Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 26 - Ellipse

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Rd Sharma Xi 2020 2021 _volume 2 Solutions for Class 11 Science Maths Chapter 26 Ellipse are provided here with simple step-by-step explanations. These solutions for Ellipse are extremely popular among Class 11 Science students for Maths Ellipse Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 2 Book of Class 11 Science Maths Chapter 26 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 2 Solutions. All Rd Sharma Xi 2020 2021 _volume 2 Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 26.22:

Question 1:

Find the equation of the ellipse whose focus is (1, −2), the directrix 3x − 2y + 5 = 0 and eccentricity equal to 1/2.

Answer:

$Let S (1, - 2) be the focus and ZZ' be the directrix . Let P (x, y) be any point on the ellipse and let PM be the perpendicular from P on the directix . Then by the definition of an ellipse, we have : S P = e . P M, where e = \frac{1}{2} \Rightarrow S P^{2} = e^{2} . P M^{2} \Rightarrow (x - 1)^{2} + (y + 2)^{2} = {(\frac{1}{2})}^{2} . {|\frac{3 x - 2 y + 5}{\sqrt{(3)^{2} + (- 2)^{2}}}|}^{2} \Rightarrow x^{2} + 1 - 2 x + y^{2} + 4 + 4 y = (\frac{1}{4}) . |\frac{9 x^{2} + 4 y^{2} + 25 - 12 x y - 20 y + 30 x}{13}| \Rightarrow 52 (x^{2} + 1 - 2 x + y^{2} + 4 + 4 y) = 9 x^{2} + 4 y^{2} + 25 - 12 x y - 20 y + 30 x \Rightarrow 52 x^{2} + 52 - 104 x + 52 y^{2} + 208 + 208 y = 9 x^{2} + 4 y^{2} + 25 - 12 x y - 20 y + 30 x \Rightarrow 43 x^{2} + 48 y^{2} - 134 x + 228 y + 12 x y + 235 = 0 This is the equation of the required ellipse .$

Page No 26.22:

Question 2:

Find the equation of the ellipse in the following cases:
(i) focus is (0, 1), directrix is x + y = 0 and e = $\frac{1}{2}$
(ii) focus is (−1, 1), directrix is x − y + 3 = 0 and e = $\frac{1}{2}$
(iii) focus is (−2, 3), directrix is 2x + 3y + 4 = 0 and e = $\frac{4}{5}$
(iv) focus is (1, 2), directrix is 3x + 4y − 5 = 0 and e = $\frac{1}{2}$ .

Answer:

(i)

$Let S (0, 1) be the focus and ZZ' be the directrix . Let P (x, y) be any point on the ellipse and let PM be the perpendicular from P on the directrix . Then by the definition, we have : SP = e \times PM \Rightarrow SP = \frac{1}{2} \times PM \Rightarrow 2 SP = PM \Rightarrow 4 {(SP)}^{2} = {PM}^{2} \Rightarrow 4 [{(x)}^{2} + {(y - 1)}^{2}] = {|\frac{x + y}{\sqrt{1^{2} + {(1)}^{2}}}|}^{2} \Rightarrow 4 [x^{2} + y^{2} + 1 - 2 y] = \frac{x^{2} + y^{2} + 2 x y}{2} \Rightarrow 8 x^{2} + 8 y^{2} + 8 - 16 y = x^{2} + y^{2} + 2 x y \Rightarrow 7 x^{2} + 7 y^{2} - 2 x y - 16 y + 8 = 0 This is the required equation of the ellipse.$

(ii)

$Let S (- 1, 1) be the focus and ZZ' be the directrix . Let P (x, y) be any point on the ellipse and let PM be the perpendicular from P on the directrix . Then by the definition, we have : SP = e \times PM \Rightarrow SP = \frac{1}{2} \times PM \Rightarrow 2 SP = PM \Rightarrow 4 {(SP)}^{2} = {PM}^{2} \Rightarrow 4 [{(x + 1)}^{2} + {(y - 1)}^{2}] = {|\frac{x - y + 3}{\sqrt{1^{2} + {(- 1)}^{2}}}|}^{2} \Rightarrow 4 [x^{2} + 1 + 2 x + y^{2} + 1 - 2 y] = \frac{x^{2} + y^{2} + 9 - 2 x y - 6 y + 6 x}{2} \Rightarrow 8 x^{2} + 8 + 16 x + 8 y^{2} + 8 - 16 y = x^{2} + y^{2} + 9 - 2 x y - 6 y + 6 x \Rightarrow 7 x^{2} + 7 y^{2} + 2 x y - 10 y + 10 x + 7 = 0 This is the required equation of the ellipse.$

(iii)

