - 0) \Rightarrow y = 0$

$Equation of normal is, \Rightarrow y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 0 = \frac{- 1}{0} (x - 0) \Rightarrow x = 0$

(iv)
$y =2 x^{2} -3 x -1 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 4 x - 3 Given (x_{1}, y_{1}) = (1, - 2) Slope of tangent, m = {(\frac{d y}{d x})}_{(1, - 2)} =4-3=1 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y + 2 = 1 (x - 1) \Rightarrow y + 2 = x - 1 \Rightarrow x - y - 3 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y + 2 = - 1 (x - 1) \Rightarrow y + 2 = - x + 1 \Rightarrow x + y + 1 = 0$

(v)
$y^{2} = \frac{x^{3}}{4 - x} Differentiating both sides w.r.t. x, 2 y \frac{d y}{d x} = \frac{(4 - x) (3 x^{2}) - x^{3} (- 1)}{{(4 - x)}^{2}} = \frac{12 x^{2} - 3 x^{3} + x^{3}}{{(4 - x)}^{2}} = \frac{12 x^{2} - 2 x^{3}}{{(4 - x)}^{2}} \frac{d y}{d x} = \frac{12 x^{2} - 2 x^{3}}{2 y {(4 - x)}^{2}} Given (x_{1}, y_{1}) = (2, - 2) Slope of tangent, m = {(\frac{d y}{d x})}_{(2, - 2)} = \frac{48 - 16}{- 16} =-2 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y + 2 = - 2 (x - 2) \Rightarrow y + 2 = - 2 x + 4 \Rightarrow 2 x + y - 2 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y + 2 = \frac{1}{2} (x - 2) \Rightarrow 2 y + 4 = x - 2 \Rightarrow x - 2 y - 6 = 0$

(vi)
$y = x^{2} + 4 x + 1 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 2 x + 4 When x =3, y = 9 + 12 + 1 = 22 So, (x_{1}, y_{1}) = (3, 22) Slope of tangent, m = {(\frac{d y}{d x})}_{x = 3} =10 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 22 = 10 (x - 3) \Rightarrow y - 22 = 10 x - 30 \Rightarrow 10 x - y - 8 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 22 = \frac{- 1}{10} (x - 3) \Rightarrow 10 y - 220 = - x + 3 \Rightarrow x + 10 y - 223 = 0$

(vii)
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \Rightarrow \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{- 2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{- x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(a \cos θ, b \sin θ)} = \frac{- a \cos θ (b^{2})}{b \sin θ (a^{2})} = \frac{- b \cos θ}{a \sin θ} Given (x_{1}, y_{1}) = (a \cos θ, b \sin θ) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \sin θ = \frac{- b \cos θ}{a \sin θ} (x - a \cos θ) \Rightarrow a y \sin θ - a b \sin^{2} θ = - b x \cos θ + a b \cos^{2} θ \Rightarrow b x \cos θ + a y \sin θ = a b Dividing by ab, \Rightarrow \frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - b \sin θ = \frac{a \sin θ}{b \cos θ} (x - a \cos θ) \Rightarrow b y \cos θ - b^{2} \sin θ \cos θ = a x \sin θ - a^{2} \sin θ \cos θ \Rightarrow a x \sin θ - b y \cos θ = (a^{2} - b^{2}) \sin θ \cos θ Dividing by \sin θ \cos θ, a x \sec θ - b y cosec θ = (a^{2} - b^{2})$

(viii)
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \Rightarrow \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(a \sec θ, b \tan θ)} = \frac{a \sec θ (b^{2})}{b \tan θ (a^{2})} = \frac{b}{a \sin θ} Given (x_{1}, y_{1}) = (a \sec θ, b \tan θ) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \tan θ = \frac{b}{a \sin θ} (x - a \sec θ) \Rightarrow a y \sin θ - a b \frac{\sin^{2} θ}{\cos θ} = b x - \frac{a b}{\cos θ} \Rightarrow \frac{a y \sin θ \cos θ - a b \sin^{2} θ}{\cos θ} = \frac{b x \cos θ - a b}{\cos θ} \Rightarrow a y \sin θ \cos θ - a b \sin^{2} θ = b x \cos θ - a b \Rightarrow b x \cos θ - a y \sin θ \cos θ = a b (1 - \sin^{2} θ) \Rightarrow b x \cos θ - a y \sin θ \cos θ = a b \cos^{2} θ Dividing by ab \cos^{2} θ, \Rightarrow \frac{x}{a} \sec θ - \frac{y}{b} \tan θ = 1$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - b \tan θ = \frac{- a \sin θ}{b} (x - a \sec θ) \Rightarrow y b - b^{2} \tan θ = - a x \sin θ + a^{2} \tan θ \Rightarrow a x \sin θ + b y = (a^{2} + b^{2}) \tan θ Dividing by tan θ, a x \cos θ + b y \cot θ = (a^{2} + b^{2})$

(ix)
$y^{2} =4 ax Differentiating both sides w.r.t. x, 2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} Given (x_{1}, y_{1}) = (\frac{a}{m^{2}}, \frac{2 a}{m}) Slope of tangent = {(\frac{d y}{d x})}_{(\frac{a}{m^{2}}, \frac{2 a}{m})} = \frac{2 a}{(\frac{2 a}{m})} = m Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{2 a}{m} = m (x - \frac{a}{m^{2}}) \Rightarrow \frac{m y - 2 a}{m} = m (\frac{m^{2} x - a}{m^{2}}) \Rightarrow m y - 2 a = m^{2} x - a \Rightarrow m^{2} x - m y + a = 0$

$Equation of normal is, y - y_{1} = \frac{1}{Slope of tangent} (x - x_{1}) \Rightarrow y - \frac{2 a}{m} = \frac{- 1}{m} (x - \frac{a}{m^{2}}) \Rightarrow \frac{m y - 2 a}{m} = \frac{- 1}{m} (\frac{m^{2} x - a}{m^{2}}) \Rightarrow m^{3} y - 2 a m^{2} = - m^{2} x + a \Rightarrow m^{2} x + m^{3} y - 2 a m^{2} - a = 0$

(x)
$c^{2} (x^{2} + y^{2}) = x^{2} y^{2} Differentiating both sides w.r.t. x, \Rightarrow 2 x c^{2} + 2 y c^{2} \frac{d y}{d x} = x^{2} 2 y \frac{d y}{d x} + 2 x y^{2} \Rightarrow \frac{d y}{d x} (2 y c^{2} - 2 x^{2} y) = 2 x y^{2} - 2 x c^{2} \Rightarrow \frac{d y}{d x} = \frac{x y^{2} - x c^{2}}{y c^{2} - x^{2} y} Slope of tangent, m = {(\frac{d y}{d x})}_{(\frac{c}{\cos θ}, \frac{c}{\sin θ})} = \frac{\frac{c^{3}}{\cos θ \sin^{2} θ} - \frac{c^{3}}{\cos θ}}{\frac{c^{3}}{\sin θ} - \frac{c^{3}}{\cos^{2} θ \sin θ}} = \frac{\frac{1 - \sin^{2} θ}{\cos θ \sin^{2} θ}}{\frac{\cos^{2} θ - 1}{\cos^{2} θ \sin θ}} = \frac{c o s^{2} θ}{\cos θ \sin^{2} θ} \times \frac{\cos^{2} θ \sin θ}{- \sin^{2} θ} = \frac{- \cos^{3} θ}{\sin^{3} θ} Given (x_{1}, y_{1}) = (\frac{c}{\cos θ}, \frac{c}{\sin θ}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{c}{\sin θ} = \frac{- \cos^{3} θ}{\sin^{3} θ} (x - \frac{c}{\cos θ}) \Rightarrow \frac{y \sin θ - c}{\sin θ} = \frac{- \cos^{3} θ}{\sin^{3} θ} (\frac{x \cos θ - c}{\cos θ}) \Rightarrow \sin^{2} θ (y \sin θ - c) = - \cos^{2} θ (x \cos θ - c) \Rightarrow y \sin^{3} θ - c \sin^{2} θ = - x \cos^{3} θ + c \cos^{2} θ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = c (s i n^{2} θ + \cos^{2} θ) \Rightarrow x \cos^{3} θ + y \sin^{3} θ = c$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{c}{\sin θ} = \frac{\sin^{3} θ}{\cos^{3} θ} (x - \frac{c}{\cos θ}) \Rightarrow \cos^{3} θ (y - \frac{c}{\sin θ}) = \sin^{3} θ (x - \frac{c}{\cos θ}) \Rightarrow y \cos^{3} θ - \frac{c \cos^{3} θ}{\sin θ} = x \sin^{3} θ - \frac{c \sin^{3} θ}{\cos θ} \Rightarrow x \sin^{3} θ - y \cos^{3} θ = \frac{c \sin^{3} θ}{\cos θ} - \frac{c \cos^{3} θ}{\sin θ} \Rightarrow x \sin^{3} θ - y \cos^{3} θ = c (\frac{\sin^{4} θ - \cos^{4} θ}{\cos θ \sin θ}) \Rightarrow x \sin^{3} θ - y \cos^{3} θ = c [\frac{(\sin^{2} θ + \cos^{2} θ) (\sin^{2} θ - \cos^{2} θ)}{\cos θ \sin θ}] \Rightarrow \sin^{3} θ - y \cos^{3} θ = 2 c [\frac{- (\cos^{2} θ - \sin^{2} θ)}{2 \cos θ \sin θ}] \Rightarrow \sin^{3} θ - y \cos^{3} θ = 2 c [\frac{- \cos (2 θ)}{\sin (2 θ)}] \Rightarrow \sin^{3} θ - y \cos^{3} θ = - 2 c c o t (2 θ) \Rightarrow \sin^{3} θ - y \cos^{3} θ + 2 c c o t (2 θ) = 0$

(xi)
${xy=c}^{2} Differentiating both sides w.r.t. x, x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} Given (x_{1}, y_{1}) = (c t, \frac{c}{t}) Slope of tangent, m = {(\frac{d y}{d x})}_{(c t, \frac{c}{t})} = \frac{- \frac{c}{t}}{c t} = \frac{- 1}{t^{2}} Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{c}{t} = \frac{- 1}{t^{2}} (x - c t) \Rightarrow \frac{y t - c}{t} = \frac{- 1}{t^{2}} (x - c t) \Rightarrow y t^{2} - c t = - x + c t \Rightarrow x + y t^{2} = 2 c t$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{c}{t} = t^{2} (x - c t) \Rightarrow y t - c = t^{3} x - c t^{4} \Rightarrow x t^{3} - y t = c t^{4} - c$

(xii)
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{- 2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{- x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- x_{1} b^{2}}{y_{1} a^{2}} Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{1} = \frac{- x_{1} b^{2}}{y_{1} a^{2}} (x - x_{1}) \Rightarrow y y_{1} a^{2} - {y_{1}}^{2} a^{2} = - x x_{1} b^{2} + {x_{1}}^{2} b^{2} \Rightarrow x x_{1} b^{2} + y y_{1} a^{2} = {x_{1}}^{2} b^{2} + {y_{1}}^{2} a^{2} . . . (1) Since (x_{1}, y_{1}) lies on the given curve.Therefore, \frac{{x_{1}}^{2}}{a^{2}} + \frac{{y_{1}}^{2}}{b^{2}} = 1 \Rightarrow \frac{{x_{1}}^{2} b^{2} + {y_{1}}^{2} a^{2}}{a^{2} b^{2}} = 1 \Rightarrow {x_{1}}^{2} b^{2} + {y_{1}}^{2} a^{2} = a^{2} b^{2} Substituting this in (1), we get x x_{1} b^{2} + y y_{1} a^{2} = a^{2} b^{2} {Dividing this by a}^{2} b^{2}, \frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1$

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{1} = \frac{y_{1} a^{2}}{x_{1} b^{2}} (x - x_{1}) \Rightarrow y x_{1} b^{2} - x_{1} y_{1} b^{2} = x y_{1} a^{2} - x_{1} y_{1} a^{2} \Rightarrow x y_{1} a^{2} - y x_{1} b^{2} = x_{1} y_{1} a^{2} - x_{1} y_{1} b^{2} \Rightarrow x y_{1} a^{2} - y x_{1} b^{2} = x_{1} y_{1} (a^{2} - b^{2}) Dividing by x_{1} y_{1} \frac{a^{2} x}{x_{1}} - \frac{b^{2} y}{y_{1}} = a^{2} - b^{2}$

(xiii)
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(x_{0}, y_{0})} = \frac{x_{0} b^{2}}{y_{0} a^{2}} Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{0} = \frac{x_{0} b^{2}}{y_{0} a^{2}} (x - x_{0}) \Rightarrow y y_{0} a^{2} - {y_{0}}^{2} a^{2} = x x_{0} b^{2} - {x_{0}}^{2} b^{2} x x_{0} b^{2} - y y_{0} a^{2} = {x_{0}}^{2} b^{2} - {y_{0}}^{2} a^{2} . . . (1) Since (x_{0}, y_{0}) lies on the given curve, \Rightarrow \frac{{x_{0}}^{2}}{a^{2}} - \frac{{y_{0}}^{2}}{b^{2}} = 1 \Rightarrow {x_{0}}^{2} b^{2} - {y_{0}}^{2} a^{2} = a^{2} b^{2} Substituting this in (1), we get \Rightarrow x x_{0} b^{2} - y y_{0} a^{2} = a^{2} b^{2} Dividing this by a^{2} b^{2} \frac{x x_{0}}{a^{2}} - \frac{y y_{0}}{b^{2}} = 1$

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{0} = \frac{- y_{0} a^{2}}{x_{0} b^{2}} (x - x_{0}) \Rightarrow y x_{0} b^{2} - x_{0} y_{0} b^{2} = - x y_{0} a^{2} + x_{0} y_{0} a^{2} \Rightarrow x y_{0} a^{2} + y x_{0} b^{2} = x_{0} y_{0} a^{2} + x_{0} y_{0} b^{2} \Rightarrow x y_{0} a^{2} + y x_{0} b^{2} = x_{0} y_{0} (a^{2} + b^{2}) Dividing by x_{0} y_{0} \frac{a^{2} x}{x_{0}} + \frac{b^{2} y}{y_{0}} = a^{2} + b^{2}$

(xiv)
$x^{\frac{2}{3}} + y^{\frac{2}{3}} = 2 Differentiating both sides w.r.t. x, \frac{2}{3} x^{\frac{- 1}{3}} + \frac{2}{3} y^{\frac{- 1}{3}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x^{\frac{- 1}{3}}}{y^{\frac{- 1}{3}}} = \frac{- y^{\frac{1}{3}}}{x^{\frac{1}{3}}} Slope of tangent, m = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 1}{1} =-1 Given (x_{1}, y_{1}) = (1, 1) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = - 1 (x - 1) \Rightarrow y - 1 = - x + 1 \Rightarrow x + y - 2 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = 1 (x - 1) \Rightarrow y - 1 = x - 1 \Rightarrow y - x = 0$

(xv)
$x^{2} = 4 y Differentiating both sides w.r.t. x, 2 x = 4 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x}{2} Slope of tangent, m = {(\frac{d y}{d x})}_{(2, 1)} = \frac{2}{2} =1 Given (x_{1}, y_{1}) = (2, 1) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = 1 (x - 2) \Rightarrow y - 1 = x - 2 \Rightarrow x - y - 1 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = - 1 (x - 2) \Rightarrow y - 1 = - x + 2 \Rightarrow x + y - 3 = 0$

(xvi)
$y^{2} = 4 x Differentiating both sides w.r.t. x, 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{2}{y} Slope of tangent, m = {(\frac{d y}{d x})}_{(1, 2)} = \frac{2}{2} =1 Given (x_{1}, y_{1}) = (1, 2) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 = 1 (x - 1) \Rightarrow y - 2 = x - 1 \Rightarrow x - y + 1 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 2 = - 1 (x - 1) \Rightarrow y - 2 = - x + 1 \Rightarrow x + y - 3 = 0$

(xvii) Equation of tangent:
$4 x^{2} + 9 y^{2} = 36 Differentiating both sides w.r.t. x, 8 x + 18 y \frac{d y}{d x} = 0 \Rightarrow 18 y \frac{d y}{d x} = - 8 x \Rightarrow \frac{d y}{d x} = \frac{- 8 x}{18 y} = \frac{- 4 x}{9 y} Slope of tangent, m = {(\frac{d y}{d x})}_{(3 \cos θ, 2 \sin θ)} = \frac{- 12 \cos θ}{18 \sin θ} = \frac{- 2 \cos θ}{3 \sin θ} Given (x_{1}, y_{1}) = (3 \cos θ, 2 \sin θ) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 \sin θ = \frac{- 2 \cos θ}{3 \sin θ} (x - 3 \cos θ) \Rightarrow 3 y \sin θ - 6 \sin^{2} θ = - 2 x \cos θ + 6 \cos^{2} θ \Rightarrow 2 x \cos θ + 3 y \sin θ = 6 (\cos^{2} θ + \sin^{2} θ) \Rightarrow 2 x \cos θ + 3 y \sin θ = 6$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 2 \sin θ = \frac{3 \sin θ}{2 \cos θ} (x - 3 \cos θ) \Rightarrow 2 y \cos θ - 4 \sin θ \cos θ = 3 x \sin θ - 9 \sin θ \cos θ \Rightarrow 3 x \sin θ - 2 y \cos θ - 5 \sin θ \cos θ = 0$

(xviii)
$y^{2} =4 ax Differentiating both sides w.r.t. x, 2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} At (x_{1}, y_{1}) Slope of tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{2 a}{y_{1}} = m Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{1} = \frac{2 a (x - x_{1})}{y_{1}} \Rightarrow y y_{1} - {y_{1}}^{2} = 2 a x - 2 a x_{1} \Rightarrow y y_{1} - 4 a x_{1} = 2 a x - 2 a x_{1} \Rightarrow y y_{1} = 2 a x + 2 a x_{1} \Rightarrow y y_{1} = 2 a (x + x_{1})$

$Equation of normal is, y - y_{1} = \frac{1}{Slope of tangent} (x - x_{1}) \Rightarrow y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - y_{1} = \frac{- y_{1}}{2 a} (x - x_{1})$

(xix)
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(\sqrt{2} a,b)} = \frac{\sqrt{2} a b^{2}}{b a^{2}} = \frac{\sqrt{2} b}{a} Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b = \frac{\sqrt{2} b}{a} (x - \sqrt{2} a) \Rightarrow a y - a b = \sqrt{2} b x - 2 a b \Rightarrow \sqrt{2} b x - a y = a b \Rightarrow \frac{\sqrt{2} x}{a} - \frac{y}{b} = 1$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - b = \frac{- a}{\sqrt{2} b} (x - \sqrt{2} a) \Rightarrow \sqrt{2} b y - \sqrt{2} b^{2} = - a x + \sqrt{2} a^{2} \Rightarrow a x + \sqrt{2} b y = \sqrt{2} b^{2} + \sqrt{2} a^{2} \Rightarrow \frac{a x}{\sqrt{2}} + b y = a^{2} + b^{2}$

Page No 15.28:

Question 4:

Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.