$Let S (- 2, 3) be the focus and ZZ' be the directrix . Let P (x, y) be any point on the ellipse and let PM be the perpendicular from P on the directrix . Then by the definition, we have : SP = e \times PM \Rightarrow SP = \frac{4}{5} \times PM \Rightarrow \frac{5}{4} S P = PM \Rightarrow \frac{25}{16} {(SP)}^{2} = {PM}^{2} \Rightarrow \frac{25}{16} [{(x + 2)}^{2} + {(y - 3)}^{2}] = {|\frac{2 x + 3 y + 4}{\sqrt{2^{2} + 3^{2}}}|}^{2} \Rightarrow \frac{25}{16} [x^{2} + 4 + 4 x + y^{2} + 9 - 6 y] = \frac{4 x^{2} + 9 y^{2} + 16 + 12 x y + 24 y + 16 x}{13} \Rightarrow 325 (x^{2} + 4 + 4 x + y^{2} + 9 - 6 y) = 16 (4 x^{2} + 9 y^{2} + 16 + 12 x y + 24 y + 16 x) \Rightarrow 325 x^{2} + 1300 + 1300 x + 325 y^{2} + 2925 - 1950 y = 64 x^{2} + 144 y^{2} + 256 + 192 x y + 384 y + 256 x \Rightarrow 261 x^{2} + 181 y^{2} + 1044 x - 2309 y - 192 x y + 3969 = 0 This is the required equation of the ellipse.$

(iv)

$Let S (1, 2) be the focus and ZZ' be the directrix . Let P (x, y) be any point on the ellipse and let PM be the perpendicular from P on the directrix . Then by the definition, we have : SP = e \times PM \Rightarrow SP = \frac{1}{2} \times PM \Rightarrow 2 SP = PM \Rightarrow 4 {(SP)}^{2} = {PM}^{2} \Rightarrow 4 [{(x - 1)}^{2} + {(y - 2)}^{2}] = {|\frac{3 x + 4 y - 5}{\sqrt{3^{2} + {(4)}^{2}}}|}^{2} \Rightarrow 4 [x^{2} + 1 - 2 x + y^{2} + 4 - 4 y] = \frac{9 x^{2} + 16 y^{2} + 25 + 24 x y - 40 y + 30 x}{25} \Rightarrow 100 x^{2} + 100 - 200 x + 100 y^{2} + 400 - 400 y = 9 x^{2} + 16 y^{2} + 25 + 24 x y - 40 y - 30 x \Rightarrow 91 x^{2} + 84 y^{2} - 24 x y - 360 y - 170 x + 475 = 0 This is the required equation of the ellipse.$

Page No 26.22:

Question 3:

Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i) 4x² + 9y² = 1
(ii) 5x² + 4y² = 1
(iii) 4x² + 3y² = 1
(iv) 25x² + 16y² = 1600.
(v) 9x² + 25y² = 225

Answer:

$(i) 4 x^{2} + 9 y^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{4}} + \frac{y^{2}}{\frac{1}{9}} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = \frac{1}{4} and b^{2} = \frac{1}{9}, i . e . a = \frac{1}{2} and b = \frac{1}{3} . Clearly a > b Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{1}{9}}{\frac{1}{4}}} \Rightarrow e = \sqrt{1 - \frac{4}{9}} \Rightarrow e = \frac{\sqrt{5}}{3} Coordinates of the foci = (\pm a e, 0) = (\pm \frac{\sqrt{5}}{6}, o) Length of the latus rectum= \frac{2 b^{2}}{a} = \frac{2 \times \frac{1}{9}}{\frac{1}{2}} = \frac{4}{9}$
$(ii) 5 x^{2} + 4 y^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{5}} + \frac{y^{2}}{\frac{1}{4}} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = \frac{1}{5} and b^{2} = \frac{1}{4}, i . e . a = \frac{1}{\sqrt{5}} and b = \frac{1}{2} . Clearly b > a Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{1}{5}}{\frac{1}{4}}} \Rightarrow e = \sqrt{1 - \frac{4}{5}} \Rightarrow e = \frac{1}{\sqrt{5}} Coordinates of the foci = (0, \pm b e) = (0, \pm \frac{1}{2 \sqrt{5}}) Length of the latus rectum= \frac{2 a^{2}}{b} = \frac{2 \times \frac{1}{5}}{\frac{1}{2}} = \frac{4}{5}$

$(iii) 4 x^{2} + 3 y^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{4}} + \frac{y^{2}}{\frac{1}{3}} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = \frac{1}{4} and b^{2} = \frac{1}{3}, i . e . a = \frac{1}{2} and b = \frac{1}{\sqrt{3}} . Clearly, b > a Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{1}{4}}{\frac{1}{3}}} \Rightarrow e = \sqrt{1 - \frac{3}{4}} \Rightarrow e = \frac{1}{2} Coordinates of the foci = (0, \pm b e) = (0, \pm \frac{1}{2 \sqrt{3}}) Length of the latus rectum= \frac{2 a^{2}}{b} = \frac{2 \times \frac{1}{4}}{\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{2}$

$(iv) 25 x^{2} + 16 y^{2} = 1600 \Rightarrow \frac{x^{2}}{64} + \frac{y^{2}}{100} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = 64 and b^{2} = 100, i . e . a = 8 and b = 10 . Clearly, b > a Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{64}{100}} \Rightarrow e = \sqrt{\frac{36}{100}} \Rightarrow e = \frac{6}{10} or \frac{3}{5} Coordinates of the foci = (0, \pm 6) Length of latus rectum= \frac{2 a^{2}}{b} = \frac{2 \times 64}{10} = \frac{64}{5}$
$(v) 9 x^{2} + 25 y^{2} = 225 \Rightarrow \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = 25 and b^{2} = 9, i . e . a = 5 and b = 3 . Clearly, a > b Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{9}{25}} \Rightarrow e = \sqrt{\frac{16}{25}} \Rightarrow e = \frac{4}{5} Coordinates of the foci = (\pm a e, 0) = (\pm 4, 0) Length of latus rectum= \frac{2 b^{2}}{a} = \frac{2 \times 9}{5} = \frac{18}{5}$

Page No 26.22:

Question 4:

Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (−3, 1) and has eccentricity $\sqrt{\frac{2}{5}}$ .

Answer:

$Let the equation of the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 ...(1) It passes through the point (-3,1) . ∴ \frac{9}{a^{2}} + \frac{1}{b^{2}} =1 ... (2) Also, e = \sqrt{\frac{2}{5}} Now, b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = a^{2} [1 - \frac{2}{5}] \Rightarrow b^{2} = \frac{3 a^{2}}{5} Substituting the value of b^{2} in eq. (2), we get: \frac{9}{a^{2}} + \frac{5}{3 a^{2}} =1 \Rightarrow \frac{27 + 5}{3 a^{2}} = 1 \Rightarrow a^{2} = \frac{32}{3} \Rightarrow b^{2} = \frac{3 \times \frac{32}{3}}{5} or \frac{32}{5} Substituting the values of a and b in eq. (1), we get: \frac{3 x^{2}}{32} + \frac{5 y^{2}}{32} = 1 \Rightarrow 3 x^{2} + 5 y^{2} = 32 This is the required equation of the ellipse.$

Page No 26.22:

Question 5:

Find the equation of the ellipse in the following cases:
(i) eccentricity e = $\frac{1}{2}$ and foci (± 2, 0)
(ii) eccentricity e = $\frac{2}{3}$ and length of latus rectum = 5
(iii) eccentricity e = $\frac{1}{2}$ and semi-major axis = 4
(iv) eccentricity e = $\frac{1}{2}$ and major axis = 12
(v) The ellipse passes through (1, 4) and (−6, 1).
(vi) Vertices (± 5, 0), foci (± 4, 0)
(vii) Vertices (0, ± 13), foci (0, ± 5)
(viii) Vertices (± 6, 0), foci (± 4, 0)
(ix) Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)
(x) Ends of major axis (0, ± $\sqrt{5}$ ), ends of minor axis (± 1, 0)
(xi) Length of major axis 26, foci (± 5, 0)
(xii) Length of minor axis 16 foci (0, ± 6)
(xiii) Foci (± 3, 0), a = 4

Answer:

$(i) e = \frac{1}{2} and foci = (\pm 2, 0) Coordinates of the foci = (\pm a e, 0) We have a e = 2 \Rightarrow a \times \frac{1}{2} = 2 \Rightarrow a = 4 Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow \frac{1}{2} = \sqrt{1 - \frac{b^{2}}{16}} On squaring both sides, we get: \frac{1}{4} = \frac{16 - b^{2}}{16} \Rightarrow 64 - 4 b^{2} = 16 \Rightarrow b^{2} = \frac{48}{4} \Rightarrow b^{2} = 12 Now, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1 \Rightarrow \frac{3 x^{2} + 4 y^{2}}{48} = 1 \Rightarrow 3 x^{2} + 4 y^{2} = 48 This is the required equation of the ellipse.$

$(ii) e = \frac{2}{3} and l ength of the latus rectum = 5 We have \frac{2 b^{2}}{a} = 5 \Rightarrow 2 b^{2} = 5 a \Rightarrow b^{2} = \frac{5 a}{2} Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow \frac{2}{3} = \sqrt{1 - \frac{\frac{5 a}{2}}{a^{2}}} On squaring both sides, we get: \frac{4}{9} = \frac{2 a - 5}{2 a} \Rightarrow 8 a = 18 a - 45 \Rightarrow a = \frac{9}{2} ∴ b^{2} = \frac{45}{4} Substituting the values of a^{2} and b^{2}, we get: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{4 x^{2}}{81} + \frac{4 y^{2}}{45} = 1 \Rightarrow \frac{20 x^{2} + 36 y^{2}}{405} = 1 \Rightarrow 20 x^{2} + 36 y^{2} = 405 This is the required equation of the ellipse.$

$(iii) e = \frac{1}{2} and semi major axis = 4 i . e . a = 4 We have e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow \frac{1}{2} = \sqrt{1 - \frac{b^{2}}{16}} On squaring both sides, we get: \frac{1}{4} = \frac{16 - b^{2}}{16} \Rightarrow 16 = 64 - 4 b^{2} \Rightarrow b^{2} = 12 Substituting the values of a^{2} and b^{2}, we get: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1 \Rightarrow \frac{3 x^{2} + 4 y^{2}}{48} = 1 \Rightarrow 3 x^{2} + 4 y^{2} = 48 This is the required equation of the ellipse.$