Answer:

$x = θ + \sin θ and y = 1 + \cos θ \frac{d x}{d θ} = 1 + \cos θ and \frac{d y}{d θ} = - \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- \sin θ}{1 + \cos θ} Slope of tangent= {(\frac{d y}{d x})}_{θ = \frac{π}{4}} = \frac{- \sin \frac{π}{4}}{1 + \cos \frac{π}{4}} = \frac{\frac{- 1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} = \frac{-1}{\sqrt{2} + 1} = \frac{-1}{\sqrt{2} + 1} × \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = 1 - \sqrt{2} (x_{1}, y_{1}) = (\frac{π}{4} + \sin \frac{π}{4}, 1 + \cos \frac{π}{4}) = (\frac{π}{4} + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - (1 + \frac{1}{\sqrt{2}}) = (1 - \sqrt{2}) [x - (\frac{π}{4} + \frac{1}{\sqrt{2}})] \Rightarrow y - 1 - \frac{1}{\sqrt{2}} = (1 - \sqrt{2}) [x - \frac{π}{4} - \frac{1}{\sqrt{2}}]$

Page No 15.28:

Question 5:

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x = θ + sinθ, y = 1 + cosθ at θ = $\frac{π}{2}$
(ii) $x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}} at t = \frac{1}{2}$
(iii) x = at², y = 2at at t = 1
(iv) x = asect, y = btant at t
(v) x = a(θ + sinθ), y = a(1 − cosθ) at θ
(vi) x = 3cosθ − cos³θ, y = 3sinθ − sin³θ [NCERT EXEMPLAR]

Answer:

(i)
$x = θ + \sin θ and y = 1 + \cos θ \frac{d x}{d θ} = 1 + \cos θ and \frac{d y}{d θ} = - \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- \sin θ}{1 + \cos θ} Slope of tangent, m = {(\frac{d y}{d x})}_{θ = \frac{π}{2}} = \frac{- \sin \frac{π}{2}}{1 + \cos \frac{π}{2}} = \frac{- 1}{1 + 0} =-1 Now, (x_{1}, y_{1}) = (\frac{π}{2} + \sin \frac{π}{2}, 1 + \cos \frac{π}{2}) = (\frac{π}{2} + 1, 1) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = - 1 (x - \frac{π}{2} - 1) \Rightarrow 2 y - 2 = - 2 x + π + 2 \Rightarrow 2 x + 2 y - π - 4 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = 1 (x - \frac{π}{2} - 1) \Rightarrow 2 y - 2 = 2 x - π - 2 \Rightarrow 2 x - 2 y = π$

(ii) Equation of tangent:
$x = \frac{2 a t^{2}}{1 + t^{2}} and y = \frac{2 a t^{3}}{1 + t^{2}} \frac{d x}{d t} = \frac{(1 + t^{2}) (4 a t) - 2 a t^{2} (2 t)}{{(1 + t^{2})}^{2}} = \frac{4 a t}{{(1 + t^{2})}^{2}} and \frac{d y}{d t} = \frac{(1 + t^{2}) 6 a t^{2} - 2 a t^{3} (2 t)}{{(1 + t^{2})}^{2}} = \frac{6 a t^{2} + 2 a t^{4}}{{(1 + t^{2})}^{2}} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{6 a t^{2} + 2 a t^{4}}{{(1 + t^{2})}^{2}}}{\frac{4 a t}{{(1 + t^{2})}^{2}}} = \frac{6 a t^{2} + 2 a t^{4}}{4 a t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = \frac{1}{2}} = \frac{\frac{3 a}{2} + \frac{a}{8}}{2 a} = \frac{\frac{12 a + a}{8}}{2 a} = \frac{13 a}{8} × \frac{1}{2 a} = \frac{13}{16} Now, (x_{1}, y_{1}) = (\frac{2 a t^{2}}{1 + t^{2}}, \frac{2 a t^{3}}{1 + t^{2}}) = (\frac{\frac{a}{2}}{1 + \frac{1}{4}}, \frac{\frac{a}{4}}{1 + \frac{1}{4}}) = (\frac{\frac{a}{2}}{\frac{5}{4}}, \frac{\frac{a}{4}}{\frac{5}{4}}) = (\frac{2 a}{5}, \frac{a}{5}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{a}{5} = \frac{13}{16} (x - \frac{2 a}{5}) \Rightarrow \frac{5 y - a}{5} = \frac{13}{16} (\frac{5 x - 2 a}{5}) \Rightarrow 5 y - a = \frac{13}{16} (5 x - 2 a) \Rightarrow 80 y - 16 a = 65 x - 26 a \Rightarrow 65 x - 80 y - 10 a = 0 \Rightarrow 13 x - 16 y - 2 a = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{a}{5} = \frac{- 16}{13} (x - \frac{2 a}{5}) \Rightarrow \frac{5 y - a}{5} = \frac{- 16}{13} (\frac{5 x - 2 a}{5}) \Rightarrow 5 y - a = \frac{- 16}{13} (5 x - 2 a) \Rightarrow 65 y - 13 a = - 80 x + 32 a \Rightarrow 80 x + 65 y - 45 a = 0 \Rightarrow 16 x + 13 y - 9 a = 0$

(iii)
$x = a t^{2} and y = 2 a t \frac{d x}{d t} = 2 a t and \frac{d y}{d t} = 2 a ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} = \frac{1}{t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = 1} = \frac{1}{1} =1 Now, (x_{1}, y_{1}) = (a, 2 a) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 a = 1 (x - a) \Rightarrow y - 2 a = x - a \Rightarrow x - y + a = 0$

Equation of normal:

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 a = - 1 (x - a) \Rightarrow y - 2 a = - x + a \Rightarrow x + y = 3 a$

(iv)
$x = a \sec t and y = b \tan t \frac{d x}{d t} = a \sec t \tan t and \frac{d y}{d t} = b \sec^{2} t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{b \sec^{2} t}{a \sec t \tan t} = \frac{b}{a} \cos e c t Slope of tangent, m = {(\frac{d y}{d x})}_{t = t} = \frac{b}{a} \cos e c t Now, (x_{1}, y_{1}) = (a \sec t, b \tan t) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \tan t = \frac{b}{a} \cos e c t (x - a s e c t) \Rightarrow y - \frac{b \sin t}{\cos t} = \frac{b}{a \sin t} (x - \frac{a}{\cos t}) \Rightarrow \frac{y \cos t - b \sin t}{\cos t} = \frac{b}{a \sin t} (\frac{x \cos t - a}{\cos t}) \Rightarrow y \cos t - b \sin t = \frac{b}{a \sin t} (x \cos t - a) \Rightarrow a y \sin t \cos t - a b \sin^{2} t = b x \cos t - a b \Rightarrow b x \cos t - a y \sin t \cos t - a b (1 - \sin^{2} t) = 0 \Rightarrow b x \cos t - a y \sin t \cos t = a b \cos^{2} t Dividing by \cos^{2} t, b x \sec t - a y \tan t = a b$

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \tan t = \frac{- a}{b} sin t (x - a \sec t) \Rightarrow y - b \frac{\sin t}{\cos t} = \frac{- a}{b} sin t (x - \frac{a}{\cos t}) \Rightarrow \frac{y \cos t - b \sin t}{\cos t} = \frac{- a}{b} sin t (\frac{x \cos t - a}{\cos t}) \Rightarrow y \cos t - b \sin t = \frac{- a}{b} s in t (x \cos t - a) \Rightarrow b y \cos t - b^{2} \sin t = - a x \sin t \cos t + a^{2} \sin t \Rightarrow a x \sin t \cos t + b y \cos t = (a^{2} + b^{2}) \sin t Dividing both sides by sin t, a x \cos t + b y \cot t = a^{2} + b^{2}$

(v)
$x = a (θ + \sin θ) and y = a (1 - \cos θ) \frac{d x}{d θ} = a (1 + \cos θ) and \frac{d y}{d θ} = a \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a \sin θ}{a (1 + \cos θ)} = \frac{\sin θ}{(1 + \cos θ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} = \tan \frac{θ}{2} . . . (1) Slope of tangent, m = {(\frac{d y}{d x})}_{θ} = \tan \frac{θ}{2} Now, (x_{1}, y_{1}) = [a (θ + \sin θ), a (1 - \cos θ)] Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - a (1 - \cos θ) = \tan \frac{θ}{2} [x - a (θ + \sin θ)] \Rightarrow y - a (2 \sin^{2} \frac{θ}{2}) = x \tan \frac{θ}{2} - a θ \tan \frac{θ}{2} - a \tan \frac{θ}{2} \sin θ \Rightarrow y - a (2 \sin^{2} \frac{θ}{2}) = x \tan \frac{θ}{2} - a θ \tan \frac{θ}{2} - a \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} 2 \sin \frac{θ}{2} \cos \frac{θ}{2} (From (1)) \Rightarrow y - 2 a \sin^{2} \frac{θ}{2} = (x - a θ) \tan \frac{θ}{2} - 2 a \sin^{2} \frac{θ}{2} \Rightarrow y = (x - a θ) \tan \frac{θ}{2}$

$Equation of normal is, y - a (1 - \cos θ) = - \cot \frac{θ}{2} [x - a (θ + \sin θ)] \Rightarrow \tan \frac{θ}{2} [y - a (2 \sin^{2} \frac{θ}{2})] = - x + a θ + a \sin θ \Rightarrow \tan \frac{θ}{2} [y - a \{2 (1 - \cos^{2} \frac{θ}{2})\}] = - x + a θ + a \sin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + a (2 \sin \frac{θ}{2} \cos \frac{θ}{2}) = - x + a θ + asin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + a \sin θ = - x + a θ + asin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) = - x + a θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + x - a θ = 0$

(vi)
$x = 3 \cos θ - \cos^{3} θ and y = 3 \sin θ - \sin^{3} θ \frac{d x}{d θ} = - 3 \sin θ + 3 \cos^{2} θ \sin θ = 3 \sin θ (\cos^{2} θ - 1) = 3 \sin θ (- \sin^{2} θ) = - 3 \sin^{3} θ and \frac{d y}{d θ} = 3 \cos θ - 3 \sin^{2} θ \cos θ = 3 \cos θ (1 - \sin^{2} θ) = 3 \cos θ (\cos^{2} θ) = 3 \cos^{3} θ ∴ \frac{d y}{d x} = \frac{3 \cos^{3} θ}{- 3 \sin^{3} θ} = - \tan^{3} θ . . . (1) Slope of tangent, m = {(\frac{d y}{d x})}_{θ} = - \tan^{3} θ Now, (x_{1}, y_{1}) = [3 \cos θ - \cos^{3} θ, 3 \sin θ - \sin^{3} θ] Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 3 \sin θ - \sin^{3} θ = - \tan^{3} θ [x - 3 \cos θ - \cos^{3} θ] \Rightarrow y - 3 \sin θ - \sin^{3} θ = \frac{- \sin^{3} θ}{\cos^{3} θ} [x - 3 \cos θ - \cos^{3} θ] \Rightarrow y \cos^{3} θ - 3 \sin θ \cos^{3} θ - \sin^{3} θ \cos^{3} θ = - x \sin^{3} θ + 3 \sin^{3} θ \cos θ + \sin^{3} θ \cos^{3} θ \Rightarrow x \sin^{3} θ + y \cos^{3} θ = 3 \sin^{3} θ \cos θ + 3 \sin θ \cos^{3} θ + 2 \sin^{3} θ \cos^{3} θ \Rightarrow x \sin^{3} θ + y \cos^{3} θ = 3 \sin^{3} θ \cos θ (\sin^{2} θ + \cos^{2} θ) + 2 \sin^{3} θ \cos^{3} θ \Rightarrow x \sin^{3} θ + y \cos^{3} θ = 3 \sin^{3} θ \cos θ + 2 \sin^{3} θ \cos^{3} θ \Rightarrow x + y \cot^{3} θ = 3 \cos θ + 2 \cos^{3} θ$

$Equation of normal is, y - (3 \cos θ - \cos^{3} θ) = \frac{- 1}{m} [x - (3 \sin θ - \sin^{3} θ)] \Rightarrow y - 3 \cos θ + \cos^{3} θ = \frac{- 1}{\tan^{3} θ} [x - 3 \sin θ + \sin^{3} θ] \Rightarrow y - 3 \cos θ + \cos^{3} θ = \frac{- \cos^{3} θ}{\sin^{3} θ} [x - 3 \sin θ + \sin^{3} θ] \Rightarrow y \sin^{3} θ - 3 \sin^{3} θ \cos θ + \sin^{3} θ \cos^{3} θ = - x \cos^{3} θ + 3 \cos^{3} θ \sin θ - \cos^{3} θ \sin^{3} θ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = 3 \sin^{3} θ \cos θ + 3 \cos^{3} θ \sin θ - 2 \cos^{3} θ \sin^{3} θ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = 3 \sin^{3} θ \cos θ (\sin^{2} θ + \cos^{2} θ) - 2 \cos^{3} θ \sin^{3} θ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = 3 \sin^{3} θ \cos θ - 2 \cos^{3} θ \sin^{3} θ \Rightarrow x co t^{3} θ + y = 3 \cos θ - 2 \cos^{3} θ$

Page No 15.28:

Question 6:

Find the equation of the normal to the curve x² + 2y² − 4x − 6y + 8 = 0 at the point whose abscissa is 2.

Answer:

Abscissa means the horizontal co-ordiante of a point.
Given that abscissa = 2.
i.e., x = 2

$x^{2} + 2 y^{2} - 4 x - 6 y + 8 = 0 . . . (1) Differentiating both sides w.r.t. x, 2 x + 4 y \frac{d y}{d x} - 4 - 6 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (4 y - 6) = 4 - 2 x \Rightarrow \frac{d y}{d x} = \frac{4 - 2 x}{4 y - 6} = \frac{2 - x}{2 y - 3} When x =2, from (1), we get 4 + 2 y^{2} - 8 - 6 y + 8 = 0 \Rightarrow 2 y^{2} - 6 y + 4 = 0 \Rightarrow y^{2} - 3 y + 2 = 0 \Rightarrow (y - 1) (y - 2) = 0 \Rightarrow y = 1 or y = 2 Case-1: y = 1 Slope of tangent = {(\frac{d y}{d x})}_{(2, 1)} = \frac{0}{- 1} =0 (x_{1}, y_{1}) = (2, 1) Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = \frac{- 1}{0} (x - 2) \Rightarrow x - 2 = 0 \Rightarrow x = 2 Case-2: y = 2 Slope of tangent = {(\frac{d y}{d x})}_{(2, 2)} = \frac{0}{1} =0 (x_{1}, y_{1}) = (2, 2) Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 2 = \frac{- 1}{0} (x - 2) \Rightarrow x - 2 = 0 \Rightarrow x = 2$

In both cases, the equation of normal is x = 2

Page No 15.28:

Question 7:

Find the equation of the normal to the curve ay² = x³ at the point (am², am³).

Answer:

$a y^{2} = x^{3} Differentiating both sides w.r.t. x, 2 a y \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{2 a y} Slope of tangent = {(\frac{d y}{d x})}_{(a m^{2}, a m^{3})} = \frac{3 a^{2} m^{4}}{2 a^{2} m^{3}} = \frac{3 m}{2} Given (x_{1}, y_{1}) = (a m^{2}, a m^{3}) Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - a m^{3} = \frac{- 2}{3 m} (x - a m^{2}) \Rightarrow 3 m y - 3 a m^{4} = - 2 x + 2 a m^{2} \Rightarrow 2 x + 3 m y - a m^{2} (2 + 3 m^{2}) = 0$

Page No 15.28:

Question 8:

The equation of the tangent at (2, 3) on the curve y² = ax³ + b is y = 4x − 5. Find the values of a and b.

Answer:

The slope of the given line y = 4x − 5 is 4
$y^{2} = a x^{3} + b . . . (1) 2 y \frac{d y}{d x} = 3 a x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 a x^{2}}{2 y} Slope of tangent= {(\frac{d y}{d x})}_{(2, 3)} = \frac{12 a}{6} =2a Given that Slope of tangent= slope of given line 2 a = 4 \Rightarrow a = 2 Substituting this and x = 2, y = 3 in (1), we get 9 = 16 + b \Rightarrow b = - 7 Hence, a = 2 and b = - 7$

Page No 15.28:

Question 9:

Find the equation of the tangent line to the curve y = x² + 4x − 16 which is parallel to the line 3x − y + 1 = 0.

Answer:

Let (x₀, y₀) be the point of intersection of both the curve and the tangent.
$y = x^{2} + 4 x - 16 Since, (x_{0}, y_{0}) lies on curve . Therefore y_{0} = {x_{0}}^{2} + 4 x_{0} - 16 . . . (1) Now, y = x^{2} + 4 x - 16 \Rightarrow \frac{d y}{d x} = 2 x + 4 Slope of tangent = {(\frac{d y}{d x})}_{(x_{0}, y_{0})} =2 x_{0} +4 Given that The tangent is parallel to the line So, Slope of tangent=slope of the given line 2 x_{0} + 4 = 3 \Rightarrow 2 x_{0} = - 1 \Rightarrow x_{0} = \frac{- 1}{2} From (1), y_{0} = \frac{1}{4} - 2 - 16 = \frac{- 71}{4} Now, slope of tangent, m =3 (x_{0}, y_{0}) = (\frac{- 1}{2}, \frac{- 71}{4}) Equation of tangent is y - y_{0} = m (x - x_{0}) \Rightarrow y + \frac{71}{4} = 3 (x + \frac{1}{2}) \Rightarrow \frac{4 y + 71}{4} = 3 (\frac{2 x + 1}{2}) \Rightarrow 4 y + 71 = 12 x + 6 \Rightarrow 12 x - 4 y - 65 = 0$

Page No 15.28:

Question 10:

Find an equation of normal line to the curve y = x³ + 2x + 6 which is parallel to the line x + 14y + 4 = 0.

Answer:

Let (x₁, y₁) be a point on the curve where we need to find the normal.
Slope of the given line = $\frac{- 1}{14}$
$Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{3} + 2 x_{1} + 6 Now, y = x^{3} + 2 x + 6 \Rightarrow \frac{d y}{d x} = 3 x^{2} + 2 Slope of the tangent= {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =3 {x_{1}}^{2} +2 Slope of the normal= \frac{- 1}{slope of the tangent} = = \frac{- 1}{3 {x_{1}}^{2} + 2} Given that, slope of the normal=slope of the given line \Rightarrow \frac{- 1}{3 {x_{1}}^{2} + 2} = \frac{- 1}{14} \Rightarrow 3 {x_{1}}^{2} + 2 = 14 \Rightarrow 3 {x_{1}}^{2} = 12 \Rightarrow {x_{1}}^{2} = 4 \Rightarrow x_{1} = \pm 2 Case-1: x_{1} = 2 y_{1} = {x_{1}}^{3} + 2 x_{1} + 6 = 8 + 4 + 6 = 18 ∴ (x_{1}, y_{1}) = (2, 18) Slope of the normal, m = \frac{- 1}{14} Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 18 = \frac{- 1}{14} (x - 2) \Rightarrow 14 y - 252 = - x + 2 \Rightarrow x + 14 y - 254 = 0 Case-2: x_{1} = - 2 y_{1} = {x_{1}}^{3} + 2 x_{1} + 6 = - 8 - 4 + 6 = - 6 ∴ (x_{1}, y_{1}) = (- 2, - 6) Slope of the normal, m = \frac{- 1}{14} Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y + 6 = \frac{- 1}{14} (x + 2) \Rightarrow 14 y + 84 = - x - 2 \Rightarrow x + 14 y + 86 = 0$

Page No 15.28:

Question 11:

Determine the equation(s) of tangent (s) line to the curve y = 4x³ − 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

Answer:

Let (x₁, y₁) be a point on the curve where we need to find the tangent(s).
Slope of the given line = $\frac{- 1}{9}$

Since, tangent is perpendicular to the given line,
Slope of the tangent = $\frac{- 1}{(\frac{- 1}{9})} = 9$
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to this curve. Since, the point lies on the curve . Hence, y_{1} = 4 {x_{1}}^{3} - 3 x_{1} + 5 Now, y = 4 x^{3} - 3 x + 5 \Rightarrow \frac{d y}{d x} = 12 x^{2} - 3 Slope of the tangent= {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 12 {x_{1}}^{2} - 3 Given that, slope of the tangent=slope of the perpendicular line \Rightarrow 12 {x_{1}}^{2} - 3 = 9 \Rightarrow 12 {x_{1}}^{2} = 12 \Rightarrow {x_{1}}^{2} = 1 \Rightarrow x_{1} = \pm 1 Case-1: x_{1} = 1 y_{1} = 4 {x_{1}}^{3} - 3 x_{1} + 5 = 4 - 3 + 5 = 6 ∴ (x_{1}, y_{1}) = (1, 6) Slope of the tangent =9 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 6 = 9 (x - 1) \Rightarrow y - 6 = 9 x - 9 \Rightarrow 9 x - y - 3 = 0 Case-2: x_{1} = - 1 y_{1} = 4 {x_{1}}^{3} - 3 x_{1} + 5 = - 4 + 3 + 5 = 4 ∴ (x_{1}, y_{1}) = (- 1, 4) Slope of the tangent =9 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 4 = 9 (x + 1) \Rightarrow y - 4 = 9 x + 9 \Rightarrow 9 x - y + 13 = 0$

Page No 15.28:

Question 12:

Find the equation of a normal to the curve y = x log_e x which is parallel to the line 2x − 2y + 3 = 0.

Answer:

Slope of the given line is 1
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to the curve. Since, the point lies on the curve . Hence, y_{1} = x_{1} \log_{e} x_{1} . . . (1) Now, y = x \log_{e} x \Rightarrow \frac{d y}{d x} = x \times \frac{1}{x} + \log_{e} x (1) = 1 + \log_{e} x Slope of tangent= 1 + \log_{e} x_{1} Slope of normal = \frac{- 1}{Slope of tangent} = \frac{- 1}{1 + \log_{e} x_{1}} Given that Slope of normal = slope of the given line \frac{- 1}{1 + \log_{e} x_{1}} = 1 \Rightarrow - 1 = 1 + \log_{e} x_{1} \Rightarrow - 2 = \log_{e} x_{1} \Rightarrow x_{1} = e^{- 2} = \frac{1}{e^{2}} Now, y_{1} = e^{- 2} (- 2) = \frac{- 2}{e^{2}} [From (1)] ∴ (x_{1}, y_{1}) = (\frac{1}{e^{2}}, \frac{- 2}{e^{2}}) Equation of normal is, y + \frac{2}{e^{2}} = 1 (x - \frac{1}{e^{2}}) \Rightarrow y + \frac{2}{e^{2}} = x - \frac{1}{e^{2}} \Rightarrow x - y = \frac{3}{e^{2}} \Rightarrow x - y = 3 e^{- 2}$

Page No 15.28:

Question 13:

Find the equation of the tangent line to the curve y = x² − 2x + 7 which is (i) parallel to the line 2x − y + 9 = 0 (ii) perpendicular to the line 5y − 15x = 13.