$(iv) e = \frac{1}{2} and major axis = 12 i . e ., 2 a = 12 o r a = 6 We have e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow \frac{1}{2} = \sqrt{1 - \frac{b^{2}}{36}} On squaring both sides, we get: \frac{1}{4} = \frac{36 - b^{2}}{36} \Rightarrow 36 = 144 - 4 b^{2} \Rightarrow b^{2} = 27 Substituting the values of a^{2} and b^{2}, we get: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x^{2}}{36} + \frac{y^{2}}{27} = 1 \Rightarrow \frac{3 x^{2} + 4 y^{2}}{108} = 1 \Rightarrow 3 x^{2} + 4 y^{2} = 108 This is the required equation of the ellipse.$

$(v) The ellipse passes through (1,4) and (- 6, 1) . \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{1}{a^{2}} + \frac{16}{b^{2}} = 1 Let \frac{1}{a^{2}} = α a n d \frac{1}{b^{2}} = β Then α + 16 β = 1 . . (1) It also passes through (- 6, 1) . \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{36}{a^{2}} + \frac{1}{b^{2}} = 1 \Rightarrow 36 α + β = 1 . . . (2) Solving eqs. (1) and (2), we get: α = \frac{3}{115} and β = \frac{7}{115} Substituting the values, we get: \frac{3 x^{2}}{115} + \frac{7 y^{2}}{115} = 1 \Rightarrow \frac{3 x^{2} + 7 y^{2}}{115} = 1 \Rightarrow 3 x^{2} + 7 y^{2} = 115 This is the required equation of ellipse.$
$(vi) Vertices (\pm 5, 0) and focus (\pm 4, 0) The coordinates of its vertices and foci are (\pm a, 0) and (\pm a e, 0), respectively. i . e . a = 5 a n d a e = 4 ∴ e = \frac{4}{5} Now, b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = 25 (1 - \frac{16}{25}) \Rightarrow b^{2} = 9 ∴ \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 This is the required equation of the ellipse.$
$(vii) Vertices (0, \pm 13) and focus (0, \pm 5) The coordinates of its vertices and foci are (0, \pm b) and (0, \pm b e), respectively. i . e . b = 13 a n d b e = 5 ∴ e = \frac{5}{13} Now, a^{2} = b^{2} (1 - e^{2}) \Rightarrow a^{2} = 169 (1 - \frac{25}{169}) \Rightarrow a^{2} = 144 ∴ \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1 This is the required equation of the ellipse.$
$(viii) Vertices (\pm 6, 0) and focus (\pm 4, 0) The coordinates of its vertices and foci are (\pm a, 0) and (\pm a e, 0), respectively. i . e ., a = 6 a n d a e = 4 ∴ e = \frac{4}{6} o r \frac{2}{3} Now, b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = 36 (1 - \frac{16}{36}) \Rightarrow b^{2} = 20 ∴ \frac{x^{2}}{36} + \frac{y^{2}}{20} = 1 This is the required equation of the ellipse. (ix) Let the equation of t h e ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . End of major axis= (\pm 3, 0) End of minor axis= (0, \pm 2) But the coordinates of the end points of the major and the minor axes are (± a,0) and (0, \pm b), respectively. ∴ a = 3 and b = 2 Then \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 This is the required equation of the ellipse. (x) Let the equation of t h e ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . End of major axis= (0, \pm \sqrt{5}) End of minor axis= (\pm 1, 0) But the coordinates of the end points of the major and the minor axes are (± a,0) and (0, \pm b), respectively. ∴ a = 1 and b = \sqrt{5} Then \frac{x^{2}}{1} + \frac{y^{2}}{5} = 1 This is the required equation of the ellipse.$
$(xi) Length of major axis=26 Foci= (\pm 5, 0) We have 2 a = 26 \Rightarrow a = 13 Also, a e = 5 \Rightarrow e = \frac{5}{13} Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow \frac{5}{13} = \sqrt{1 - \frac{b^{2}}{169}} On squaring both sides, we get: \frac{25}{169} = \frac{169 - b^{2}}{169} \Rightarrow b^{2} = 144 Now, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x^{2}}{169} + \frac{y^{2}}{144} = 1 This is the required equation of the ellipse. (xii) Length of minor axis=16 and foci= (0, \pm 6) i . e . 2 b = 16 \Rightarrow b = 8 and b e = 6 \Rightarrow e = \frac{6}{8} Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow \frac{6}{8} = \sqrt{1 - \frac{a^{2}}{64}} On squaring both sides, we get: \frac{36}{64} = \frac{64 - a^{2}}{64} \Rightarrow a^{2} = 28 \frac{x^{2}}{64} + \frac{y^{2}}{28} = 1 This is the required equation of the ellipse. (xiii) F oci= (\pm 3, 0) and a =4 i . e . a e = 3 \Rightarrow e = \frac{3}{4} Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow \frac{3}{4} = \sqrt{1 - \frac{b^{2}}{16}} On squaring both sides, we get: \frac{9}{16} = \frac{16 - b^{2}}{16} \Rightarrow b^{2} = 7 ∴ \frac{x^{2}}{16} + \frac{y^{2}}{7} = 1 This is the required equation of the ellipse.$

Page No 26.23:

Question 6:

Find the equation of the ellipse whose foci are (4, 0) and (−4, 0), eccentricity = 1/3.

Answer:

$Let the equation of the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (1) Then the coordinates of the foci are (\pm a e, 0) . Also, e = \frac{1}{3} We have a e = 4 \Rightarrow a . \frac{1}{3} = 4 \Rightarrow a = 12 Now, b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = 12^{2} {(1 - (\frac{1}{3}))}^{2} \Rightarrow b^{2} = 144 (\frac{8}{9}) \Rightarrow b^{2} = 128 Substituting the values of a^{2} and b^{2} in eq . (1), we get : \frac{x^{2}}{144} + \frac{y^{2}}{128} = 1 This is the required equation of the ellipse .$

Page No 26.23:

Question 7:

Find the equation of the ellipse in the standard form whose minor axis is equal to the distance between foci and whose latus-rectum is 10.

Answer:

$According to the question, the minor axis is equal to the distance between the foci. i . e . 2 b = 2 a e and \frac{{2b}^{2}}{a} = 10 or b^{2} = 5 a \Rightarrow b = a e \Rightarrow b^{2} = a^{2} e^{2} \Rightarrow b^{2} = a^{2} (1 - \frac{b^{2}}{a^{2}}) (∵ e = \sqrt{1 - \frac{b^{2}}{a^{2}}}) \Rightarrow b^{2} = a^{2} - b^{2} \Rightarrow a^{2} = 2 b^{2} \Rightarrow a^{2} = 10 a (∵ b^{2} = 5 a) \Rightarrow a = 10 \Rightarrow b^{2} = 5 a \Rightarrow b^{2} = 50 Substituting the values of a and b in the equation of an ellipse, we get: \frac{x^{2}}{100} + \frac{y^{2}}{50} = 1 ∴ x^{2} + 2 y^{2} = 100 This is the required equation of the ellipse.$

Page No 26.23:

Question 8:

Find the equation of the ellipse whose centre is (−2, 3) and whose semi-axis are 3 and 2 when major axis is (i) parallel to x-axis (ii) parallel to y-axis.

Answer:

$(i) When the major axis is parallel to the x -axis. Let \frac{(x - x_{1})^{2}}{a^{2}} + \frac{(y - y_{1})^{2}}{b^{2}} = 1 . . . (1) {Here, x}_{1} {and y}_{1} are -2 and 3, respectively, and 3 and 2 are the lengths of the axes. Substituting the value in eq. (1), we get: \frac{(x + 2)^{2}}{9} + \frac{(y - 3)^{2}}{4} = 1 \Rightarrow \frac{4 (x^{2} + 4 + 4 x) + 9 (y^{2} + 9 - 6 y)}{36} = 1 \Rightarrow 4 x^{2} + 16 + 16 x + 9 y^{2} + 81 - 54 y = 36 \Rightarrow 4 x^{2} + 9 y^{2} + 16 x - 54 y + 61 = 0 This is the required equation of the ellipse. (ii) When the major axis is parallel to the y -axis. Let \frac{(x - x_{1})^{2}}{b^{2}} + \frac{(y - y_{1})^{2}}{a^{2}} = 1 . . . (1) Here, x_{1} {and y}_{1} are -2 and 3, respectively, and 3 and 2 are the lengths of the axes. Substituting the value in eq. (1), we get : \frac{(x + 2)^{2}}{4} + \frac{(y - 3)^{2}}{9} = 1 \Rightarrow \frac{9 (x^{2} + 4 + 4 x) + 4 (y^{2} + 9 - 6 y)}{36} = 1 \Rightarrow 9 x^{2} + 36 + 36 x + 4 y^{2} + 36 - 24 y = 36 \Rightarrow 9 x^{2} + 4 y^{2} + 36 x - 24 y + 36 = 0 This is the required equation of the ellipse.$

Page No 26.23:

Question 9:

Find the eccentricity of an ellipse whose latus rectum is
(i) half of its minor axis
(ii) half of its major axis.

Answer:

$(i) According to the question, the latus rectum is half its minor axis. i . e . \frac{2 b^{2}}{a} = \frac{1}{2} \times (2 b) \Rightarrow 2 b^{2} = a b \Rightarrow 2 b = a Now, e = \sqrt{1 - \frac{b^{2}}{4 b^{2}}} \Rightarrow e = \sqrt{1 - \frac{1}{4}} \Rightarrow e = \frac{\sqrt{3}}{2} 2) According to the question, the latus rectum is half its minor axis. i . e . \frac{2 b^{2}}{a} = \frac{1}{2} \times (2 a) \Rightarrow 2 b^{2} = a^{2} \Rightarrow a^{2} = 2 b^{2} Now, e = \sqrt{1 - \frac{b^{2}}{2 b^{2}}} \Rightarrow e = \sqrt{1 - \frac{1}{2}} \Rightarrow e = \frac{1}{\sqrt{2}}$

Page No 26.23:

Question 10:

Find the centre, the lengths of the axes, eccentricity, foci of the following ellipse:
(i) x² + 2y² − 2x + 12y + 10 = 0
(ii) x² + 4y² − 4x + 24y + 31 = 0
(iii) 4x² + y² − 8x + 2y + 1 = 0
(iv) 3x² + 4y² − 12x − 8y + 4 = 0
(v) 4x² + 16y² − 24x − 32y − 12 = 0
(vi) x² + 4y² − 2x = 0

Answer:

$(i) x^{2} + 2 y^{2} - 2 x + 12 y + 10 = 0 \Rightarrow (x^{2} - 2 x) + 2 (y^{2} + 6 y) = - 10 \Rightarrow (x^{2} - 2 x + 1) + 2 (y^{2} + 6 y + 9) = - 10 + 18 + 1 \Rightarrow {(x - 1)}^{2} + 2 {(y + 3)}^{2} = 9 \Rightarrow \frac{{(x - 1)}^{2}}{9} + \frac{{(y + 3)}^{2}}{\frac{9}{2}} = 9 Here, x_{1} = 1 a n d y_{1} = - 3 Also, a = 3 a n d b = \frac{3}{\sqrt{2}} Centre= (1, - 3) Major axis=2 a \Rightarrow 2 \times 3 = 6 Minor axis=2 b \Rightarrow 2 \times \frac{3}{\sqrt{2}} = 3 \sqrt{2} e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{9}{2}}{9}} \Rightarrow e = \frac{1}{\sqrt{2}} Foci = (x_{1} \pm a e, y_{1}) = (1 \pm \frac{3}{\sqrt{2}}, - 3)$
$(ii) x^{2} + 4 y^{2} - 4 x + 24 y + 31 = 0 \Rightarrow (x^{2} - 4 x) + 4 (y^{2} + 6 y) = - 31 \Rightarrow (x^{2} - 4 x + 4) + 4 (y^{2} + 6 y + 9) = - 31 + 36 + 4 \Rightarrow {(x - 2)}^{2} + 4 {(y + 3)}^{2} = 9 \Rightarrow \frac{{(x - 2)}^{2}}{9} + \frac{{(y + 3)}^{2}}{\frac{9}{4}} = 9 Here, x_{1} = 2 and y_{1} = - 3 A l s o, a = 3 a n d b = \frac{3}{2} Centre= (x_{1}, y_{1}) = (2, - 3) Major axis=2 a = 2 \times 3 = 6 Minor axis=2 b = 2 \times \frac{3}{2} = 3 e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{9}{4}}{9}} \Rightarrow e = \frac{\sqrt{3}}{2} Foci = (x_{1} \pm a e, y_{1}) = (2 \pm \frac{3 \sqrt{3}}{2}, - 3)$
$(iii) 4 x^{2} + y^{2} - 8 x + 2 y + 1 = 0 \Rightarrow 4 (x^{2} - 2 x) + (y^{2} + 2 y) = - 1 \Rightarrow 4 (x^{2} - 2 x + 1) + (y^{2} + 2 y + 1) = - 1 + 4 + 1 \Rightarrow 4 {(x - 1)}^{2} + {(y + 1)}^{2} = 4 \Rightarrow \frac{{(x - 1)}^{2}}{1} + \frac{{(y + 1)}^{2}}{4} = 1 Here, x_{1} = 1 a n d y_{1} = - 1 A l s o, a = 1 a n d b = 2 Centre= (x_{1}, y_{1}) = (1, - 1) Major axis=2 b = 2 \times 2 = 4 Minor axis=2 a = 2 \times 1 = 2 e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{1}{4}} \Rightarrow e = \frac{\sqrt{3}}{2} Foci = (x_{1}, y_{1} \pm b e) = (1, - 1 \pm \sqrt{3})$
$(iv) 3 x^{2} + 4 y^{2} - 12 x - 8 y + 4 = 0 \Rightarrow 3 (x^{2} - 4 x) + 4 (y^{2} - 2 y) = - 4 \Rightarrow 3 (x^{2} - 4 x + 4) + 4 (y^{2} - 2 y + 1) = - 4 + 12 + 4 \Rightarrow 3 {(x - 2)}^{2} + 4 {(y - 1)}^{2} = 12 \Rightarrow \frac{{(x - 2)}^{2}}{4} + \frac{{(y - 1)}^{2}}{3} = 1 Here, x_{1} = 2, y_{1} = 1 A l s o, a = 2 a n d b = \sqrt{3} Centre= (x_{1}, y_{1}) = (2, 1) Major axis=2 a = 2 \times 2 = 4 Minor axis=2 b = 2 \times \sqrt{3} = 2 \sqrt{3} e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{3}{4}} \Rightarrow e = \frac{1}{2} Foci = (x_{1} \pm a e, y_{1}) = (2 \pm 1, 1)$
$(v) 4 x^{2} + 16 y^{2} - 24 x - 32 y - 12 = 0 \Rightarrow 4 (x^{2} - 6 x) + 16 (y^{2} - 2 y) = 12 \Rightarrow 4 (x^{2} - 6 x + 9) + 16 (y^{2} - 2 y + 1) = 12 + 36 + 16 \Rightarrow 4 {(x - 3)}^{2} + 16 {(y - 1)}^{2} = 64 \Rightarrow \frac{{(x - 3)}^{2}}{16} + \frac{{(y - 1)}^{2}}{4} = 9 Centre= (3, 1) M ajor axis=2 a = 2 \times 4 = 8 Minor axis=2 b = 2 \times 2 = 4 e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{4}{16}} \Rightarrow e = \frac{\sqrt{3}}{2} Foci = (x \pm a e, y) = (3 \pm 2 \sqrt{3}, 1) (vi) x^{2} - 2 x + 4 y^{2} = 0 \Rightarrow (x^{2} - 2 x) + 4 (y^{2}) = 0 \Rightarrow (x^{2} - 2 x + 1) + 4 (y^{2}) = 1 \Rightarrow {(x - 1)}^{2} + 4 {(y)}^{2} = 1 \Rightarrow \frac{{(x - 1)}^{2}}{1} + \frac{{(y)}^{2}}{\frac{1}{4}} = 9 Centre= (1, 0) Major axis=2 a = 2 \times 1 = 2 Minor axis=2 b = 2 \times \frac{1}{2} = 1 e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{1}{4}}{1}} \Rightarrow e = \frac{\sqrt{3}}{2} Foci = (x \pm a e, y) = (1 \pm \frac{3}{\sqrt{2}}, 0)$