Answer:

(i) Slope of the given line is 2
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to the curve y = x^{2} - 2 x + 7 Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{2} - 2 x_{1} + 7 . . . (1) Now, y = x^{2} - 2 x + 7 \Rightarrow \frac{d y}{d x} = 2 x - 2 Slope of tangent at point (x_{1}, y_{1}) = 2 x_{1} - 2 Given that Slope of tangent= Slope of the given line \Rightarrow 2 x_{1} - 2 = 2 \Rightarrow 2 x_{1} = 4 \Rightarrow x_{1} = 2 Now, y_{1} = 4 - 4 + 7 = 7 ∴ (x_{1}, y_{1}) = (2, 7) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 7 = 2 (x - 2) \Rightarrow y - 7 = 2 x - 4 \Rightarrow 2 x - y + 3 = 0$

(ii) Slope of the given line is 3
Slope of the line perpendicular to this line = $\frac{- 1}{3}$

$Let (x_{1}, y_{1}) be the point where the tangent is drawn to the curve. Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{2} - 2 x_{1} + 7 . . . (1) Now, y = x^{2} - 2 x + 7 \Rightarrow \frac{d y}{d x} = 2 x - 2 Slope of tangent at (x_{1}, y_{1}) = 2 x_{1} - 2 Given that Slope of tangent at (x_{1}, y_{1}) = Slope of the perpendicular line \Rightarrow 2 x_{1} - 2 = \frac{- 1}{3} \Rightarrow 6 x_{1} - 6 = - 1 \Rightarrow 6 x_{1} = 5 \Rightarrow x_{1} = \frac{5}{6} Now, y_{1} = \frac{25}{36} - \frac{10}{6} + 7 = \frac{25 - 60 + 252}{36} = \frac{217}{36} ∴ (x_{1}, y_{1}) = (\frac{5}{6}, \frac{217}{36}) Equation of tangent is, y - \frac{217}{36} = \frac{- 1}{3} (x - \frac{5}{6}) \Rightarrow \frac{36 y - 217}{36} = \frac{- 6 x + 5}{18} \Rightarrow 36 y - 217 = - 12 x + 10 \Rightarrow 12 x + 36 y - 227 = 0$

Page No 15.28:

Question 14:

Find the equations of all lines having slope 2 and that are tangent to the curve $y = \frac{1}{x - 3}, x \neq 3$ .

Answer:

Slope of given tangent = 2
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to this curve. Since, the point lies on the curve . Hence, y_{1} = \frac{1}{x_{1} - 3} Now, y = \frac{1}{x - 3} ⇒ \frac{d y}{d x} = \frac{- 1}{{(x - 3)}^{2}} Slope of tangent = (\frac{d y}{d x}) = \frac{- 1}{{(x_{1} - 3)}^{2}} Given that Slope of the tangent = 2 \Rightarrow \frac{- 1}{{(x_{1} - 3)}^{2}} = 2 \Rightarrow {(x_{1} - 3)}^{2} = - 2 \Rightarrow x_{1} - 3 = \sqrt{- 2}, which does not exist because 2 is negative.$
So, there does not exist any such tangent.

Page No 15.28:

Question 15:

Find the equations of all lines of slope zero and that are tangent to the curve $y = \frac{1}{x^{2} - 2 x + 3}$ .

Answer:

Slope of the given tangent is 0.

$Let (x_{1}, y_{1}) be a point where the tangent is drawn to the curve (1). Since, the point lies on the curve . Hence, y_{1} = \frac{1}{{x_{1}}^{2} - 2 x_{1} + 3} . . . (1) Now, y = \frac{1}{x^{2} - 2 x + 3} \Rightarrow \frac{d y}{d x} = \frac{(x^{2} - 2 x + 3) (0) - (2 x - 2) 1}{{(x^{2} - 2 x + 3)}^{2}} = \frac{- 2 x + 2}{{(x^{2} - 2 x + 3)}^{2}} Slope of tangent= \frac{- 2 x_{1} + 2}{{({x_{1}}^{2} - 2 x_{1} + 3)}^{2}} Given that Slope of tangent = slope of the given line \Rightarrow \frac{- 2 x_{1} + 2}{{({x_{1}}^{2} - 2 x_{1} + 3)}^{2}} = 0 \Rightarrow - 2 x_{1} + 2 = 0 \Rightarrow 2 x_{1} = 2 \Rightarrow x_{1} = 1 Now, y = \frac{1}{1 - 2 + 3} = \frac{1}{2} [From (1)] ∴ (x_{1}, y_{1}) = (1, \frac{1}{2}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{1}{2} = 0 (x - 1) \Rightarrow y = \frac{1}{2}$

Page No 15.28:

Question 16:

Find the equation of the tangent to the curve $y = \sqrt{3 x - 2}$ which is parallel to the 4x − 2y + 5 = 0.

Answer:

Slope of the given line is 2
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to the curve y = \sqrt{3 x - 2} Since, the point lies on the curve . Hence, y_{1} = \sqrt{3 x_{1} - 2} . . . (1) Now, y = \sqrt{3 x - 2} \Rightarrow \frac{d y}{d x} = \frac{3}{2 \sqrt{3 x - 2}} Slope of tangent at (x_{1}, y_{1}) = \frac{3}{2 \sqrt{3 x_{1} - 2}} Given that Slope of tangent = slope of the given line \Rightarrow \frac{3}{2 \sqrt{3 x_{1} - 2}} = 2 \Rightarrow 3 = 4 \sqrt{3 x_{1} - 2} \Rightarrow 9 = 16 (3 x_{1} - 2) \Rightarrow \frac{9}{16} = 3 x_{1} - 2 \Rightarrow 3 x_{1} = \frac{9}{16} + 2 = \frac{9 + 32}{16} = \frac{41}{16} \Rightarrow x_{1} = \frac{41}{48} Now, y_{1} = \sqrt{\frac{123}{48} - 2} = \sqrt{\frac{27}{48}} = \sqrt{\frac{9}{16}} = \frac{3}{4} [From (1)] ∴ (x_{1}, y_{1}) = (\frac{41}{48}, \frac{3}{4}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{3}{4} = 2 (x - \frac{41}{48}) \Rightarrow \frac{4 y - 3}{4} = 2 (\frac{48 x - 41}{48}) \Rightarrow 24 y - 18 = 48 x - 41 \Rightarrow 48 x - 24 y - 23 = 0$

Page No 15.28:

Question 17:

Find the equation of the tangent to the curve x² + 3y − 3 = 0, which is parallel to the line y = 4x − 5.

Answer:

Suppose (x₁, y₁) be the point of contact of tangent.
We can find the slope of the given line by differentiating the equation w.r.t x
So, Slope of the line = 4

$Since, (x_{1}, y_{1}) lies on the curve . Therefore, {x_{1}}^{2} + 3 y_{1} - 3 = 0 . . . (1) Now, x^{2} + 3 y - 3 = 0 \Rightarrow 2 x + 3 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{3} Slope of tangent, m = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 2 x_{1}}{3} Given that tangent is parallel to the line, So Slope of tangent, m = slope of the given line \frac{- 2 x_{1}}{3} = 4 \Rightarrow x_{1} = - 6 36 + 3 y_{1} - 3 = 0 (From (1)) \Rightarrow 3 y_{1} = - 33 \Rightarrow y_{1} = - 11 (x_{1}, y_{1}) = (- 6, - 11) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y + 11 = 4 (x + 6) \Rightarrow y + 11 = 4 x + 24 \Rightarrow 4 x - y + 13 = 0$

Page No 15.29:

Question 18:

Prove that ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2$ touches the straight line $\frac{x}{a} + \frac{y}{b} = 2$ for all n ∈ N, at the point (a, b).

Answer:

$Now, {(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2 \frac{n}{a} {(\frac{x}{a})}^{n - 1} + \frac{n}{b} {(\frac{y}{b})}^{n - 1} \frac{d y}{d x} = 0 \frac{n}{b} {(\frac{y}{b})}^{n - 1} \frac{d y}{d x} = \frac{- n}{a} {(\frac{x}{a})}^{n - 1} \frac{d y}{d x} = \frac{- n}{a} {(\frac{x}{a})}^{n - 1} \times \frac{b}{n} {(\frac{b}{y})}^{n - 1} = \frac{- b}{a} {(\frac{b x}{a y})}^{n - 1} Slope of tangent= {(\frac{d y}{d x})}_{(a, b)} = \frac{- b}{a} {(\frac{b * a}{a * b})}^{n - 1} = \frac{- b}{a} ... (2) The equation of tangent is y - b = \frac{- b}{a} (x - a) \Rightarrow y a - a b = - x b + a b \Rightarrow x b + y a = 2 a b \Rightarrow \frac{x}{a} + \frac{y}{b} = 2$

So, the given line touches the given curve at the given point.

Page No 15.29:

Question 19:

Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at $t = \frac{π}{4}$ .

Answer:

$x = \sin 3 t and y = \cos 2 t \frac{d x}{d t} = 3 \cos 3 t and \frac{d y}{d t} = - 2 \sin 2 t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 2 \sin 2 t}{3 \cos 3 t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = \frac{π}{4}} =- \frac{- 2 \sin (\frac{π}{2})}{3 \cos (\frac{3 π}{4})} = \frac{- 2}{\frac{- 3}{\sqrt{2}}} = \frac{2 \sqrt{2}}{3} x_{1} = \sin (3 \times \frac{π}{4}) = \frac{1}{\sqrt{2}} and y_{1} = \cos (2 \times \frac{π}{4}) = 0 So, (x_{1}, y_{1}) = (\frac{1}{\sqrt{2}}, 0) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = \frac{2 \sqrt{2}}{3} (x - \frac{1}{\sqrt{2}}) \Rightarrow 3 y = 2 \sqrt{2} x - 2 \Rightarrow 2 \sqrt{2} x - 3 y - 2 = 0$

Page No 15.29:

Question 20:

At what points will be tangents to the curve y = 2x³ − 15x² + 36x − 21 be parallel to x-axis? Also, find the equations of the tangents to the curve at these points.

Answer:

Slope of x - axis is 0
Let (x₁, y₁) be the required point.
$y = 2 x^{3} - 15 x^{2} + 36 x - 21 Since (x_{1}, y_{1}) lies on the curve . Therefore y_{1} = 2 {x_{1}}^{3} - 15 {x_{1}}^{2} + 36 x_{1} - 21 . . . (1) Now, y = 2 x^{3} - 15 x^{2} + 36 x - 21 \Rightarrow \frac{d y}{d x} = 6 x^{2} - 30 x + 36 Slope of tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 6 {x_{1}}^{2} - 30 x_{1} + 36 Given that Slope of tangent at (x, y) = slope of the x -axis 6 {x_{1}}^{2} - 30 x_{1} + 36 = 0 \Rightarrow {x_{1}}^{2} - 5 x_{1} + 6 = 0 \Rightarrow (x_{1} - 2) (x_{1} - 3) = 0 \Rightarrow x_{1} = 2 or x_{1} = 3 Case-1: x_{1} = 2 y_{1} = 16 - 60 + 72 - 21 = 7 (From (1)) (x_{1}, y_{1}) = (2, 7) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 7 = 0 (x - 2) \Rightarrow y = 7 Case-2: x_{1} = 3 y_{1} = 54 - 135 + 108 - 21 = 6 (From (1)) (x_{1}, y_{1}) = (3, 6) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 6 = 0 (x - 3) \Rightarrow y = 6$

Page No 15.29:

Question 21:

Find the equation of the tangents to the curve 3x² – y² = 8, which passes through the point (4/3, 0).

Answer:

We have,

3x² – y² = 8 ...(i)

Differentiating both sides w.r.t x, we get

$6 x - 2 y \frac{d y}{d x} = 0 \Rightarrow 2 y \frac{d y}{d x} = 6 x \Rightarrow \frac{d y}{d x} = \frac{6 x}{2 y} \Rightarrow \frac{d y}{d x} = \frac{3 x}{y}$

Let tangent at (h, k) pass through $(\frac{4}{3}, 0)$ .

Since, (h, k) lies on (i), we get

$3 h^{2} - k^{2} = 8 . . . (ii)$

Slope of tangent at (h, k) = $\frac{3 h}{k}$

The equation of tangent at (h, k) is given by,

$(y - k) = \frac{3 h}{k} (x - h) . . . (iii)$

Since, the tangent passess through $(\frac{4}{3}, 0)$ .

$∴ (0 - k) = \frac{3 h}{k} (\frac{4}{3} - h) \Rightarrow - k = \frac{4 h}{k} - \frac{3 h^{2}}{k} \Rightarrow - k^{2} = 4 h - 3 h^{2}$

$\Rightarrow 8 - 3 h^{2} = 4 h - 3 h^{2} [From (ii)] \Rightarrow 8 = 4 h \Rightarrow h = 2$

Using (ii), we get

$12 - k^{2} = 8 \Rightarrow k^{2} = 4 \Rightarrow k = \pm 2$

So, the points on curve (i) at which tangents pass through $(\frac{4}{3}, 0)$ are $(2, \pm 2)$ .

Now, from (iii), the equation of tangents are
$(y - 2) = \frac{6}{2} (x - 2), or, 3 x - y - 4 = 0, and (y + 2) = \frac{6}{- 2} (x - 2), or, 3 x + y - 4 = 0$

Page No 15.40:

Question 1:

Find the angle of intersection of the following curves:

(i) y² = x and x² = y [NCERT EXEMPLAR]
(ii) y = x² and x² + y² = 20
(iii) 2y²= x³ and y² = 32x
(iv) x²+ y² − 4x − 1 = 0 and x² + y² − 2y − 9 = 0
(v) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and x² + y² = ab
(vi) x² + 4y² = 8 and x² − 2y² = 2
(vii) x² = 27y and y² = 8x
(viii) x² + y² = 2x and y² = x
(ix) y = 4 − x² and y = x² [NCERT EXEMPLAR]

Answer:

$(i) Given curves are, y^{2} = x . . . (1) x^{2} = y . . . (2) From these two equations, we get {(x^{2})}^{2} = x \Rightarrow x^{4} - x = 0 \Rightarrow x (x^{3} - 1) = 0 \Rightarrow x = 0 or x = 1 Substituting the values of x in (2) we get, y = 0 or y = 1 ∴ (x, y) = (0, 0) or (1, 1) Differenntiating (1) w.r.t. x, 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} . . . (3) Differenntiating (2) w.r.t . x, 2 x = \frac{d y}{d x} . . . (4) Case -1: (x, y) = (0, 0) The tangent to curve is parallel to x - axis . Hence, the angle between the tangents to two curve at (0, 0) is a right angle . ∴ θ = \frac{π}{2} Case -2 : (x, y) = (1, 1) From (3) we have, m_{1} = \frac{1}{2} From (4) we have, m_{2} = 2 (1) = 2 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} - 2}{1 + \frac{1}{2} \times 2}| = \frac{3}{4} \Rightarrow θ = \tan^{- 1} (\frac{3}{4})$

$(i i) Given curves are, y = x^{2} . . . (1) x^{2} + y^{2} = 20 . . . (2) From these two equations we get y + y^{2} = 20 \Rightarrow y^{2} + y - 20 = 0 \Rightarrow (y + 5) (y - 4) = 0 \Rightarrow y = - 5 or y = 4 Substituting the values of y in (1) we get, x^{2} = - 5 or x^{2} = 4 \Rightarrow x = \pm 2 and x^{2} = -5  has no real solution So, (x, y) = (2, 4) or (- 2, 4) Differenntiating (1) w.r.t. x, \frac{d y}{d x} = 2 x . . . (3) Differenntiating (2) w.r.t . x, 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} . . . (4) Case -1: (x, y) = (2, 4) From (3) we have, m_{1} = 2 (2) = 4 From (4) we have, m_{2} = \frac{- 2}{4} = \frac{- 1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{4 + \frac{1}{2}}{1 + 4 (\frac{- 1}{2})}| = \frac{9}{2} \Rightarrow θ = \tan^{- 1} (\frac{9}{2}) Case -2: (x, y) = (- 2, 4) From (3) we have, m_{1} = 2 (- 2) = - 4 From (4) we have, m_{2} = \frac{2}{4} = \frac{1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 4 - \frac{1}{2}}{1 - 4 (\frac{1}{2})}| = \frac{9}{2} \Rightarrow θ = \tan^{- 1} (\frac{9}{2})$

$(iii) Given curves are, 2 y^{2} = x^{3} . . . (1) y^{2} = 32 x . . . (2) From these two equations we get 2 (32 x) = x^{3} \Rightarrow 64 x = x^{3} \Rightarrow x (x^{2} - 64) = 0 \Rightarrow x = 0, 8, - 8 Substituting the value of x in (2) we get, y_{1} = 0, 16, - 16 ∴ (x_{1}, y_{1}) = (0, 0), (8, 16) or (8, - 16) Differentiating (1) w.r.t. x, 4 y \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{4 y} . . . (3) Differenntiating (2) w.r.t. x, 2 y \frac{d y}{d x} = 32 \Rightarrow \frac{d y}{d x} = \frac{16}{y} . . . (4) Case - 1: (x, y) = (0, 0) From (3) we have, m_{1} = \frac{0}{0} ∴ We cannot determine θ in this case. Case - 2 : (x, y) = (8, 16) From (3) we have, m_{1} = \frac{192}{64} = 3 From (4) we have, m_{2} = \frac{16}{16} = 1 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{3 - 1}{1 + 3}| = \frac{2}{4} = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2}) Case- 3 : (x_{1}, y_{1}) = (8, - 16) From (3) we have, m_{1} = \frac{192}{- 64} = - 3 From (4) we have, m_{2} = \frac{16}{- 16} = - 1 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 3 + 1}{1 + 3}| = \frac{2}{4} = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2})$

$(i v) Given curves are, x^{2} + y^{2} - 4 x - 1 = 0 . . . (1) x^{2} + y^{2} - 2 y - 9 = 0 . . . (2) From (3)  we get x^{2} + y^{2} = 4 x + 1 Substituting this in (2), 4 x + 1 - 2 y - 9 = 0 \Rightarrow 4 x - 2 y = 8 \Rightarrow 2 x - y = 4 \Rightarrow y = 2 x - 4 . . . (3) Substituting this in (1), x^{2} + {(2 x - 4)}^{2} - 4 x - 1 = 0 \Rightarrow x^{2} + 4 x^{2} + 16 - 16 x - 4 x - 1 = 0 \Rightarrow 5 x^{2} - 20 x + 15 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow (x - 3) (x - 1) = 0 \Rightarrow x = 3 or x = 1 Substituting the values of x in (3), we get, y = 2 or y = - 2 ∴ (x, y) = (3, 2), (1, - 2) Differentiating (1) w.r.t. x, 2 x + 2 y \frac{d y}{d x} - 4 = 0 \Rightarrow \frac{d y}{d x} = \frac{4 - 2 x}{2 y} = \frac{2 - x}{y} . . . (4) Differenntiating (2) w.r.t . x, 2 x + 2 y \frac{d y}{d x} - 2 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (2 y - 2) = - 2 x \Rightarrow \frac{d y}{d x} = \frac{2 x}{2 - 2 y} = \frac{x}{1 - y} . . . (5) Case - 1: (x, y) = (3, 2) From (4), we get, m_{1} = \frac{2 - 3}{2} = \frac{- 1}{2} From (5), we get, m_{2} = \frac{3}{1 - 2} = - 3 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} + 3}{1 + \frac{3}{2}}| = 1 \Rightarrow θ = \tan^{- 1} (1) = \frac{π}{4} Case - 2: (x, y) = (1, - 2) From (4), we get, m_{1} = \frac{2 - 1}{- 2} = \frac{- 1}{2} From (5), we get, m_{2} = \frac{1}{1 + 2} = \frac{1}{3} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - \frac{1}{3}}{1 - \frac{1}{6}}| = 1 \Rightarrow θ = \tan^{- 1} (1) = \frac{π}{4}$