Page No 26.23:

Question 11:

Find the equation of an ellipse whose foci are at (± 3, 0) and which passes through (4, 1).

Answer:

$Let the equation of the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (1) Then a e = 3 Also x = 4 and y = 1 [Ellipse passing through (4, 1)] Substituting the values of x and y in eq . (1), we get : \frac{4^{2}}{a^{2}} + \frac{1^{2}}{b^{2}} = 1 \Rightarrow \frac{16}{a^{2}} + \frac{1}{b^{2}} = 1 Now, b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = a^{2} - a^{2} e^{2} \Rightarrow b^{2} = a^{2} - 9 o r a^{2} = b^{2} + 9 . . . (2) \Rightarrow \frac{16}{a^{2}} + \frac{1}{b^{2}} = 1 \Rightarrow 16 b^{2} + a^{2} = a^{2} b^{2} \Rightarrow 16 b^{2} + b^{2} + 9 = b^{2} (b^{2} + 9) \Rightarrow 17 b^{2} + 9 = b^{4} + 9 b^{2} \Rightarrow b^{4} - 8 b^{2} - 9 = 0 \Rightarrow (b^{2} - 9) (b^{2} + 1) \Rightarrow b = \pm 3 Substituting the value of b in eq. (2), we get: a = 3 \sqrt{2} ∴ \frac{x^{2}}{18} + \frac{y^{2}}{9} = 1 This is the required equation of the ellipse.$

Page No 26.23:

Question 12:

Find the equation of an ellipse whose eccentricity is 2/3, the latus-rectum is 5 and the centre is at the origin.

Answer:

$Let the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1. ... (1) e = \frac{2}{3} and l atus rectum=5 (Given) Now, \frac{2 b^{2}}{a} = 5 \Rightarrow 2 b^{2} = 5 a . . . (2) \Rightarrow 2 a^{2} (1 - e^{2}) = 5 a [∵ b^{2} = a^{2} (1 - e^{2})] \Rightarrow 2 a^{2} [1 - \frac{4}{9}] = 5 a \Rightarrow 2 a^{2} \times \frac{5}{9} = 5 a \Rightarrow 10 a^{2} = 45 a \Rightarrow a = \frac{9}{2} Substituting the value of a in eq. (2), we get: 2 b^{2} = 5 \times \frac{9}{2} \Rightarrow b^{2} = \frac{45}{4} Substituting the values of a^{2} and b^{2} in eq. (1), we get: \frac{x^{2}}{\frac{81}{4}} + \frac{y^{2}}{\frac{45}{4}} =1 \Rightarrow \frac{4 x^{2}}{81} + \frac{4 y^{2}}{45} = 1 This is the required equation of the ellipse.$

Page No 26.23:

Question 13:

Find the equation of an ellipse with its foci on y-axis, eccentricity 3/4, centre at the origin and passing through (6, 4).

Answer:

$Let the equation of the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 and let e= \frac{3}{4} . Also, let the foci be on the y- axis. ∴ a^{2} = b^{2} (1 - e^{2}) \Rightarrow a^{2} = b^{2} (1 - \frac{9}{16}) \Rightarrow a^{2} = \frac{7}{16} b^{2} \Rightarrow \frac{x^{2}}{a^{2}} + \frac{7 y^{2}}{16 a^{2}} =1 . . . (1) It passes through (6, 4) . Then \frac{36}{a^{2}} + \frac{112}{16 a^{2}} =1 \Rightarrow \frac{576 + 112}{16 a^{2}} = 1 \Rightarrow 16 a^{2} = 688 \Rightarrow a^{2} = 43 Now, b^{2} = \frac{16 a^{2}}{7} \Rightarrow b^{2} = \frac{688}{7} Substituting the values of a^{2} and b^{2} in eq. (1), we get: \frac{x^{2}}{43} + \frac{7 y^{2}}{688} =1 This is the required equation of the ellipse.$

Page No 26.23:

Question 14:

Find the equation of an ellipse whose axes lie along coordinate axes and which passes through (4, 3) and (−1, 4).