$(v) Given curves are, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (1) x^{2} + y^{2} = a b . . . (2) Multiplying (2) by \frac{1}{a^{2}}, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} = \frac{b}{a} . . . (3) Subtracting (1) from (3), we get \frac{y^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = \frac{b}{a} - 1 \Rightarrow y^{2} (\frac{b^{2} - a^{2}}{a^{2} b^{2}}) = \frac{b - a}{a} \Rightarrow y^{2} = \frac{b - a}{a} \times \frac{a^{2} b^{2}}{(b + a) (b - a)} = \frac{a b^{2}}{b + a} \Rightarrow y = \pm b \sqrt{\frac{a}{b + a}} Substituting this in (3), \frac{x^{2}}{a^{2}} + \frac{a b^{2}}{(b + a) (a^{2})} = \frac{b}{a} \Rightarrow (a + b) x^{2} + a b^{2} = a b^{2} + a^{2} b \Rightarrow x^{2} = \frac{a^{2} b}{a + b} \Rightarrow x = \pm a \sqrt{\frac{b}{a + b}} ∴ (x, y) = (\pm a \sqrt{\frac{b}{a + b}}, \pm b \sqrt{\frac{a}{b + a}}) Now, (x, y) = (a \sqrt{\frac{b}{a + b}}, b \sqrt{\frac{a}{b + a}}) Differentiating (1) w.r.t. x, we get, \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x b^{2}}{a^{2} y} \Rightarrow m_{1} = \frac{- a b^{2} \sqrt{\frac{b}{a + b}}}{a^{2} b \sqrt{\frac{a}{b + a}}} = \frac{- b \sqrt{b}}{a \sqrt{a}} Differenntiating (2) w.r.t . x, we get, 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} \Rightarrow m_{2} = \frac{- a \sqrt{\frac{b}{a + b}}}{b \sqrt{\frac{a}{b + a}}} = \frac{- a \sqrt{b}}{b \sqrt{a}} We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- b \sqrt{b}}{a \sqrt{a}} + \frac{a \sqrt{b}}{b \sqrt{a}}}{1 + (\frac{b \sqrt{b}}{a \sqrt{a}}) (\frac{a \sqrt{b}}{b \sqrt{a}})}| = \frac{\frac{- b^{2} \sqrt{a b} + a^{2} \sqrt{a b}}{a^{2} b}}{\frac{a^{2} b + a b^{2}}{a^{2} b}} = \frac{\sqrt{a b} (a + b) (a - b)}{a^{2} b} \times \frac{a^{2} b}{a b (a + b)} = \frac{a - b}{\sqrt{a b}} \Rightarrow θ = \tan^{- 1} (\frac{a - b}{\sqrt{a b}}) {Similarly,  we can prove that θ=tan}^{- 1} (\frac{a - b}{\sqrt{a b}}) for all possibilities of (x, y)$

$(v i) Given curves are, x^{2} + 4 y^{2} = 8 . . . (1) x^{2} - 2 y^{2} = 2 . . . (2) From (1) and (2) we get 6 y^{2} = 6 \Rightarrow y = 1 or y_{1} = - 1 Substituting the values of y in (1) x = 2, - 2 or x = 2, - 2 So, (x, y) = (2, 1), (2, - 1), (- 2, 1), (- 2, - 1) Differenntiating (1) w.r.t. x, 2 x + 8 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{4 y} . . . (3) Differenntiating (2) w.r.t. x, 2 x - 4 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x}{2 y} . . . (4) Case -1: (x, y) = (2, 1) From (3), we get, m_{1} = \frac{- 1}{2} From (4), we get, m_{2} = 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -2: (x, y) = (2, - 1) From (3), we get, m_{1} = \frac{1}{2} From (4), we get, m_{2} = - 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} + 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -3 : (x, y) = (- 2, 1) From (3), we get, m_{1} = \frac{1}{2} From (4), we get, m_{2} = - 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} + 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -4 : (x, y) = (- 2, - 1) From (3), we get, m_{1} = \frac{- 1}{2} From (4), we get, m_{2} = 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3)$

$(vii) Given curves are, x^{2} = 27 y . . . (1) y^{2} = 8 x . . . (2) From (2)  we get x = \frac{y^{2}}{8} Substituting this in (1), {(\frac{y^{2}}{8})}^{2} = 27 y \Rightarrow y^{4} = 1728 y \Rightarrow y (y^{3} - 12^{3}) = 0 \Rightarrow y = 0 or y = 12 Substituting the v a l u e s o f y in (2), w e g e t, \Rightarrow x = 0 or x = 18 \Rightarrow (x, y) = (0, 0), (18, 12) Differentiating (1) w.r.t. x, 2 x = 27 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{2 x}{27} . . . (3) Differenntiating (2) w.r.t . x, 2 y \frac{d y}{d x} = 8 \Rightarrow \frac{d y}{d x} = \frac{4}{y} . . . (4) Case - 1: (x, y) = (0, 0) From (4) we have, m_{2} is undefined ∴ We cannot find θ Case - 2: (x, y) = (18, 12) From (3) we have, m_{1} = \frac{36}{27} = \frac{4}{3} From (4) we have, m_{2} = \frac{4}{12} = \frac{1}{3} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{9}}| = \frac{9}{13} \Rightarrow θ = \tan^{- 1} (\frac{9}{13})$

$(v i i i) Given curves are, x^{2} + y^{2} = 2 x . . . (1) y^{2} = x . . . (2) F rom these two equations we get x^{2} + x = 2 x \Rightarrow x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0 or x = 1 Substituting the values of x in (2) we get, y = 0 or y =±1 ∴ (x, y) = (0, 0), (1, 1), (1, - 1) Differenntiating (1) w.r.t. x, we get, 2 x + 2 y \frac{d y}{d x} = 2 \Rightarrow \frac{d y}{d x} = \frac{1 - x}{y} . . . (3) Differenntiating (2) w.r.t. x,we get, 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} . . . (4) Case -1: (x, y) = (0, 0) From (3) we get, m_{1} is undefined . ∴ We can not find θ Case -2: Let (x, y) = (1, 1) From (3) we get, m_{1} = 0 From (4) we get, m_{2} = \frac{1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{0 - \frac{1}{2}}{1 + 0}| = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2}) Case -3: Let (x, y) = (1, - 1) From (3) we get, m_{1} = 0 From (4) we get, m_{2} = \frac{- 1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{0 + \frac{1}{2}}{1}| = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2})$

$(ix) Given curves are, y = 4 - x^{2} . . . . . (1) y = x^{2} . . . . . (2) From (1) and (2), we get 4 - x^{2} = x^{2} \Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt{2} Substituting the v a l u e s o f x in (2), w e g e t, \Rightarrow y = 2 \Rightarrow (x, y) = (\sqrt{2},2), (- \sqrt{2}, 2) Differentiating (1) w.r.t. x, \frac{d y}{d x} = - 2 x . . . . . (3) Differentiating (2) w.r.t . x, \frac{d y}{d x} = 2 x . . . . . (4) Case 1: (x, y) = (\sqrt{2}, 2) From (3), we have, m_{1} = - 2 \sqrt{2} From (4) we have, m_{2} = 2 \sqrt{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 2 \sqrt{2} - 2 \sqrt{2}}{1 - 8}| = \frac{4 \sqrt{2}}{7} \Rightarrow θ = \tan^{- 1} (\frac{4 \sqrt{2}}{7}) Case 1: (x, y) = (- \sqrt{2}, 2) From (3), we have, m_{1} = 2 \sqrt{2} From (4) we have, m_{2} = - 2 \sqrt{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{2 \sqrt{2} + 2 \sqrt{2}}{1 - 8}| = \frac{4 \sqrt{2}}{7} \Rightarrow θ = \tan^{- 1} (\frac{4 \sqrt{2}}{7})$

Page No 15.40:

Question 2:

Show that the following set of curves intersect orthogonally.

(i) y = x³ and 6y = 7 − x²
(ii) x³ − 3xy² = −2 and 3x²y − y³ = 2
(iii) x² + 4y² = 8 and x² − 2y² = 4

Answer:

$(i) y = x^{3} . . . (1) 6 y = 7 - x^{2} . . . (2) From (1) and (2) we get 6 x^{3} = 7 - x^{2} \Rightarrow 6 x^{3} + x^{2} - 7 = 0 x =1 satisfies this. Dividing this by x -1 ,we get 6 x^{2} + 7 x + 7 = 0, {Discriminant = 7}^{2} -4 (6) (7) =-119<0 So, this has no real roots. When x =1, y = x^{3} =1 (From (1)) So, (x, y) = (1, 1) Differentiating (1) w.r.t. x, \frac{d y}{d x} = 3 x^{2} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 1)} = 3 Differenntiating (2) w.r.t . x, 6 \frac{d x}{d x} = - 2 x \Rightarrow \frac{d x}{d x} = \frac{- 2 x}{6} = \frac{- x}{3} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 1}{3} Now, m_{1} \times m_{2} = 3 \times \frac{- 1}{3} \Rightarrow m_{1} \times m_{2} = - 1 Since, m_{1} \times m_{2} = - 1$
So, the given curves intersect orthogonally.

$(i i) Let the given curves intersect at (x_{1}, y_{1}) x^{3} - 3 x y^{2} = - 2 . . . (1) 3 x^{2} y - y^{3} = 2 . . . (2) Differentiating (1) w.r.t. x, 3 x^{2} - 3 y^{2} - 6 x y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{3 x^{2} - 3 y^{2}}{6 x y} = \frac{x^{2} - y^{2}}{2 x y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{{x_{1}}^{2} - {y_{1}}^{2}}{2 x_{1} y_{1}} Differenntiating (2) w.r.t . x, 3 x^{2} \frac{d y}{d x} + 6 x y - 3 y^{2} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (3 x^{2} - 3 y^{2}) = - 6 x y \Rightarrow \frac{d y}{d x} = \frac{- 6 x y}{3 x^{2} - 3 y^{2}} = \frac{- 2 x y}{x^{2} - y^{2}} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 2 x_{1} y_{1}}{{x_{1}}^{2} - {y_{1}}^{2}} Now, m_{1} \times m_{2} = \frac{{x_{1}}^{2} - {y_{1}}^{2}}{2 x_{1} y_{1}} \times \frac{- 2 x_{1} y_{1}}{{x_{1}}^{2} - {y_{1}}^{2}} \Rightarrow m_{1} \times m_{2} = - 1 Since, m_{1} \times m_{2} = - 1$
So, the given curves intersect orthogonally.

$(iii) x^{2} + 4 y^{2} = 8 . . . (1) x^{2} - 2 y^{2} = 4 . . . (2) From (1) and (2) we get 6 y^{2} = 4 \Rightarrow y^{2} = \frac{2}{3} \Rightarrow y = \frac{\sqrt{2}}{\sqrt{3}} or y = \frac{- \sqrt{2}}{\sqrt{3}} From (1), x^{2} + \frac{8}{3} = 8 \Rightarrow x^{2} = \frac{16}{3} \Rightarrow x = \pm \frac{4}{\sqrt{3}} So, (x, y) = (\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}), (\frac{4}{\sqrt{3}}, \frac{- \sqrt{2}}{\sqrt{3}}), (\frac{- 4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}), (\frac{- 4}{\sqrt{3}}, - \frac{\sqrt{2}}{\sqrt{3}}) Consider point (x_{1}, y_{1}) = (\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}) Differentiating (1) w.r.t. x, 2 x + 8 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{4 y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}})} = \frac{- \frac{4}{\sqrt{3}}}{4 \frac{\sqrt{2}}{\sqrt{3}}} = \frac{- 1}{\sqrt{2}} Differenntiating (2) w.r.t . x, 2 x - 4 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x}{2 y} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}})} = \frac{\frac{4}{\sqrt{3}}}{2 \frac{\sqrt{2}}{\sqrt{3}}} = \sqrt{2} Now, m_{1} \times m_{2} = \frac{- 1}{\sqrt{2}} \times \sqrt{2} \Rightarrow m_{1} \times m_{2} = - 1 Since, m_{1} \times m_{2} = - 1 Hence,, the curves are orthogonal at (\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}) . Similarly, we can see that the curves are orthogonal in each possibility of (x_{1}, y_{1}) .$

Page No 15.40:

Question 3:

Show that the following curves intersect orthogonally at the indicated points:

(i) x² = 4y and 4y + x² = 8 at (2, 1)
(ii) x² = y and x³ + 6y = 7 at (1, 1)
(iii) y² = 8x and 2x² + y² = 10 at $(1, 2 \sqrt{2})$

Answer:

$(i) x^{2} = 4 y . . . (1) 4 y + x^{2} = 8 . . . (2) Given point is (2, 1) Differentiating (1) w.r.t. x, 2 x = 4 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x}{2} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(2, 1)} = \frac{2}{2} = 1 Differenntiating (2) w.r.t . x, 4 \frac{d y}{d x} + 2 x = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{2} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(2, 1)} = \frac{- 2}{2} = - 1 Since, m_{1} \times m_{2} = - 1$
Hence, the given curves intersect orthogonally at the given point.

$(i i) x^{2} = y . . . (1) x^{3} + 6 y = 7 . . . (2) Given point is (1, 1) Differentiating (1) w.r.t. x, 2 x = \frac{d y}{d x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 1)} = 2 (1) = 2 Differenntiating (2) w.r.t . x, 3 x^{2} + 6 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x^{2}}{2} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 1}{2} Since, m_{1} \times m_{2} = - 1$
Hence, the given curves intersect orthogonally at the given point.

$(i i i) y^{2} = 8 x . . . (1) 2 x^{2} + y^{2} = 10 . . . (2) Given point is (1, 2 \sqrt{2}) Differentiating (1) w.r.t. x, 2 y \frac{d y}{d x} = 8 \Rightarrow \frac{d y}{d x} = \frac{4}{y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 2 \sqrt{2})} = \frac{4}{2 \sqrt{2}} = \sqrt{2} Differenntiating (2) w.r.t . x, 4 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{y} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 2 \sqrt{2})} = \frac{- 2}{2 \sqrt{2}} = \frac{- 1}{\sqrt{2}} Since, m_{1} \times m_{2} = - 1$
Hence, the given curves intersect orthogonally at the given point.

Page No 15.40:

Question 4:

Show that the curves 4x = y² and 4xy = k cut at right angles, if k² = 512.

Answer:

$Given : 4 x = y^{2} . . . (1) 4 x y = k . . . (2) From (1) and (2), we get y^{3} = k \Rightarrow y = k^{\frac{1}{3}} From (1), we get 4 x = k^{\frac{2}{3}} \Rightarrow x = \frac{k^{\frac{2}{3}}}{4} On differentiating (1) w.r.t. x, we get 4 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{2}{y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(\frac{k^{\frac{2}{3}}}{4}, k^{\frac{1}{3}})} = \frac{2}{k^{\frac{1}{3}}} = 2 k^{\frac{- 1}{3}} On differentiating (2) w.r.t . x, we get 4 x \frac{d y}{d x} + 4 y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(\frac{k^{\frac{2}{3}}}{4}, k^{\frac{1}{3}})} = \frac{- k^{\frac{1}{3}}}{(\frac{k^{\frac{2}{3}}}{4})} = - 4 k^{\frac{- 1}{3}} It is given that the curves intersect at right angles. ∴ m_{1} \times m_{2} = - 1 \Rightarrow 2 k^{\frac{- 1}{3}} \times - 4 k^{\frac{- 1}{3}} = - 1 \Rightarrow 8 k^{\frac{- 2}{3}} = 1 \Rightarrow k^{\frac{- 2}{3}} = \frac{1}{8} \Rightarrow k^{\frac{2}{3}} = 8 Cubing on both sides, we get k^{2} = 512$

Page No 15.40:

Question 5:

Show that the curves 2x = y² and 2xy = k cut at right angles, if k² = 8.

Answer:

$Given : 2 x = y^{2} . . . (1) 2 x y = k . . . (2) From (1) and (2), we get y^{3} = k \Rightarrow y = k^{\frac{1}{3}} From (1), we get 2 x = k^{\frac{2}{3}} \Rightarrow x = \frac{k^{\frac{2}{3}}}{2} On differentiating (1) w.r.t. x, we get 2 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(\frac{k^{\frac{2}{3}}}{2}, k^{\frac{1}{3}})} = \frac{1}{k^{\frac{1}{3}}} = k^{\frac{- 1}{3}} On differentiating (2) w.r.t . x, we get 2 x \frac{d y}{d x} + 2 y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(\frac{k^{\frac{2}{3}}}{2}, k^{\frac{1}{3}})} = \frac{- k^{\frac{1}{3}}}{(\frac{k^{\frac{2}{3}}}{2})} = - 2 k^{\frac{- 1}{3}} It is given that the curves intersect orthogonally . ∴ m_{1} \times m_{2} = - 1 \Rightarrow k^{\frac{- 1}{3}} \times - 2 k^{\frac{- 1}{3}} = - 1 \Rightarrow 2 k^{\frac{- 2}{3}} = 1 \Rightarrow k^{\frac{- 2}{3}} = \frac{1}{2} \Rightarrow k^{\frac{2}{3}} = 2 Cubing on both sides, we get k^{2} = 8$

Page No 15.40:

Question 6:

Prove that the curves xy = 4 and x² + y² = 8 touch each other. [NCERT EXEMPLAR]

Answer:

$Given : x y = 4 . . . . . (1) x^{2} + y^{2} = 8 . . . . . (2) From (1), we get x = \frac{4}{y} Substituting x = \frac{4}{y} in (2), we get {(\frac{4}{y})}^{2} + y^{2} = 8 \Rightarrow \frac{16}{y^{2}} + y^{2} = 8 \Rightarrow 16 + y^{4} = 8 y^{2} \Rightarrow y^{4} - 8 y^{2} + 16 = 0 \Rightarrow {(y^{2} - 4)}^{2} = 0 \Rightarrow y^{2} - 4 = 0 \Rightarrow y^{2} = 4 \Rightarrow y = \pm 2 Substituting y = \pm 2, we get x = \pm 2 So, the given curves touch each other at two points (2, 2) and (- 2, - 2) .$

Page No 15.40:

Question 7:

Prove that the curves y² = 4x and x² + y² $-$ 6x + 1 = 0 touch each other at the point (1, 2). [NCERT EXEMPLAR]

Answer:

$Given : y^{2} = 4 x . . . . . (1) and x^{2} + y^{2} - 6 x + 1 = 0 . . . . . (2) From (1) and (2), we get x^{2} + 4 x - 6 x + 1 = 0 \Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow {(x - 1)}^{2} = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1 Substititing x = 1 in (1), we get y^{2} = 4 \Rightarrow y = \pm 2 So, the two given curves touch each other at two points (1, 2) and (1, - 2) .$

Page No 15.41:

Question 8:

Find the condition for the following set of curves to intersect orthogonally:

(i) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 and x y = c^{2}$
(ii) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 and \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$

Answer:

$(i) \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . . . (1) x y = c^{2} . . . (2) Let the curves intersect orthogonally at (x_{1}, y_{1}) . On d ifferentiating (1) on both sides w.r.t. x, we get \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{a^{2} y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{x_{1} b^{2}}{a^{2} y_{1}} On d ifferentiating (2) on both sides w.r.t. x, we get x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- y_{1}}{x_{1}} It is given that the curves intersect orhtogonally at (x_{1}, y_{1}) . ∴ m_{1} \times m_{2} = - 1 \Rightarrow \frac{x_{1} b^{2}}{a^{2} y_{1}} \times \frac{- y_{1}}{x_{1}} = - 1 \Rightarrow a^{2} = b^{2}$

(ii) The condition for the curves $a x^{2} + b y^{2} = 1 and a' x^{2} + b' y^{2} = 1$ to intersect orthogonally is given below:
$\frac{1}{a} - \frac{1}{b} = \frac{1}{a'} - \frac{1}{b'} So, the condition for the curves \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 and \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1 to intersect orthogonally is \frac{1}{\frac{1}{a^{2}}} - \frac{1}{\frac{1}{b^{2}}} = \frac{1}{\frac{1}{A^{2}}} - \frac{1}{\frac{- 1}{B^{2}}} \Rightarrow a^{2} - b^{2} = A^{2} + B^{2}$

Page No 15.41:

Question 9:

Show that the curves $\frac{x^{2}}{a^{2} + λ_{1}} + \frac{y^{2}}{b^{2} + λ_{1}} = 1 and \frac{x^{2}}{a^{2} + λ_{2}} + \frac{y^{2}}{b^{2} + λ_{2}} = 1$ intersect at right angles.

Answer:

$We have, \frac{x^{2}}{a^{2} + λ_{1}} + \frac{y^{2}}{b^{2} + λ_{1}} = 1 . . . (1) and \frac{x^{2}}{a^{2} + λ_{2}} + \frac{y^{2}}{b^{2} + λ_{2}} = 1 . . . (2) Now we can find the slope of both the curve by differentiating w . r . t x \Rightarrow \frac{2 x}{a^{2} + λ_{1}} + \frac{2 y \frac{d y}{d x}}{b^{2} + λ_{1}} = 0 and \frac{2 x}{a^{2} + λ_{2}} + \frac{2 y \frac{d y}{d x}}{b^{2} + λ_{2}} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} and \frac{d y}{d x} = - \frac{x}{y} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} \Rightarrow m_{1} = - \frac{x}{y} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} and m_{2} = - \frac{x}{y} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} Subtracting (2) from (1), we get, x^{2} (\frac{1}{a^{2} + λ_{1}} - \frac{1}{a^{2} + λ_{2}}) + y^{2} (\frac{1}{b^{2} + λ_{1}} - \frac{1}{b^{2} + λ_{2}}) = 0 \Rightarrow \frac{x^{2}}{y^{2}} = \frac{λ_{2} - λ_{1}}{(b^{2} + λ_{1}) (b^{2} + λ_{2})} \times \frac{1}{\frac{λ_{1} - λ_{2}}{(a^{2} + λ_{1}) (a^{2} + λ_{2})}} N o w, m_{1} \times \times m_{2} = \frac{x^{2}}{y^{2}} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} = \frac{λ_{2} - λ_{1}}{(b^{2} + λ_{1}) (b^{2} + λ_{2})} \times \frac{(a^{2} + λ_{1}) (a^{2} + λ_{2})}{λ_{1} - λ_{2}} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} = - 1 hence, (1) and (2) cuts orthogonally .$

Page No 15.41:

Question 10:

If the straight line xcos $α$ + ysin $α$ = p touches the curve $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , then prove that a²cos² $α$ $-$ b²sin² $α$ = p².

Answer:

$We have, x \cos α + y \sin α = p . . . . . (i) \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . . . . . (ii) As, the straight line (i) touches the curve (ii) . So, the straight line (i) is tangent to the curve (ii) . Also, the slope of the straight line, m = \frac{- \cos α}{\sin α} And, the slope of the tangent to the curve = \frac{d y}{d x} = \frac{b^{2}}{a^{2}} \times \frac{x}{y} So, \frac{b^{2}}{a^{2}} \times \frac{x}{y} = \frac{- \cos α}{\sin α} \Rightarrow x b^{2} \sin α = - y a^{2} \cos α \Rightarrow x = \frac{- y a^{2} \cos α}{b^{2} \sin α} . . . . . (iii) Substituting the value of x in (i), we get x \cos α + y \sin α = p \Rightarrow \frac{- y a^{2} \cos^{2} α}{b^{2} \sin α} + y \sin α = p \Rightarrow \frac{- y a^{2} \cos^{2} α + y b^{2} \sin α}{b^{2} \sin α} = p \Rightarrow y (- a^{2} \cos^{2} α + b^{2} \sin α) = p b^{2} \sin α \Rightarrow y = \frac{p b^{2} \sin α}{(b^{2} \sin α - a^{2} \cos^{2} α)} So, from (iii), we get x = \frac{- y a^{2} \cos α}{b^{2} \sin α} = \frac{- y a^{2} \cos α}{b^{2} \sin α} = \frac{- a^{2} \cos α}{b^{2} \sin α} \times \frac{p b^{2} \sin α}{(b^{2} \sin α - a^{2} \cos^{2} α)} \Rightarrow x = \frac{- p a^{2} \cos α}{(b^{2} \sin α - a^{2} \cos^{2} α)} Substituting the values x and y in (ii), we get \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{1}{a^{2}} \times {(\frac{- p a^{2} \cos α}{(b^{2} \sin α - a^{2} \cos^{2} α)})}^{2} - \frac{1}{b^{2}} \times {(\frac{p b^{2} \sin α}{(b^{2} \sin α - a^{2} \cos^{2} α)})}^{2} = 1 \Rightarrow \frac{p^{2} a^{2} \cos^{2} α}{{(b^{2} \sin α - a^{2} \cos^{2} α)}^{2}} - \frac{p^{2} b^{2} \sin^{2} α}{{(b^{2} \sin α - a^{2} \cos^{2} α)}^{2}} = 1 \Rightarrow \frac{p^{2} a^{2} \cos^{2} α - p^{2} b^{2} \sin^{2} α}{{(b^{2} \sin α - a^{2} \cos^{2} α)}^{2}} = 1 \Rightarrow \frac{p^{2} (a^{2} \cos^{2} α - b^{2} \sin^{2} α)}{{(b^{2} \sin α - a^{2} \cos^{2} α)}^{2}} = 1 \Rightarrow \frac{p^{2} (a^{2} \cos^{2} α - b^{2} \sin^{2} α)}{{(a^{2} \cos^{2} α - b^{2} \sin^{2} α)}^{2}} = 1 \Rightarrow \frac{p^{2}}{(a^{2} \cos^{2} α - b^{2} \sin^{2} α)} = 1 ∴ p^{2} = a^{2} \cos^{2} α - b^{2} \sin^{2} α$

Page No 15.41:

Question 1:

The equation to the normal to the curve y = sin x at (0, 0) is

(a) x = 0
(b) y = 0
(c) x + y = 0
(d) x − y = 0

Answer:

(c) x + y = 0

$Given : y = \sin x On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = \cos x Slope of the tangent = {(\frac{d y}{d x})}_{(0, 0)} = cos 0 =1 Slope of the normal, m = \frac{- 1}{1} =-1 Given : (x_{1}, y_{1}) = (0, 0) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = - 1 (x - 0) \Rightarrow y = - x \Rightarrow x + y = 0$

Page No 15.41:

Question 2:

The equation of the normal to the curve y = x + sin x cos x at x = π/2 is

(a) x = 2
(b) x = π
(c) x + π = 0
(d) 2x = π

Answer:

(d) 2x = π

$Given : y = x + \sin x \cos x On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = 1 + \cos^{2} x - \sin^{2} x Slope of the tangent= {(\frac{d y}{d x})}_{x = \frac{π}{2}} {=1+cos}^{2} \frac{π}{2} {-sin}^{2} \frac{π}{2} =1-1=0 Slope of the normal, m = \frac{- 1}{0} When x = \frac{π}{2}, y= \frac{π}{2} +cos \frac{π}{2} sin \frac{π}{2} = \frac{π}{2} Now, (x_{1}, y_{1}) = (\frac{π}{2}, \frac{π}{2}) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{π}{2} = \frac{- 1}{0} (x - \frac{π}{2}) \Rightarrow x = \frac{π}{2} \Rightarrow 2 x = π$

Page No 15.41:

Question 3:

The equation of the normal to the curve y = x(2 − x) at the point (2, 0) is

(a) x − 2y = 2
(b) x − 2y + 2 = 0
(c) 2x + y = 4
(d) 2x + y − 4 = 0

Answer:

(a) x − 2y = 2

$Here, y = x (2 - x) = 2 x - x^{2} \Rightarrow \frac{d y}{d x} = 2 - 2 x Slope of the tangent= {(\frac{d y}{d x})}_{(2, 0)} = 2 - 4 = - 2 S lope of the normal, m = \frac{- 1}{- 2} = \frac{1}{2} Given : (x_{1}, y_{1}) = (2, 0) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = \frac{1}{2} (x - 2) \Rightarrow 2 y = x - 2 \Rightarrow x - 2 y = 2$

Page No 15.41:

Question 4:

The point on the curve y² = x where tangent makes 45° angle with x-axis is

(a) (1/2, 1/4)
(b) (1/4, 1/2)
(c) (4, 2)
(d) (1, 1)

Answer:

(b) (1/4, 1/2)

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = x_{1} Now, y^{2} = x \Rightarrow 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1}{2 y_{1}} Given: \frac{1}{2 y_{1}} = 1 \Rightarrow 2 y_{1} = 1 \Rightarrow y_{1} = \frac{1}{2} Now, x_{1} = {y_{1}}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} ∴ (x_{1}, y_{1}) = (\frac{1}{4}, \frac{1}{2})$

Page No 15.41:

Question 5:

If the tangent to the curve x = a t², y = 2 at is perpendicular to x-axis, then its point of contact is

(a) (a, a)
(b) (0, a)
(c) (0, 0)
(d) (a, 0)

Answer:

(c) (0, 0)

Let the required point be (x₁, y₁).

$Since, the point lies on the curve . Hence, x_{1} = a t^{2} and y_{1} = 2 a t Now, x = a t^{2} and y = 2 a t \Rightarrow \frac{d x}{d t} = 2 a t and \frac{d y}{d t} = 2 a \Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} = \frac{1}{t} = \frac{2 a}{y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{2 a}{y_{1}} It is given that the tangent is perpendicular to the y -axis. It means that it is parallel to the x -axis. ∴ Slope of the tangent = Slope of the x -axis \frac{2 a}{y_{1}} = 0 \Rightarrow a = 0 Now, x_{1} = a t^{2} = 0 and y_{1} = 2 a t = 0 ∴ (x_{1}, y_{1}) = (0, 0)$

Page No 15.41:

Question 6:

The point on the curve y = x² − 3x + 2 where tangent is perpendicular to y = x is

(a) (0, 2)
(b) (1, 0)
(c) (−1, 6)
(d) (2, −2)

Answer:

(b) (1, 0)

y = x
$\Rightarrow \frac{d y}{d x} = 1$

$Let (x_{1}, y_{1}) be the required point. Since, the point lies on the curve, Hence, y_{1} = {x_{1}}^{2} - 3 x_{1} + 2 Now, y = x^{2} - 3 x + 2 ∴ \frac{d y}{d x} = 2 x - 3 Slope of the tangent= {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 2 x_{1} - 3$
The tangent is perpendicular to this line.
∴Slope of the tangent = $\frac{- 1}{Slope of the line} = \frac{- 1}{1} = - 1$

Now,
$2 x_{1} - 3 = - 1 \Rightarrow 2 x_{1} = 2 \Rightarrow x_{1} = 1 and y_{1} = {x_{1}}^{2} - 3 x_{1} + 2 = 1 - 3 + 2 = 0 ∴ (x_{1}, y_{1}) = (1, 0)$

Page No 15.41:

Question 7:

The point on the curve y² = x where tangent makes 45° angle with x-axis is

(a) (1/2, 1/4)
(b) (1/4, 1/2)
(c) (4, 2)
(d) (1, 1)

Answer:

(b) (1/4, 1/2)

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = x_{1} Now, y^{2} = x \Rightarrow 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1}{2 y_{1}} Given: \frac{1}{2 y_{1}} = 1 \Rightarrow 2 y_{1} = 1 \Rightarrow y_{1} = \frac{1}{2} Now, x_{1} = {y_{1}}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} ∴ (x_{1}, y_{1}) = (\frac{1}{4}, \frac{1}{2})$

Page No 15.41:

Question 8:

The point at the curve y = 12x − x² where the slope of the tangent is zero will be

(a) (0, 0)
(b) (2, 16)
(c) (3, 9)
(d) none of these

Answer:

(d) none of these

$Let (x_{1}, y_{1}) be the required point. Since, the point lies on the curve, Hence, y_{1} = 12 x_{1} - {x_{1}}^{2} Now, y = 12 x - x^{2} \Rightarrow \frac{d y}{d x} = 12 - 2 x S lope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 12 - 2 x_{1} Given: Slope of the tangent=0 12 - 2 x_{1} = 0 \Rightarrow x_{1} = 6 Now, y_{1} = 12 x_{1} - {x_{1}}^{2} = 72 - 36 = 36 ∴ (x_{1}, y_{1}) = (6, 36)$

Page No 15.41:

Question 9:

The angle between the curves y² = x and x²= y at (1, 1) is

(a) $\tan^{- 1} \frac{4}{3}$
(b) $\tan^{- 1} \frac{3}{4}$
(c) 90°
(d) 45°

Answer:

(b) $\tan^{- 1} \frac{3}{4}$

$Given : y^{2} = x . . . (1) x^{2} = y . . . (2) Point = (1, 1) On d ifferentiating (1) w.r.t. x, we get 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} \Rightarrow m_{1} = \frac{1}{2} On d ifferentiating (2) w.r.t. x, we get 2 x = \frac{d y}{d x} \Rightarrow m_{2} = 2 (1) = 2 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} - 2}{1 + \frac{1}{2} \times 2}| = \frac{3}{4} \Rightarrow θ = \tan^{- 1} (\frac{3}{4})$

Page No 15.42:

Question 10:

The equation of the normal to the curve 3x² − y² = 8 which is parallel to x + 3y = 8 is

(a) x + 3y = 8
(b) x + 3y + 8 = 0
(c) x + 3y ± 8 = 0
(d) x + 3y = 0

Answer:

(c) x + 3y ± 8 = 0

The slope of line x + 3y = 8 is $\frac{- 1}{3}$ .

$Since the normal is parallel to the given line, the equation of normal will be of the given form. x + 3 y = k 3 x^{2} - y^{2} = 8 Let (x_{1}, y_{1}) be the point of intersection of the two curves . Then, x_{1} + 3 y_{1} = k . . . (1) 3 {x_{1}}^{2} - {y_{1}}^{2} = 8 . . . (2) Now, 3 x^{2} - y^{2} = 8 On d ifferentiating both sides w.r.t. x, we get 6 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{6 x}{2 y} = \frac{3 x}{y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{3 x_{1}}{y_{1}} Slope of the normal, m = \frac{- 1}{(\frac{3 x}{y})} = \frac{- y_{1}}{3 x_{1}} Given: Slope of the normal = Slope of the given line \Rightarrow \frac{- y_{1}}{3 x_{1}} = \frac{- 1}{3} \Rightarrow y_{1} = x_{1} . . . (3) From (2), we get 3 {x_{1}}^{2} - {x_{1}}^{2} = 8 \Rightarrow 2 {x_{1}}^{2} = 8 \Rightarrow {x_{1}}^{2} = 4 \Rightarrow x_{1} = \pm 2 Case 1: When x_{1} =2 From (3), we get y_{1} = x_{1} = 2 ∴ (x_{1}, y) = (2, 2) From (1), we get 2 + 3 (2) = k \Rightarrow 2 + 6 = k \Rightarrow k = 8 ∴ Equation of the normal from (1) \Rightarrow x + 3 y = 8 \Rightarrow x + 3 y - 8 = 0 Case 2: When x_{1} =-2 From (3), we get y_{1} = x_{1} = - 2, ∴ (x_{1}, y) = (- 2, - 2) From (1), we get - 2 + 3 (- 2) = k \Rightarrow - 2 - 6 = k \Rightarrow k = - 8 ∴ Equation of the normal from (1) \Rightarrow x + 3 y = - 8 \Rightarrow x + 3 y + 8 = 0 From both the cases, we get the equation of the normal as: x + 3 y \pm 8 = 0$

Page No 15.42:

Question 11:

The equations of tangent at those points where the curve y = x² − 3x + 2 meets x-axis are

(a) x − y + 2 = 0 = x − y − 1
(b) x + y − 1 = 0 = x − y − 2
(c) x − y − 1 = 0 = x − y
(d) x − y = 0 = x + y

Answer:

(b) x + y − 1 = 0 = x − y − 2

Let the tangent meet the x-axis at point (x, 0).
Now,
$y = x^{2} - 3 x + 2 \Rightarrow \frac{d y}{d x} = 2 x - 3 The tangent passes through point (x, 0). ∴ 0 = x^{2} - 3 x + 2 \Rightarrow (x - 2) (x - 1) = 0 \Rightarrow x = 2 or x = 1 Case 1: When x =2: Slope of the tangent, m = {(\frac{d y}{d x})}_{(2, 0)} =4-3=1 ∴ (x_{1}, y_{1}) = (2, 0) Equation of the tangent: y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = 1 (x - 2) \Rightarrow x - y - 2 = 0 Case 2: When x =1: Slope of the tangent, m = {(\frac{d y}{d x})}_{(2, 0)} =2-3=-1 ∴ (x_{1}, y_{1}) = (1, 0) Equation of the tangent: y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = - 1 (x - 1) \Rightarrow x + y - 1 = 0$

Page No 15.42:

Question 12:

The slope of the tangent to the curve x = t² + 3 t − 8, y = 2t² − 2t − 5 at point (2, −1) is

(a) 22/7
(b) 6/7
(c) −6
(d) none of these

Answer:

(b) 6/7

$x = t^{2} + 3 t - 8 and y = 2 t^{2} - 2 t - 5 \frac{d x}{d t} = 2 t + 3 and \frac{d y}{d t} = 4 t - 2 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} The g iven point is (2, -1). ∴ x =2 and y =-1 Now, t^{2} + 3 t - 8 = 2 and 2 t^{2} - 2 t - 5 = - 1 Let us solve one of these to get the value of t. t^{2} + 3 t - 10 = 0 and 2 t^{2} - 2 t - 4 = 0 \Rightarrow (t + 5) (t - 2) = 0 and (2 t + 2) (t - 2) = 0 \Rightarrow t = - 5 or t= 2 and t =-1 or t =2 These two have t = 2 as a common solution . ∴ Slope of the tangent = {(\frac{d y}{d x})}_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$

Page No 15.42:

Question 13:

At what point the slope of the tangent to the curve x² + y² − 2x − 3 = 0 is zero

(a) (3, 0), (−1, 0)
(b) (3, 0), (1, 2)
(c) (−1, 0), (1, 2)
(d) (1, 2), (1, −2)

Answer:

(d) (1, 2), (1, −2)

Let (x₁, y₁) be the required point.

$Since, the point lie on the curve . Hence, {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 3 = 0 . . . (1) Now, x^{2} + y^{2} - 2 x - 3 = 0 \Rightarrow 2 x + 2 y \frac{d y}{d x} - 2 = 0 ∴ \frac{d y}{d x} = \frac{2 - 2 x}{2 y} = \frac{1 - x}{y} Now, Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1}} S lope of the tangent=0 (Given) ∴ \frac{1 - x_{1}}{y_{1}} = 0 \Rightarrow 1 - x_{1} = 0 \Rightarrow x_{1} = 1 From (1), we get {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 3 = 0 \Rightarrow 1 + {y_{1}}^{2} - 2 - 3 = 0 \Rightarrow {y_{1}}^{2} - 4 = 0 \Rightarrow y_{1} = \pm 2 So, the points are (1, 2) and (1, - 2) .$

Page No 15.42:

Question 14:

The angle of intersection of the curves xy = a² and x² − y² = 2a²is

(a) 0°
(b) 45°
(c) 90°
(d) none of these

Answer:

(c) 90°

$Given : x y = a^{2} . . . (1) x^{2} - y^{2} = 2 a^{2} . . . (2) Let (x_{1}, y_{1}) be the point of intersection. On d ifferentiating (1) w.r.t. x, we get x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- y_{1}}{x_{1}} On d ifferentiating (2) w.r.t. x, we get 2 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x}{y} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{x_{1}}{y_{1}} Now, m_{1} \times m_{2} = \frac{- y_{1}}{x_{1}} \times \frac{x_{1}}{y_{1}} \Rightarrow m_{1} \times m_{2} = - 1 ∵ m_{1} \times m = - 1 So, the angle between the curves is 90°.$

Page No 15.42:

Question 15:

If the curve ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then a is equal to

(a) 1
(b) −6
(c) 6
(d) 0

Answer:

(c) 6

$Given : a y + x^{2} = 7 . . . (1) x^{3} = y . . . (2) Point= (1, 1) On d ifferentiating (1) w.r.t. x, we get a \frac{d y}{d x} + 2 x = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{a} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 2}{a} Again, on d ifferentiating (2) w.r.t. x, we get 3 x^{2} = \frac{d y}{d x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 1)} = 3 It is given that the curves are orthogonal at the given point. ∴ m_{1} \times m_{2} = - 1 \Rightarrow \frac{- 2}{a} \times 3 = - 1 \Rightarrow a = 6$

Page No 15.42:

Question 16:

If the line y = x touches the curve y = x² + bx + c at a point (1, 1) then

(a) b = 1, c = 2
(b) b = −1, c = 1
(c) b = 2, c = 1
(d) b = −2, c = 1

Answer:

(b) b = −1, c = 1

We can find the slope of the line by differentiating w.r.t. x.
Slope of the given line = 1
Now,
$y = x^{2} + b x + c . . . (1) \Rightarrow \frac{d y}{d x} = 2 x + b Slope of the tangent= {(\frac{d y}{d x})}_{(1, 1)} =2+ b Given: Slope of the tangent = 1 \Rightarrow 2 + b = 1 \Rightarrow b = - 1 On s ubstituting b = - 1, x =1 and y =1 in (1), we get \Rightarrow 1 = 1 - 1 + c \Rightarrow c = 1$

Page No 15.42:

Question 17:

The slope of the tangent to the curve x = 3t² + 1, y = t³ −1 at x = 1 is

(a) 1/2
(b) 0
(c) −2
(d) ∞

Answer:

(b) 0

$Given : x = 3 t^{2} + 1 y = t^{3} - 1 x = 1 Now, 3 t^{2} + 1 = 1 \Rightarrow 3 t^{2} = 0 \Rightarrow t = 0 \frac{d x}{d t} = 6 t and \frac{d y}{d t} = 3 t^{2} ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{3 t^{2}}{6 t} = \frac{t}{2} Thus, we get Slope of the tangent= {(\frac{d y}{d x})}_{t = 0} = \frac{0}{2} =0$

Page No 15.42:

Question 18:

The curves y = ae^x and y = be^−x cut orthogonally, if

(a) a = b
(b) a = −b
(c) ab = 1
(d) ab = 2

Answer:

(c) ab = 1

$Given : y = a e^{x} . . . (1) y = b e^{- x} . . . (2) Let the point of intersection of these two curves be (x_{1}, y_{1}) . Now, On d ifferentiating (1) w.r.t. x, we get \frac{d y}{d x} = a e^{x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = a e^{x_{1}} Again, on d ifferentiating (2) w.r.t. x, we get \frac{d y}{d x} = - b e^{- x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = - b e^{- x_{1}} It is given that the curves cut orthogonally. ∴ m_{1} \times m_{2} = - 1 \Rightarrow a e^{x_{1}} \times (- b e^{- x_{1}}) = - 1 \Rightarrow a b = 1$

Page No 15.42:

Question 19:

The equation of the normal to the curve x = a cos³ θ, y = a sin³ θ at the point θ = π/4 is

(a) x = 0
(b) y = 0
(c) c = y
(d) x + y = a

Answer:

(c) x = y

$Here, x = a \cos^{3} θ and y = a \sin^{3} θ \frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ and \frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 a \sin^{2} θ \cos θ}{- 3 a \cos^{2} θ \sin θ} = - \tan θ Now, Slope of the tangent = {(\frac{d y}{d x})}_{θ = \frac{π}{4}} =-tan \frac{π}{4} =-1 (x_{1}, y_{1}) = (a \cos^{3} \frac{π}{4}, a \sin^{3} \frac{π}{4}) = (\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}) ∴ Equation of the normal = y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{a}{2 \sqrt{2}} = 1 (x - \frac{a}{2 \sqrt{2}}) \Rightarrow x = y$

Page No 15.42:

Question 20:

If the curves y = 2 e^x and y = ae^−x intersect orthogonally, then a =

(a) 1/2
(b) −1/2
(c) 2
(d) 2e²

Answer:

(a) 1/2

$Given : y = 2 e^{x} . . . (1) y = a e^{- x} . . . (2) Let the point of intersection of these curves be (x_{1}, y_{1}) . On d ifferentiating (1) w.r.t. x, we get \frac{d y}{d x} = 2 e^{x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 2 e^{x_{1}} Now, on d ifferentiating (2) w.r.t. x, we get \frac{d y}{d x} = - a e^{- x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = - a e^{- x_{1}} It is given that the curves are orthogonal. ∴ m_{1} \times m_{2} = - 1 \Rightarrow 2 e^{x_{1}} \times (- a e^{- x_{1}}) = - 1 \Rightarrow 2 a = 1 \Rightarrow a = \frac{1}{2}$

Page No 15.42:

Question 21:

The point on the curve y = 6x − x² at which the tangent to the curve is inclined at π/4 to the line x + y = 0 is

(a) (−3, −27)
(b) (3, 9)
(c) (7/2, 35/4)
(d) (0, 0)

Answer:

(b) (3, 9)

Let (x₁, y₁) be the point where the given curve intersect the given line at the given angle.
$Since, the point lie on the curve . Hence, y_{1} = 6 x_{1} - {x_{1}}^{2} Now, y = 6 x - x^{2} \Rightarrow \frac{d y}{d x} = 6 - 2 x \Rightarrow m_{1} = 6 - 2 x_{1} and x + y = 0 \Rightarrow 1 + \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - 1 \Rightarrow m_{2} = - 1 It is given that the angle between them is \frac{π}{4} . ∴ \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| \Rightarrow \tan \frac{π}{4} = |\frac{6 - 2 x_{1} + 1}{1 - 6 + 2 x_{1}}| \Rightarrow 1 = |\frac{7 - 2 x_{1}}{2 x_{1} - 5}| \Rightarrow \frac{7 - 2 x_{1}}{2 x_{1} - 5} = \pm 1 \Rightarrow \frac{7 - 2 x_{1}}{2 x_{1} - 5} = 1 or \frac{7 - 2 x_{1}}{2 x_{1} - 5} =-1 \Rightarrow 7 - 2 x_{1} = 2 x_{1} - 5 or 7 - 2 x_{1} = - 2 x_{1} + 5 \Rightarrow 4 x_{1} = 12 or 2 = 0 (It is not true.) \Rightarrow x_{1} = 3 and y_{1} = 6 x_{1} - {x_{1}}^{2} = 18 - 9 = 9 ∴ (x_{1}, y_{1}) = (3, 9)$

Page No 15.42:

Question 22:

The angle of intersection of the parabolas y² = 4 ax and x² = 4ay at the origin is

(a) π/6
(b) π/3
(c) π/2
(d) π/4

Answer:

(c) π/2

$Given : y^{2} = 4 a x . . . (1) x^{2} = 4 a y . . . (2) Point = (0, 0) On d ifferentiating (1) w.r.t. x, we get 2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} \Rightarrow m_{1} = \infty Now, on d ifferentiating (2) w.r.t. x, we get 2 x = 4 a \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x}{2 a} = 0 ∴ \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\infty}{1 + 0}| = \infty \Rightarrow θ = \tan^{- 1} \infty = \frac{π}{2}$

Page No 15.42:

Question 23:

The angle of intersection of the curves y = 2 sin² x and y = cos 2 x at $x = \frac{π}{6}$ is

(a) π/4
(b) π/2
(c) π/3
(d) none of these

Answer:

(c) π/3

$Given: x = \frac{π}{6} Now, y = 2 \sin^{2} x \Rightarrow \frac{d y}{d x} = 4 \sin x \cos x \Rightarrow \frac{d y}{d x} = 2 \sin 2 x \Rightarrow m_{1} = {(\frac{d y}{d x})}_{x = \frac{π}{6}} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} Also, y = \cos 2 x \Rightarrow \frac{d y}{d x} = - 2 \sin 2 x \Rightarrow m_{2} = {(\frac{d y}{d x})}_{x = \frac{π}{6}} = - 2 \times \frac{\sqrt{3}}{2} = - \sqrt{3} ∴ \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\sqrt{3} + \sqrt{3}}{1 - \sqrt{3} \sqrt{3}}| = |\frac{2 \sqrt{3}}{- 2}| = \sqrt{3} \Rightarrow θ = \tan^{- 1} (\sqrt{3}) = \frac{π}{3}$

Page No 15.42:

Question 24:

Any tangent to the curve y = 2x⁷ + 3x + 5

(a) is parallel to x-axis
(b) is parallel to y-axis
(c) makes an acute angle with x-axis
(d) makes an obtuse angle with x-axis

Answer:

(c) makes an acute angle with x-axis

We have, y = 2x⁷ + 3x + 5
$\frac{d y}{d x} = 14 x^{6} + 3 \Rightarrow \frac{d y}{d x} > 3 (∵ x^{6} is always positive for any real value of x) \Rightarrow \frac{d y}{d x} > 0 S o, \tan θ > 0 Hence, θ lies in first quadrant . Thus, the tangent to the curve makes an acute angle with x - axis$

Page No 15.42:

Question 25:

The point on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes is

(a) $(4, \frac{8}{3})$
(b) $(- 4, \frac{8}{3})$
(c) $(4, - \frac{8}{3})$
(d) none of these

Answer:

(a) $(4, \frac{8}{3})$ and (c) $(4, - \frac{8}{3})$

Let (x₁, y₁) be the required point.
$Since, (x_{1}, y_{1}) lies on the given curve ∴ 9 {y_{1}}^{2} = {x_{1}}^{3} . . . (1) Now, 9 y^{2} = x^{3} 18 y \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{18 y} = \frac{x^{2}}{6 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{{x_{1}}^{2}}{6 y_{1}} Slope of the normal = \frac{- 1}{\frac{{x_{1}}^{2}}{6 y_{1}}} = \frac{- 6 y_{1}}{{x_{1}}^{2}} It is given that the normal makes equal intercepts with the axes. ∴ Slope of the normal = ±1 Now, \frac{- 6 y_{1}}{{x_{1}}^{2}} = \pm 1 \Rightarrow \frac{- 6 y_{1}}{{x_{1}}^{2}} = 1 or \frac{- 6 y_{1}}{{x_{1}}^{2}} =-1 \Rightarrow y_{1} = \frac{- {x_{1}}^{2}}{6} or y_{1} = \frac{{x_{1}}^{2}}{6} . . . (2) Case 1: When y_{1} = \frac{- {x_{1}}^{2}}{6} From (1), we have 9 (\frac{{x_{1}}^{4}}{36}) = {x_{1}}^{3} \Rightarrow {x_{1}}^{4} = 4 {x_{1}}^{3} \Rightarrow {x_{1}}^{4} - 4 {x_{1}}^{3} = 0 \Rightarrow {x_{1}}^{3} (x_{1} - 4) = 0 \Rightarrow x_{1} = 0, 4 Putting x_{1} = 0 in (1), we get, 9 {y_{1}}^{2} = 0 \Rightarrow y_{1} = 0 Putting x_{1} = 4 in (1), we get, 9 {y_{1}}^{2} = 4^{3} \Rightarrow y_{1} = \pm \frac{8}{3} But, the line making the equal intercepts with the coordinate axes can not pass through the origin . So, the points are (4, \pm \frac{8}{3})$

Page No 15.42:

Question 26:

The slope of the tangent to the curve x = t² + 3t − 8, y = 2t² − 2t − 5 at the point (2, −1) is

(a) $\frac{22}{7}$
(b) $\frac{6}{7}$
(c) $\frac{7}{6}$
(d) $- \frac{6}{7}$

Answer:

(b) $\frac{6}{7}$

Given:
x = t² + 3t − 8 and y = 2t² − 2t − 5

$\Rightarrow \frac{d x}{d t} = 2 t + 3 and \frac{d y}{d t} = 4 t - 2 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} The given point is (2, -1). ∴ x = 2 and y = -1 Now, t^{2} + 3 t - 8 = 2 and 2 t^{2} - 2 t - 5 = - 1 \Rightarrow t^{2} + 3 t - 10 = 0 and t^{2} - t - 2 = 0 \Rightarrow (t + 5) (t - 2) = 0 and (t + 1) (t - 2) = 0 \Rightarrow t = - 5, 2 and t = - 1, 2 We can see that t = 2 satisfies both of these. ∴ Slope of the tangent = {(\frac{d y}{d x})}_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$
Thus, the correct option is (b).

Page No 15.43:

Question 27:

The line y = mx + 1 is a tangent to the curve y² = 4x, if the value of m is

(a) 1
(b) 2
(c) 3
(d) $\frac{1}{2}$

Answer:

(a) 1
Let (x₁, y₁) be the required point.
The slope of the given line is m.
We have
$y^{2} = 4 x \Rightarrow 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{4}{2 y} = \frac{2}{y} Slope of the tangent = (\frac{d y}{d x})_{(x_{1}, y_{1})} = \frac{2}{y_{1}} Given: Slope of the tangent = m Now, \frac{2}{y_{1}} = m . . . (1)$

Because the given line is a tangent to the given curve at point (x₁, y₁), this point lies on both the line and the curve.

$∴ y_{1} = m x_{1} + 1 and {y_{1}}^{2} = 4 x_{1} \Rightarrow x_{1} = \frac{y_{1} - 1}{m} and x_{1} = \frac{{y_{1}}^{2}}{4} So, \frac{y_{1} - 1}{m} = \frac{{y_{1}}^{2}}{4} \Rightarrow \frac{y_{1} - 1}{(\frac{2}{y_{1}})} = \frac{{y_{1}}^{2}}{4} [From (1)] \Rightarrow \frac{y_{1} (y_{1} - 1)}{2} = \frac{{y_{1}}^{2}}{4} \Rightarrow 2 {y_{1}}^{2} - 2 y_{1} = {y_{1}}^{2} \Rightarrow {y_{1}}^{2} - 2 y_{1} = 0 \Rightarrow {y_{1}}^{2} - 2 y_{1} = 0 \Rightarrow y_{1} (y_{1} - 2) = 0 \Rightarrow y_{1} = 0, 2 So, For y_{1} =0, m = \frac{2}{0} = \infty For y_{1} =2, m = \frac{2}{2} = 1$

Page No 15.43:

Question 28:

The normal at the point (1, 1) on the curve 2y + x² = 3 is

(a) x + y = 0
(b) x − y = 0
(c) x + y + 1 = 0
(d) x − y = 1

Answer:

(b) x − y = 0

$Given : 2 y + x^{2} = 3 \Rightarrow 2 \frac{d y}{d x} + 2 x = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{2} = - x Slope of the tangent = {(\frac{d y}{d x})}_{(1, 1)} =-1 Slope of the normal, m = \frac{- 1}{Slope of the tangent} = \frac{- 1}{- 1} =1 Now, (x_{1}, y_{1}) = (1, 1) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = 1 (x - 1) \Rightarrow x - y = 0$

Page No 15.43:

Question 29:

The normal to the curve x² = 4y passing through (1, 2) is

(a) x + y = 3
(b) x − y = 3
(c) x + y = 1
(d) x − y = 1

Answer:

Disclaimer: None of the given options is correct.

$Given : x^{2} = 4 y \Rightarrow 2 x = 4 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{2 x}{4} = \frac{x}{2} Slope of the tangent = {(\frac{d y}{d x})}_{(1, 2)} = \frac{1}{2} Slope of the normal, m = \frac{- 1}{Slope of the tangent} = \frac{- 1}{\frac{1}{2}} =-2 Also, (x_{1}, y_{1}) = (1, 2) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 = - 2 (x - 1) \Rightarrow y - 2 = - 2 x + 2 \Rightarrow 2 x + y = 4$

Page No 15.43:

Question 30:

The abscissa of the point on the curve 3y = 6x - 5x³, the normal at which passes through the origin is
(a) 1 (b) $\frac{1}{3}$ (c) 2 (d) $\frac{1}{2}$

Answer:

Let (h, k) be the point on the curve 3y = 6x − 5x³, the normal at which passes through the origin.

∴ 3k = 6h − 5h³ .....(1)

3y = 6x − 5x³

Differentiating both sides with respect to x, we get

$3 \frac{d y}{d x} = 6 - 15 x^{2}$

$\Rightarrow \frac{d y}{d x} = 2 - 5 x^{2}$

∴ Slope of tangent to the given curve at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = 2 - 5 h^{2}$

Slope of normal to the given curve at (h, k) = $- \frac{1}{{(\frac{d y}{d x})}_{(h, k)}} = - \frac{1}{(2 - 5 h^{2})}$

Equation of the normal to the given curve at (h, k) is

$y - k = - \frac{1}{(2 - 5 h^{2})} (x - h)$

This equation passes through the origin.

$∴ 0 - k = - \frac{1}{(2 - 5 h^{2})} (0 - h)$

$\Rightarrow k = - \frac{h}{2 - 5 h^{2}}$ .....(2)

From (1) and (2), we have

$- \frac{h}{2 - 5 h^{2}} = \frac{6 h - 5 h^{3}}{3}$

$\Rightarrow \frac{1}{5 h^{2} - 2} = \frac{6 - 5 h^{2}}{3}$

$\Rightarrow 30 h^{2} - 25 h^{4} - 12 + 10 h^{2} = 3$

$\Rightarrow 25 h^{4} - 40 h^{2} + 15 = 0$

$\Rightarrow 5 h^{4} - 8 h^{2} + 3 = 0$

$\Rightarrow 5 h^{4} - 5 h^{2} - 3 h^{2} + 3 = 0$

$\Rightarrow 5 h^{2} (h^{2} - 1) - 3 (h^{2} - 1) = 0$

$\Rightarrow (h^{2} - 1) (5 h^{2} - 3) = 0$

⇒ h² − 1 = 0 or 5h² − 3 = 0

⇒ h² = 1 or $h^{2} = \frac{3}{5}$

⇒ h = ±1 or $h = \pm \sqrt{\frac{3}{5}}$

Thus, the abscissa of the points on the given curve, the normal at which passes through the origin are $- \sqrt{\frac{3}{5}}$ , −1, 1 and $\sqrt{\frac{3}{5}}$ .

Hence, the correct answer is option (a).

Page No 15.43:

Question 31:

Two curves x³ - 3xy² + 2 = 0 and 3x²y - y³ = 2
(a) touch each other (b) cut at right angle
(c) cut an angle $\frac{π}{3}$ (d) cut at an angle $\frac{π}{4}$

Answer:

The given curves are x³ − 3xy² + 2 = 0 and 3x²y − y³ = 2.

We know that the angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection.

Let (x₁, y₁) be the point of intersection of two curves,

Consider the first curve C₁≡ x³ − 3xy² + 2 = 0.

x³ − 3xy² + 2 = 0

Differentiating both sides with respect to x, we get

$3 x^{2} - 3 (x \times 2 y \frac{d y}{d x} + y^{2} \times 1) + 0 = 0$

$\Rightarrow 3 x^{2} - 6 x y \frac{d y}{d x} - 3 y^{2} = 0$

$\Rightarrow \frac{d y}{d x} = \frac{3 x^{2} - 3 y^{2}}{6 x y}$

∴ Slope of tangent to the first curve at (x₁, y₁) = ${(\frac{d y}{d x})}_{C_{1}} = \frac{x_{1}^{2} - y_{1}^{2}}{2 x_{1} y_{1}}$

Now, consider the second curve C₂≡ 3x²y − y³ = 2.

3x²y − y³ = 2

Differentiating both sides with respect to x, we get

$3 (x^{2} \times \frac{d y}{d x} + y \times 2 x) - 3 y^{2} \frac{d y}{d x} = 0$

$\Rightarrow (3 x^{2} - 3 y^{2}) \frac{d y}{d x} = - 6 x y$

$\Rightarrow \frac{d y}{d x} = - \frac{6 x y}{3 x^{2} - 3 y^{2}}$

∴ Slope of tangent to the second curve at (x₁, y₁) = ${(\frac{d y}{d x})}_{C_{2}} = - \frac{2 x_{1} y_{1}}{x_{1}^{2} - y_{1}^{2}}$

Now, ${(\frac{d y}{d x})}_{C_{1}} \times {(\frac{d y}{d x})}_{C_{2}} = (\frac{x_{1}^{2} - y_{1}^{2}}{2 x_{1} y_{1}}) \times (- \frac{2 x_{1} y_{1}}{x_{1}^{2} - y_{1}^{2}}) = - 1$

The tangents to the two curves are perpendicular to each other.

Thus, the two curves cut at right angle.

Hence, the correct answer is option (b).

Page No 15.43:

Question 32:

The tangent to the curve x = e^t cost, y = e^t sin t at t = $\frac{π}{4}$ makes with x-axis an angle
(a) 0 (b) $\frac{π}{4} (c) \frac{π}{3} (d) \frac{π}{2}$

Answer:

The given curve is x = e^t cost, y = e^t sint.

x = e^t cost

Differentiating both sides with respect to t, we get

$\frac{d x}{d t} = e^{t} \times (- \sin t) + \cos t \times e^{t}$

$\Rightarrow \frac{d x}{d t} = e^{t} (\cos t - \sin t)$

y = e^t sint

Differentiating both sides with respect to t, we get

$\frac{d y}{d t} = e^{t} \times \cos t + \sin t \times e^{t}$

$\Rightarrow \frac{d y}{d t} = e^{t} (\cos t + \sin t)$

Now,

Slope of tangent to the given curve $= \frac{d y}{d x}$ $= \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $= \frac{e^{t} (\cos t + \sin t)}{e^{t} (\cos t - \sin t)}$ $= \frac{\cos t + \sin t}{\cos t - \sin t}$

∴ Slope of tangent to the given curve at $t = \frac{π}{4}$ $= {(\frac{d y}{d x})}_{t = \frac{π}{4}}$ $= \frac{\cos \frac{π}{4} + \sin \frac{π}{4}}{\cos \frac{π}{4} - \sin \frac{π}{4}}$ $= \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}$ $= \frac{\sqrt{2}}{0} = \infty$

Let the angle made by the tangent to the given curve at t = $\frac{π}{4}$ with the x-axis be θ.

∴ tanθ = Slope of tangent to the given curve at $t = \frac{π}{4}$ = $\infty$

$\Rightarrow \tan θ = \tan \frac{π}{2}$

$\Rightarrow θ = \frac{π}{2}$

Thus, the tangent to the curve x = e^t cost, y = e^t sint at t = $\frac{π}{4}$ makes with x-axis an angle $\frac{π}{2}$ .

Hence, the correct answer is option (d).

Page No 15.43:

Question 33:

The tangent to the curve y = e^2x at the point (0,1) meets x-axis at :
(a) (0, 1) (b) $(- \frac{1}{2}, 0)$ (c) (2, 0) (d) (0, 2)

Answer:

The given curve is y = e^2x.

y = e^2x

$\Rightarrow \frac{d y}{d x} = 2 e^{2 x}$

$\Rightarrow {(\frac{d y}{d x})}_{(0, 1)} = 2 e^{2 \times 0} = 2 \times 1 = 2$         .....(1)

So, the equation of tangent at (0, 1) is

$y - 1 = {(\frac{d y}{d x})}_{(0, 1)} (x - 0)$

$\Rightarrow y - 1 = 2 x$               [Using (1)]

$\Rightarrow 2 x - y + 1 = 0$    .....(2)

This tangent meet the x-axis where y = 0.

Putting y = 0 in (2), we get

$2 x - 0 + 1 = 0$

$\Rightarrow x = - \frac{1}{2}$

Thus, the tangent to the given curve y = e^2x at the point (0,1) meets x-axis at $(- \frac{1}{2}, 0)$ .

Hence, the correct answer is option (b).

Page No 15.43:

Question 34:

The equation of tangent to the curve y(1 + x²) = 2 - x, where it crosses x-axis is
(a) x + 5y = 2 (b) x - 5y = 2 (c) 5x - y = 2 (d) 5x + y = 2

Answer:

The equation of given curve is

y(1 + x²) = 2 − x .....(1)

This cuts the x-axis at the point, where y = 0.

Putting y = 0 in (1), we get

0 × (1 + x²) = 2 − x

⇒ 2 − x = 0

⇒ x = 2

So, the given curve intersects the x-axis at (2, 0).

y(1 + x²) = 2 − x

Differentiating both sides with respect to x, we get

$y \times 2 x + (1 + x^{2}) \times \frac{d y}{d x} = - 1$

$\Rightarrow (1 + x^{2}) \frac{d y}{d x} = - 1 - 2 x y$

$\Rightarrow \frac{d y}{d x} = - \frac{1 + 2 x y}{1 + x^{2}}$

∴ Slope of tangent to the given curve at (2, 0) = ${(\frac{d y}{d x})}_{(2, 0)} = - \frac{1 + 2 \times 2 \times 0}{1 + {(2)}^{2}} = - \frac{1}{5}$

So, the equation of tangent at (2, 0) is

$y - 0 = {(\frac{d y}{d x})}_{(2, 0)} (x - 2)$

$\Rightarrow y = - \frac{1}{5} (x - 2)$

$\Rightarrow 5 y = - x + 2$

$\Rightarrow x + 5 y = 2$

Thus, the equation of tangent to the given curve where it crosses x-axis is x + 5y = 2.

Hence, the correct answer is option (a).

Page No 15.43:

Question 35:

The points at which the tangents to the curve y = x³ - 12x + 18 are parallel to x-axis are
(a) (2, -2)(-2, -34) (b) (2, 34)(-2, 0)
(c) (0, 34)(-2, 0) (d) (2, 2)(-2, 34)

Answer:

Let (x₁, y₁) be the point of contact of tangent with the given curve y = x³ − 12x + 18.

∴ $y_{1} = x_{1}^{3} - 12 x + 18$ .....(1)

y = x³ − 12x + 18

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 3 x^{2} - 12$

∴ Slope of tangent at (x₁, y₁) = ${(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 3 x_{1}^{2} - 12$

It is given that, the tangent is parallel to the x-axis.

$∴ {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 0$

$\Rightarrow 3 x_{1}^{2} - 12 = 0$

$\Rightarrow x_{1}^{2} = 4$

⇒ x₁ = −2 or x₁ = 2

Putting x₁ = −2 in (1), we get

$y_{1} = {(- 2)}^{3} - 12 \times (- 2) + 18 = - 8 + 24 + 18 = 34$

Putting x₁ = 2 in (1), we get

$y_{1} = {(2)}^{3} - 12 \times 2 + 18 = 8 - 24 + 18 = 2$

So, the coordinates of the point of contact are (−2, 34) and (2, 2).

Thus, the points at which the tangents to the given curve are parallel to x-axis are (−2, 34) and (2, 2).

Hence, the correct answer is option (d).

Page No 15.43:

Question 36:

The curve y = x^1/5 has at (0, 0)
(a) a vertical tangent (b) a horizontal tangent
(c) an oblique tangent (d) no tangent

Answer:

The given curve is

$y = x^{\frac{1}{5}}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{1}{5} x^{- \frac{4}{5}}$

At (0, 0), we have

${(\frac{d y}{d x})}_{(0, 0)} = \frac{1}{5} \times {(0)}^{- \frac{4}{5}} = \frac{1}{5} \times \infty = \infty$

We know that, if $\frac{d y}{d x} = \infty$ , then the tangent is parallel to the y-axis.

Now, ${(\frac{d y}{d x})}_{(0, 0)} = \infty$

So, the tangent to the given curve at (0, 0) is parallel to the y-axis.

Thus, the curve $y = x^{\frac{1}{5}}$ has at (0, 0) a vertical tangent.

Hence, the correct answer is option (a).

Page No 15.44:

Question 1:

The equation of the normal to the curve y = tan x at (0, 0) is ______________.

Answer:

The given curve is

y = tanx

Differentiating both sides with respect to x, we have

$\frac{d y}{d x} = \sec^{2} x$

∴ Slope of tangent at (0, 0) = ${(\frac{d y}{d x})}_{(0, 0)} = \sec^{2} 0 = {(1)}^{2} = 1$

⇒ Slope of normal at (0, 0) = $- \frac{1}{{(\frac{d y}{d x})}_{(0, 0)}} = - \frac{1}{1} = - 1$

So, the equation of normal at (0, 0) is

$y - 0 = - \frac{1}{{(\frac{d y}{d x})}_{(0, 0)}} (x - 0)$

$\Rightarrow y = - 1 \times x$

$\Rightarrow x + y = 0$

Thus, the equation of the normal to the given curve at (0, 0) is x + y = 0.

The equation of the normal to the curve y = tanx at (0, 0) is ___x + y = 0___.

Page No 15.44:

Question 2:

The value of 'a' for which y = x² + ax + 25 touches the axis of x are ______________.

Answer:

The given curve is y = x² + ax + 25.

Suppose this curve touches the x-axis at (h, 0).

∴ 0 = h² + ah + 25 .....(1)

y = x² + ax + 25

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 2 x + a$

∴ Slope of tangent at (h, 0) = ${(\frac{d y}{d x})}_{(h, 0)} = 2 h + a$

The given curve touches the x-axis at (h, 0). Therefore,

${(\frac{d y}{d x})}_{(h, 0)} = 0$ (Slope of the x-axis = 0)

$\Rightarrow 2 h + a = 0$

$\Rightarrow h = - \frac{a}{2}$

Putting $h = - \frac{a}{2}$ in (1), we get

$0 = {(- \frac{a}{2})}^{2} + a \times (- \frac{a}{2}) + 25$

$\Rightarrow \frac{a^{2}}{4} - \frac{a^{2}}{2} + 25 = 0$

$\Rightarrow \frac{a^{2}}{4} = 25$

$\Rightarrow a^{2} = 100$

$\Rightarrow a = \pm 10$

So, the values of 'a' are −10 and 10.

Thus, the values of 'a' for which y = x² + ax + 25 touches the x-axis are −10 and 10.

The value of 'a' for which y = x² + ax + 25 touches the axis of x are __−10 and 10__.

Page No 15.44:

Question 3:

The points on the curve y = 12x - x³ at which the gradient is zero are _______________.

Answer:

Let (h, k) be the point on the curve y = 12x − x³ at which the gradient (or slope) of the tangent is zero.

∴ k = 12h − h³ .....(1)

y = 12x − x³

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 12 - 3 x^{2}$

$∴ {(\frac{d y}{d x})}_{(h, k)} = 12 - 3 h^{2} = 0$ (Given)

$\Rightarrow 12 - 3 h^{2} = 0$

$\Rightarrow h^{2} = 4$

⇒ h = −2 or h = 2

Putting h = −2 in (1), we get

k = 12 × (−2) − (−2)³ = −24 + 8 = −16

Putting h = 2 in (1), we get

k = 12 × 2 − (2)³ = 24 − 8 = 16

So, the coordinates of the required point are (−2, −16) and (2, 16).

Thus, the points on the curve y = 12x − x³ at which the gradient is zero are (−2, −16) and (2, 16).

The points on the curve y = 12x − x³ at which the gradient is zero are ___(−2, −16) and (2, 16)___.

Page No 15.44:

Question 4:

The coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are_______________.

Answer:

Let (h, k) be the point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3.

y = x log_ex

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = x \times \frac{1}{x} + \log_{e} x \times 1$

$\Rightarrow \frac{d y}{d x} = 1 + \log_{e} x$

It is given that, the normal is parallel to the line 2x − 2y = 3.

∴ Slope of normal at (h, k) = Slope of the given line

$\Rightarrow - \frac{1}{{(\frac{d y}{d x})}_{(h, k)}} = - \frac{2}{(- 2)}$

$\Rightarrow - \frac{1}{1 + \log_{e} h} = 1$

$\Rightarrow 1 + \log_{e} h = - 1$

$\Rightarrow \log_{e} h = - 2$

$\Rightarrow h = e^{- 2} = \frac{1}{e^{2}}$

Now, (h, k) lies on the given curve.

$∴ k = h \log_{e} h$ .....(1)

Putting $h = \frac{1}{e^{2}}$ in (1), we get

$k = \frac{1}{e^{2}} \times \log_{e} \frac{1}{e^{2}}$

$\Rightarrow k = \frac{1}{e^{2}} \times \log_{e} e^{- 2}$

$\Rightarrow k = - \frac{2}{e^{2}} (\log a^{b} = b \log a)$

So, the coordinates of the required points are $(\frac{1}{e^{2}}, - \frac{2}{e^{2}})$ .

Thus, the coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are $(\frac{1}{e^{2}}, - \frac{2}{e^{2}})$ .

The coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are $(\frac{1}{e^{2}}, - \frac{2}{e^{2}})$ .

Page No 15.44:

Question 5:

The coordinates of the point on the curve y = 2 + $\sqrt{4 x + 1}$ where tangent has slope $\frac{2}{5}$ are _________________.

Answer:

Let (h, k) be the point on the curve $y = 2 + \sqrt{4 x + 1}$ where tangent has slope $\frac{2}{5}$ .

$y = 2 + \sqrt{4 x + 1}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 0 + \frac{1}{2 \sqrt{4 x + 1}} \times 4$

$\Rightarrow \frac{d y}{d x} = \frac{2}{\sqrt{4 x + 1}}$

∴ Slope of tangent at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = \frac{2}{5}$ (Given)

$\Rightarrow \frac{2}{\sqrt{4 h + 1}} = \frac{2}{5}$

$\Rightarrow \sqrt{4 h + 1} = 5$

$\Rightarrow 4 h + 1 = 25$

$\Rightarrow 4 h = 24$

$\Rightarrow h = 6$

Now, (h, k) lies on the given curve $y = 2 + \sqrt{4 x + 1}$ .

$∴ k = 2 + \sqrt{4 h + 1}$ .....(1)

Putting h = 6 in (1), we get

$k = 2 + \sqrt{4 \times 6 + 1}$

$\Rightarrow k = 2 + \sqrt{25}$

$\Rightarrow k = 2 + 5 = 7$

So, the coordinates of the required point are (6, 7).

Thus, the coordinates of the point on the curve $y = 2 + \sqrt{4 x + 1}$ where tangent has slope $\frac{2}{5}$ are (6, 7).

The coordinates of the point on the curve y = 2 + $\sqrt{4 x + 1}$ where tangent has slope $\frac{2}{5}$ are ___(6, 7)___.

Page No 15.44:

Question 6:

The slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is ________________.

Answer:

The given curve is x = 3t² + 1, y = t³ − 1.

When x = 1, we have

3t² + 1 = 1

⇒ 3t² = 0

⇒ t = 0

Now,

x = 3t² + 1

Differentiating both sides with respect to t, we get

$\frac{d x}{d t} = 6 t$        .....(1)

y = t³ − 1

Differentiating both sides with respect to t, we get

$\frac{d y}{d t} = 3 t^{2}$       .....(2)

$∴ \frac{d y}{d x}$ $= \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $= \frac{3 t^{2}}{6 t}$ $= \frac{t}{2}$       [Using (1) and (2)]

At t = 0, slope of the tangent = ${(\frac{d y}{d x})}_{t = 0}$ = $\frac{0}{2}$ = 0 $(\frac{d y}{d x} = \frac{t}{2})$

So, the slope of the tangent to the given curve at t = 0 (or x = 1) is 0.

Thus, the slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is 0.

The slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is ___0___.

Page No 15.44:

Question 7:

The angle of intersection of the curves y = x² and x = y² at (0, 0) is __________________.

Answer:

The given curves are y = x² and x = y².

Let C₁ represents the curve y = x² and C₂ represents the curve x = y².

y = x²

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 2 x$

∴ Slope of the tangent to the curve C₁ at (0, 0) = ${(\frac{d y}{d x})}_{(0, 0)}$ = 2 × 0 = 0

So, the the tangent to the curve y = x² at (0, 0) is parallel to the x-axis.

x = y²

Differentiating both sides with respect to x, we get

$1 = 2 y \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 y}$

∴ Slope of the tangent to the curve C₂ at (0, 0) = ${(\frac{d y}{d x})}_{(0, 0)}$ $= \frac{1}{2 \times 0} = \frac{1}{0} = \infty$

So, the tangent to the curve x = y² at (0, 0) is parallel to the y-axis.

Now, the tangent to the curve y = x² at (0, 0) is parallel to the x-axis and tangent to the curve x = y² at (0, 0) is parallel to the y-axis. So, the angle between the tangents to the given curves at (0, 0) is $\frac{π}{2}$ .

Thus, the angle of intersection of the given curves = x² and x = y² at (0, 0) is $\frac{π}{2}$ .

The angle of intersection of the curves y = x² and x = y² at (0, 0) is $\frac{π}{2}$ .

Page No 15.44:

Question 8:

The slope of the tangent to the curve y = be^−x/a where it crosses y-axis is ______________.

Answer:

The equation of given curve is

$y = b e^{- \frac{x}{a}}$ .....(1)

It crosses the y-axis at the point, where x = 0.

Putting x = 0 in (1), we get

$y = b e^{- \frac{0}{a}}$

$\Rightarrow y = b e^{0}$

$\Rightarrow y = b$

So, the coordinates of the point where the given curve crosses y-axis is (0, b).

$y = b e^{- \frac{x}{a}}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = b e^{- \frac{x}{a}} \times (- \frac{1}{a})$

$\Rightarrow \frac{d y}{d x} = - \frac{b}{a} e^{- \frac{x}{a}}$

∴ Slope of the tangent at (0, b) $= {(\frac{d y}{d x})}_{(0, b)}$ $= - \frac{b}{a} e^{- \frac{0}{a}}$ $= - \frac{b}{a} e^{0}$ $= - \frac{b}{a}$

Thus, the slope of tangent to the given curve $y = b e^{- \frac{x}{a}}$ where it crosses y-axis i.e. (0, b) is $- \frac{b}{a}$ .

The slope of the tangent to the curve $y = b e^{- \frac{x}{a}}$ where it crosses y-axis is $- \frac{b}{a}$ .

Page No 15.44:

Question 9:

The tangent to the curve y = e^2x at (0, 1) cuts x-axis at the point __________________.

Answer:

The given curve is

y = e^2x

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 2 e^{2 x}$

∴ Slope of tangent at (0, 1) = ${(\frac{d y}{d x})}_{(0, 1)} = 2 e^{2 \times 0} = 2 \times 1 = 2$         .....(1)

So, the equation of tangent at (0, 1) is

$y - 1 = {(\frac{d y}{d x})}_{(0, 1)} (x - 0)$

$\Rightarrow y - 1 = 2 x$               [Using (1)]

$\Rightarrow 2 x - y + 1 = 0$    .....(2)

This tangent cuts the x-axis where y = 0.

Putting y = 0 in (2), we get

$2 x - 0 + 1 = 0$

$\Rightarrow x = - \frac{1}{2}$

So, the coordinates of the required point are $(- \frac{1}{2}, 0)$ .

Thus, the tangent to the given curve y = e^2x at (0, 1) cuts the x-axis at $(- \frac{1}{2}, 0)$ .

The tangent to the curve y = e^2x at (0, 1) cuts x-axis at the point $(- \frac{1}{2}, 0)$ .

Page No 15.44:

Question 10:

The slope of the normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is equal to _________________.

Answer:

The equation of given curve is

y³ − xy − 8 = 0

Differentiating both sides with respect to x, we get

$3 y^{2} \frac{d y}{d x} - (x \frac{d y}{d x} + y) - 0 = 0$

$\Rightarrow (3 y^{2} - x) \frac{d y}{d x} = y$

$\Rightarrow \frac{d y}{d x} = \frac{y}{3 y^{2} - x}$

Now, slope of tangent at (0, 2) = ${(\frac{d y}{d x})}_{(0, 2)} = \frac{2}{3 \times {(2)}^{2} - 0} = \frac{2}{12} = \frac{1}{6}$ .....(1)

∴ Slope of the normal at (0, 2)

$= - \frac{1}{{(\frac{d y}{d x})}_{(0, 2)}}$

$= - \frac{1}{(\frac{1}{6})}$ [From (1)]

$= - 6$

Thus, the slope of normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is −6.

The slope of the normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is equal to ___−6___.

Page No 15.44:

Question 11:

If the normal to the curve y² = 5x -1, at the point(1, -2) is of the form ax - 5y + b = 0, then a + b = _________________.

Answer:

The equation of given curve is

y² = 5x −1

Differentiating both sides with respect to x, we get

$2 y \frac{d y}{d x} = 5$

$\Rightarrow \frac{d y}{d x} = \frac{5}{2 y}$

∴ Slope of normal at (1, −2)

$= - \frac{1}{{(\frac{d y}{d x})}_{(1, - 2)}}$

$= - \frac{1}{\frac{5}{2 \times (- 2)}}$

$= \frac{4}{5}$

So, the equation of normal at (1, −2) is

$y - (- 2) = - \frac{1}{{(\frac{d y}{d x})}_{(1, - 2)}} (x - 1)$

$\Rightarrow y - (- 2) = \frac{4}{5} (x - 1)$

$\Rightarrow 5 y + 10 = 4 x - 4$

$\Rightarrow 4 x - 5 y - 14 = 0$

Comparing with the given equation of normal ax − 5y + b = 0, we get

a = 4 and b = −14

∴ a + b = 4 + (−14) = −10

Thus, the value of a + b is −10.

If the normal to the curve y² = 5x −1, at the point(1, −2) is of the form ax − 5y + b = 0, then a + b = ___−10___.

Page No 15.44:

Question 12:

If the line ax + by + c = 0 is normal to the curve xy = 1, then the set of values of $\frac{a}{b}$ , is _____________.

Answer:

The equation of given curve is xy = 1.

Let (x₁, y₁) be the point of contact of normal with the curve.

$∴ x_{1} y_{1} = 1$ .....(1)

Now,

xy = 1

Differentiating both sides with respect to x, we get

$x \frac{d y}{d x} + y = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{y}{x}$

∴ Slope of the normal at (x₁, y₁)

$= - \frac{1}{{(\frac{d y}{d x})}_{(x_{1}, y_{1})}}$

$= - \frac{1}{(- \frac{y_{1}}{x_{1}})}$

$= \frac{x_{1}}{y_{1}}$

$= x_{1}^{2}$ [Using (1)]

The given equation of normal to the curve is ax + by + c = 0.

Slope of this line = $- \frac{a}{b}$

$∴ x_{1}^{2} = - \frac{a}{b}$

$\Rightarrow \frac{a}{b} = - x_{1}^{2}$

Now, $x_{1}^{2}$ is always positive.

∴ $\frac{a}{b}$ < 0 ∀ x ∈ R

Thus, the set of values of $\frac{a}{b}$ is the set of all negative real numbers i.e. (−∞, 0).

If the line ax + by + c = 0 is normal to the curve xy = 1, then the set of values of $\frac{a}{b}$ , is __(−∞, 0)__.

Page No 15.44:

Question 13:

If the normal to the curve y = f(x) at (3, 4) makes an angle $\frac{3 π}{4}$ with positive x-axis, then f '(3) is equal to ________________.

Answer:

It is given that, the normal to the curve y = f(x) at (3,4) makes an angle $\frac{3 π}{4}$ with positive x-axis.

∴ Slope of the normal = $\tan \frac{3 π}{4} = \tan (π - \frac{π}{4}) = - \tan \frac{π}{4}$ = −1 .....(1)

Now, y = f(x)

$\Rightarrow \frac{d y}{d x} = f' (x)$

∴ Slope of the normal at (3, 4) $= - \frac{1}{{(\frac{d y}{d x})}_{(3, 4)}} = - \frac{1}{f' (3)}$ .....(2)

From (1) and (2), we have

$- \frac{1}{f' (3)} = - 1$

$\Rightarrow f' (3) = 1$

Thus, the value of $f' (3)$ is 1.

If the normal to the curve y = f(x) at (3, 4) makes an angle $\frac{3 π}{4}$ with positive x-axis, then f '(3) is equal to ___1___.

Page No 15.44:

Question 14:

The equation of the tangent to the curve y = x + $\frac{4}{x^{2}}$ , that is parallel to x-axis, is ________________.

Answer:

The equation of the given curve is $y = x + \frac{4}{x^{2}}$ .

Let (h, k) be the point of contact of tangent with the curve.

$∴ k = h + \frac{4}{h^{2}}$ .....(1)

$y = x + \frac{4}{x^{2}}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 1 - \frac{8}{x^{3}}$

∴ Slope of tangent at (h, k) $= {(\frac{d y}{d x})}_{(h, k)} = 1 - \frac{8}{h^{3}}$

It is given that, the tangent to the given curve is parallel to the x-axis.

∴ Slope of tangent = 0

$\Rightarrow {(\frac{d y}{d x})}_{(h, k)} = 0$

$\Rightarrow 1 - \frac{8}{h^{3}} = 0$

$\Rightarrow h^{3} = 8$

$\Rightarrow h = 2$

Putting h = 2 in (1), we get

$k = 2 + \frac{4}{{(2)}^{2}} = 2 + 1 = 3$

Thus, the point of contact of tangent with the curve is (2, 3).

So, the equation of the tangent at (2, 3) is

$y - 3 = {(\frac{d y}{d x})}_{(2, 3)} (x - 2)$

$\Rightarrow y - 3 = 0 [{(\frac{d y}{d x})}_{(2, 3)} = 0]$

$\Rightarrow y = 3$

Thus, the equation of tangent to the given curve $y = x + \frac{4}{x^{2}}$ , that is parallel to x-axis, is y = 3.

The equation of the tangent to the curve y = x + $\frac{4}{x^{2}}$ , that is parallel to x-axis, is ___y = 3___.

Page No 15.44:

Question 16:

If slope of tangent to curve y = x³ at a point is equal to ordinate of point, then the point is _________________.

Answer:

The equation of given curve is y = x³.

Let (h, k) be the point on the curve at which slope of tangent is equal to ordinate of the point.

∴ k = h³        .....(1)

Now,

y = x³

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 3 x^{2}$

∴ Slope of tangent at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = 3 h^{2}$

It is given that,

Slope of tangent at (h, k) = Ordinate of (h, k)

⇒ 3h² = k       .....(2)

From (1) and (2), we get

$3 h^{2} = h^{3}$

$\Rightarrow h^{3} - 3 h^{2} = 0$

$\Rightarrow h^{2} (h - 3) = 0$

⇒ h = 0 or h − 3 = 0

⇒ h = 0 or h = 3

When h = 0,

k = (0)³ = 0           [From (1)]

When h = 3,

k = (3)³ = 27           [From (1)]

So, the coordinates of the required points are (0, 0) and (3, 27).

Thus, if slope of tangent to curve y = x³ at a point is equal to ordinate of point, then the points are (0, 0) and (3, 27).

If slope of tangent to curve y = x³ at a point is equal to ordinate of point, then the point is ___(0, 0) and (3, 27)___.

Page No 15.44:

Question 17:

The slope of the normal to the curve x² + y² − 2x + 4y − 5 = 0 at (2, 1) is _________________.

Answer:

The equation of given curve is

x² + y² − 2x + 4y − 5 = 0

Differentiating both sides with respect to x, we get

$2 x + 2 y \frac{d y}{d x} - 2 + 4 \frac{d y}{d x} - 0 = 0$

$\Rightarrow (2 y + 4) \frac{d y}{d x} = 2 - 2 x$

$\Rightarrow \frac{d y}{d x} = \frac{1 - x}{y + 2}$

∴ Slope of normal at (2, 1)

$= - \frac{1}{{(\frac{d y}{d x})}_{(2, 1)}}$

$= - \frac{1}{(\frac{1 - 2}{1 + 2})}$

$= - \frac{1}{(- \frac{1}{3})}$

= 3

Thus, the slope of normal to the given curve x² + y² − 2x + 4y − 5 = 0 at (2, 1) is 3.

The slope of the normal to the curve x² + y² − 2x + 4y − 5 = 0 at (2, 1) is ___3___.

Page No 15.44:

Question 18:

The point on the curve y² = x, the tangent at which makes an angle of 45^o with x-axis is ____________________.

Answer:

Let (h, k) be the point on the curve y² = x, the tangent at which makes an angle of 45º with x-axis.

y² = x

Differentiating both sides with respect to x, we get

$2 y \frac{d y}{d x} = 1$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 y}$

∴ Slope of tangent at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = \frac{1}{2 k}$      .....(1)

It is given that, the tangent makes an angle of 45º with x-axis.

∴ Slope of the tangent = tan45º = 1        .....(2)

From (1) and (2), we have

$\frac{1}{2 k} = 1$

$\Rightarrow k = \frac{1}{2}$

Now, (h, k) lies on the curve.

∴ k² = h     .....(3)

Putting $k = \frac{1}{2}$ in (3), we get

$h = {(\frac{1}{2})}^{2} = \frac{1}{4}$

So, the coordinates of the required point are $(\frac{1}{4}, \frac{1}{2})$ .

Thus, the point on the curve y² = x, the tangent at which makes an angle of 45º with x-axis is $(\frac{1}{4}, \frac{1}{2})$ .

The point on the curve y² = x, the tangent at which makes an angle of 45^o with x-axis is $(\frac{1}{4}, \frac{1}{2})$ .

Page No 15.44:

Question 19:

The curve y = 4x² + 2x − 8 and y = x³ − x + 13 touch each other at the point _________________.

Answer:

Disclaimer: The solution has been provided for the following question.

The curve y = 4x² + 2x − 8 and y = x³ − x + 10 touch each other at the point _________________.

Solution:

Let the given curves each other at the point (h, k).

∴ k = 4h² + 2h − 8 .....(1)

k = h³ − h + 10 .....(2)

Consider the first curve C₁ ≡ y = 4x² + 2x − 8.

y = 4x² + 2x − 8

$\Rightarrow \frac{d y}{d x} = 8 x + 2$

∴ Slope of tangent to the curve C₁ at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = 8 h + 2$

Consider the second curve C₂ ≡ y = x³ − x + 10.

y = x³ − x + 10

$\Rightarrow \frac{d y}{d x} = 3 x^{2} - 1$

∴ Slope of tangent to the curve C₂ at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = 3 h^{2} - 1$

It is given that, the two curves touch each other at (h, k).

∴ Slope of tangent to the curve C₁ at (h, k) = Slope of tangent to the curve C₂ at (h, k)

$\Rightarrow 8 h + 2 = 3 h^{2} - 1$

$\Rightarrow 3 h^{2} - 8 h - 3 = 0$

$\Rightarrow 3 h^{2} - 9 h + h - 3 = 0$

$\Rightarrow 3 h (h - 3) + 1 (h - 3) = 0$

$\Rightarrow (3 h + 1) (h - 3) = 0$

⇒ 3h + 1 = 0 or h − 3 = 0

⇒ h = $- \frac{1}{3}$ or h = 3

Putting h = $- \frac{1}{3}$ in (1), we get

$k = 4 {(- \frac{1}{3})}^{2} + 2 \times (- \frac{1}{3}) - 8 = \frac{4}{9} - \frac{2}{3} - 8 = - \frac{74}{9}$

Putting h = $- \frac{1}{3}$ in (2), we get

$k = {(- \frac{1}{3})}^{3} - (- \frac{1}{3}) + 10 = - \frac{1}{27} + \frac{1}{3} + 10 = \frac{278}{27}$

Here, the values of k are not same for the two curves for h = $- \frac{1}{3}$ .

So, the two curves do not touch each other when h = $- \frac{1}{3}$ . However, the tangents are parallel to each other at h = $- \frac{1}{3}$ .

Putting h = 3 in (1), we get

$k = 4 \times {(3)}^{2} + 2 \times 3 - 8 = 36 + 6 - 8 = 34$

Putting h = 3 in (2), we get

$k = {(3)}^{3} - 3 + 10 = 27 - 3 + 10 = 34$

Here, the values of k are same for the two curves when h = 3.

So, the two curves touch each other at (3, 34).

Thus, the curves y = 4x² + 2x − 8 and y = x³ − x + 10 touch each other at the point (3, 34).

The curve y = 4x² + 2x − 8 and y = x³ − x + 10 touch each other at the point ___(3, 34)___.

Page No 15.45:

Question 1:

Find the point on the curve y = x² − 2x + 3, where the tangent is parallel to x-axis.

Answer:

The slope of the x-axis is 0.
Now, let (x₁, y₁) be the required point.
Here,
$Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{2} - 2 x_{1} + 3 . . . (1) Now, y = x^{2} - 2 x + 3 \frac{d y}{d x} = 2 x - 2 Slope of the tangent at (x, y) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 2 x_{1} - 2 Given: Slope of the tangent at (x_{1}, y_{1}) =Slope of the x -axis = 2 x_{1} - 2 = 0 \Rightarrow x_{1} = 1 and y_{1} = 1 - 2 + 3 = 2 [From (1)] ∴ Required point= (x_{1}, y_{1}) = (1, 2)$

Page No 15.45:

Question 2:

Find the slope of the tangent to the curve x = t² + 3t − 8, y = 2t² − 2t − 5 at t = 2.

Answer:

$Here, x = t^{2} + 3 t - 8 and y = 2 t^{2} - 2 t - 5 \frac{d x}{d t} = 2 t + 3 and \frac{d y}{d t} = 4 t - 2 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} Now, Slope of the tangent= {(\frac{d y}{d x})}_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$

Page No 15.45:

Question 3:

If the tangent line at a point (x, y) on the curve y = f(x) is parallel to x-axis, then write the value of $\frac{d y}{d x}$ .

Answer:

The slope of the x-axis is 0.
Also, the tangent at a point (x, y) on the curve y = f(x) is parallel to the x-axis.
∴ Slope of the tangent $(\frac{d y}{d x})$ = Slope of the x-axis = 0

Page No 15.45:

Question 4:

Write the value of $\frac{d y}{d x}$ , if the normal to the curve y = f(x) at (x, y) is parallel to y-axis.

Answer:

The slope of the y-axis is $\infty$ .
Also, the normal at (x, y) on the curve y = f(x) is parallel to the y-axis.
∴ Slope of the normal = Slope of the y-axis = $\infty$
$⇒ \frac{d y}{d x} =Slope of the tangent= \frac{- 1}{Slope of the normal} = \frac{- 1}{\infty} =0$

Page No 15.45:

Question 5:

If the tangent to a curve at a point (x, y) is equally inclined to the coordinates axes then write the value of $\frac{d y}{d x}$ .

Answer:

Because the tangent to the curve at (x, y) is equally inclined to the coordinate axes, the angle made by the tangent with the axes can be $\pm 45 °$ .
$∴ \frac{d y}{d x} =Slope of the tangent=tan (\pm 45) =±1$

Page No 15.45:

Question 6:

If the tangent line at a point (x, y) on the curve y = f(x) is parallel to y-axis, find the value of $\frac{d x}{d y}$ .

Answer:

Slope of the y-axis is $\infty$ .
Also, the tangent at (x, y) on the curve y = f(x) is parallel to the y-axis,
∴ Slope of the tangent, $\frac{d y}{d x}$ = Slope of the y-axis = $\infty$
$\frac{d x}{d y} = \frac{1}{(\frac{d y}{d x})} = \frac{1}{\infty} = 0$

Page No 15.45:

Question 7:

Find the slope of the normal at the point 't' on the curve $x = \frac{1}{t}, y = t$ .

Answer:

$Here, x = \frac{1}{t} and y = t \frac{d x}{d t} = \frac{- 1}{t^{2}} and \frac{d y}{d t} = 1 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{1}{(\frac{- 1}{t^{2}})} = - t^{2} Now, Slope of the tangent= {(\frac{d y}{d x})}_{} = - t^{2} Slope of the normal= \frac{- 1}{Slope of the tangent} = \frac{- 1}{- t^{2}} = \frac{1}{t^{2}}$

Page No 15.45:

Question 8:

Write the coordinates of the point on the curve y² = x where the tangent line makes an angle $\frac{π}{4}$ with x-axis.

Answer:

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = x_{1} Now, y^{2} = x \Rightarrow 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1}{2 y_{1}} Given: \frac{1}{2 y_{1}} = 1 \Rightarrow 2 y_{1} = 1 \Rightarrow y_{1} = \frac{1}{2} Now, x_{1} = {y_{1}}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} ∴ (x_{1}, y_{1}) = (\frac{1}{4}, \frac{1}{2})$

Page No 15.45:

Question 9:

Write the angle made by the tangent to the curve x = e^t cos t, y = e^t sin t at $t = \frac{π}{4}$ with the x-axis.

Answer:

$Here, x = e^{t} \cos t and y = e^{t} \sin t \frac{d x}{d t} = e^{t} c os t - e^{t} \sin t and \frac{d y}{d t} = e^{t} \sin t + e^{t} \cos t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{e^{t} \sin t + e^{t} \cos t}{e^{t} c os t - e^{t} \sin t} = \frac{\sin t + \cos t}{\cos t - \sin t} Now, Slope of the tangent = {(\frac{d y}{d x})}_{t = \frac{π}{4}} = \frac{\sin \frac{π}{4} + \cos \frac{π}{4}}{\cos \frac{π}{4} - \sin \frac{π}{4}} = \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}} = \frac{\frac{2}{\sqrt{2}}}{0} =∞ Let θ be the angle made by the tangent with the x -axis. ∴ tan θ =∞ \Rightarrow θ = \frac{π}{2}$

Page No 15.45:

Question 10:

Write the equation of the normal to the curve y = x + sin x cos x at $x = \frac{π}{2}$ .

Answer:

$Here, y = x + \sin x \cos x On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = 1 + \cos^{2} x - \sin^{2} x Now, Slope of the tangent = {(\frac{d y}{d x})}_{x = \frac{π}{2}} {= 1+cos}^{2} \frac{π}{2} {-sin}^{2} \frac{π}{2} = 1-1=0 When x = \frac{π}{2}, y = \frac{π}{2} +sin \frac{π}{2} cos \frac{π}{2} = \frac{π}{2} ∴ (x_{1}, y_{1}) = (\frac{π}{2}, \frac{π}{2}) Equation of the normal = y - y_{1} = \frac{- 1}{Slope of the tangent} (x - x_{1}) \Rightarrow y - \frac{π}{2} = \frac{- 1}{0} (x - \frac{π}{2}) \Rightarrow x = \frac{π}{2} \Rightarrow 2 x = π$

Page No 15.45:

Question 11:

Find the coordinates of the point on the curve y² = 3 − 4x where tangent is parallel to the line 2x + y − 2 = 0.

Answer:

Let (x₁, y₁) be the required point.
Slope of the given line = $-$ 2

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = 3 - 4 x_{1} . . . (1) Now, y^{2} = 3 - 4 x \Rightarrow 2 y \frac{d y}{d x} = - 4 ∴ \frac{d y}{d x} = \frac{- 4}{2 y} = \frac{- 2}{y} S lope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 2}{y_{1}} Given: Slope of the tangent = Slope of the line \Rightarrow \frac{- 2}{y_{1}} = - 2 \Rightarrow y_{1} = 1 From (1), we get 1 = 3 - 4 x_{1} \Rightarrow - 2 = - 4 x_{1} \Rightarrow x_{1} = \frac{1}{2} ∴ (x_{1}, y_{1}) = (\frac{1}{2}, 1)$

Page No 15.45:

Question 12:

Write the equation on the tangent to the curve y = x² − x + 2 at the point where it crosses the y-axis.

Answer:

When the curve crosses the y-axis, the point on the curve is of the form (0, y).
Here,
$y = x^{2} - x + 2 \Rightarrow y = 0 - 0 + 2 = 2 So, the point where the curve crosses the y -axis is (0, 2). Now, y = x^{2} - x + 2 \Rightarrow \frac{d y}{d x} = 2 x - 1 Slope of the tangent, m = {(\frac{d y}{d x})}_{(0, 2)} =2 (0) -1=-1 ∴ (x_{1}, y_{1}) = (0, 2) and Equation of tangent = y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 = - 1 (x - 0) \Rightarrow y - 2 = - x \Rightarrow x + y - 2 = 0$

Page No 15.45:

Question 13:

Write the angle between the curves y² = 4x and x² = 2y − 3 at the point (1, 2).

Answer:

$Given : y^{2} = 4 x . . . (1) x^{2} = 2 y - 3 . . . (2) On d ifferentiating (1) w.r.t. x, we get 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{2}{y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 2)} = \frac{2}{2} = 1 On d ifferentiating (2) w.r.t. x, we get 2 x = 2 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = x \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 2)} = 1 Thus, we get \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| \Rightarrow \tan θ = |\frac{1 - 1}{1 + 1}| \Rightarrow \tan θ = 0 \Rightarrow θ = 0^{o}$

Page No 15.45:

Question 14:

Write the angle between the curves y = e^−x and y = e^x at their point of intersections.

Answer:

$Given : y = e^{- x} . . . (1) y = e^{x} . . . (2) On s ubstituting the value of y in (1), we get e^{- x} = e^{x} \Rightarrow x = 0 and y = 1 [From (2)] On d ifferentiating (1) w.r.t. x,we get \frac{d y}{d x} = - e^{- x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(0, 1)} = - 1 On d ifferentiating (2) w.r.t . x, we get \frac{d y}{d x} = e^{x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(0, 1)} = 1 ∵ m_{1} \times m_{2} = - 1 Since the multiplication of the slopes is - 1 so the slopes are perpendicular to each other . ∴ Required angle = \frac{π}{2}$

Page No 15.45:

Question 15:

Write the slope of the normal to the curve $y = \frac{1}{x}$ at the point $(3, \frac{1}{3})$ .

Answer:

$Given : y = \frac{1}{x} On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = \frac{- 1}{x^{2}} Now, Slope of the tangent = {(\frac{d y}{d x})}_{(3, \frac{1}{3})} = \frac{- 1}{9} Slope of the normal = \frac{- 1}{Slope of tangent} = \frac{- 1}{(\frac{- 1}{9})} = 9$

Page No 15.45:

Question 16:

Write the coordinates of the point at which the tangent to the curve y = 2x² − x + 1 is parallel to the line y = 3x + 9.

Answer:

Let (x₁, y₁) be the required point.
Slope of the given line = 3
$Since, the point lies on the curve . Hence, y_{1} = 2 {x_{1}}^{2} - x_{1} + 1 . . . (1) Now, y = 2 x^{2} - x + 1 ∴ \frac{d y}{d x} = 4 x - 1 Now, S lope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 4 x_{1} - 1 Given: Slope of the tangent = Slope of line ∴ 4 x_{1} - 1 = 3 \Rightarrow x_{1} = 1 From (1), we get y_{1} = 2 - 1 + 1 = 2 ∴ (x_{1}, y_{1}) = (1, 2)$

Page No 15.46:

Question 17:

Write the equation of the normal to the curve y = cos x at (0, 1).

Answer:

$Given : y = \cos x On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = - \sin x Now, Slope of the tangent= {(\frac{d y}{d x})}_{(0, 1)} =-sin 0=0 and Equation of the normal = y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = \frac{- 1}{0} (x - 0) \Rightarrow x = 0$

Page No 15.46:

Question 18:

Write the equation of the tangent drawn to the curve $y = \sin x$ at the point (0,0).

Answer:

We have,

$y = \sin x$

$\Rightarrow \frac{d y}{d x} = \cos x$

Slope at (0, 0) = m = ${[\frac{d y}{d x}]}_{x = 0} = \cos 0 = 1$

So, the equation of the tangent at (0,0) is given by,

y = mx

Putting m = 1, we get

The equation of the tangent is y = x.

View NCERT Solutions for all chapters of Class 12

Login or Create a free account