Answer:

$Let the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 and it passes through the points (4,3) and (-1,4). ∴ \frac{16}{a^{2}} + \frac{9}{b^{2}} =1 and \frac{1}{a^{2}} + \frac{16}{b^{2}} =1 Let α= \frac{1}{a^{2}} and β= \frac{1}{b^{2}} Then 16 α + 9 β = 1 and α + 16 β = 1 Solving these two equations, we get: α = \frac{7}{247} and β = \frac{15}{247} \Rightarrow \frac{1}{a^{2}} = \frac{7}{247} and \frac{1}{b^{2}} = \frac{15}{247} . . . (1) Substituting eq. (1) in the equation of an ellipse, we get: \frac{7 x^{2}}{247} + \frac{15 y^{2}}{247} = 1 This is the required equation of the ellipse.$

Page No 26.23:

Question 15:

Find the equation of an ellipse whose axes lie along the coordinate axes, which passes through the point (−3, 1) and has eccentricity equal to $\sqrt{2 / 5}$ .

Answer:

$Let the equation of the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 ...(1) It passes through the point (-3,1) . ∴ \frac{9}{a^{2}} + \frac{1}{b^{2}} =1 ...( 2) and e = \sqrt{\frac{2}{5}} Now, b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = a^{2} (1 - \frac{2}{5}) \Rightarrow b^{2} = a^{2} \times \frac{3}{5} or \frac{3 a^{2}}{5} Substituting the value of b^{2} in eq. (2), we get: \frac{9}{a^{2}} + \frac{5}{3 a^{2}} =1 \Rightarrow \frac{27 + 5}{3 a^{2}} = 1 \Rightarrow a^{2} = \frac{32}{3} \Rightarrow b^{2} = \frac{3 \times \frac{32}{3}}{5} or \frac{32}{5} Substituting the values of a and b in eq. (1), we get: \frac{3 x^{2}}{32} + \frac{5 y^{2}}{32} = 1 \Rightarrow 3 x^{2} + 5 y^{2} = 32 This is the required equation of the ellipse.$

Page No 26.23:

Question 16:

Find the equation of an ellipse, the distance between the foci is 8 units and the distance between the directrices is 18 units.

Answer:

$The distance between the foci is 8 units. i . e . 2 a e = 8 . . . (1) The d istance between the directrices is 18 units. i . e . \frac{2 a}{e} = 18 . . . (2) Comparing eqs. (1) and (2), we get: e = \frac{8}{2 a} Substituting the value of e in eq. (2), we get: \frac{2 a}{\frac{8}{2 a}} = 18 \Rightarrow 4 a^{2} = 18 \times 8 \Rightarrow a^{2} = 36 or a = 6 Now, 2 a e = 8 \Rightarrow 12 e = 8 or e = \frac{2}{3} Then b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = 36 (1 - \frac{4}{9}) \Rightarrow b^{2} = 36 \times \frac{5}{9} \Rightarrow b^{2} = 20 ∴ \frac{x^{2}}{36} + \frac{y^{2}}{20} = 1 This is the required equation of ellipse.$

Page No 26.23:

Question 17:

Find the equation of an ellipse whose vertices are (0, ± 10) and eccentricity e = $\frac{4}{5}$ .

Answer:

$Let the equation of the required ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (1) Since the vertices of the ellipse are on the y - axis, the coordinates of the vertices are (0, \pm b) . ∴ b = 10 Now, a^{2} = b^{2} (1 - e^{2}) \Rightarrow a^{2} = 100 {(1 - \frac{4}{5})}^{2} \Rightarrow a^{2} = 100 \times (\frac{9}{25}) \Rightarrow a^{2} = 36 Substituting the values of a^{2} and b^{2} in equation (1), we get : \frac{x^{2}}{36} + \frac{y^{2}}{100} = 1 \Rightarrow \frac{100 x^{2} + 36 y^{2}}{3600} = 1 ∴ 100 x^{2} + 36 y^{2} = 3600 This is the required equation of the ellipse .$

Page No 26.23:

Question 18:

A rod of length 12 m moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with x-axis.

Answer:

Let AB be the rod making an angle θ with OX and let P (x, y) be the point on it such that AP = 3 cm.
Then, PB = AB – AP = (12 – 3) cm = 9 cm [∵ AB = 12 cm]
From P, draw PQ⊥OY and PR⊥OX.

$In △ P B Q, we have : \cos θ = \frac{P Q}{P B} = \frac{x}{9} In △ P R A, we have : \sin θ = \frac{P R}{P A} = \frac{y}{3} Since \sin^{2} θ + \cos^{2} θ = 1, we have : {(\frac{y}{3})}^{2} + {(\frac{x}{9})}^{2} = 1 \Rightarrow \frac{x^{2}}{81} + \frac{y^{2}}{9} = 1 Thus, the locus of a point P on the rod is \frac{x^{2}}{81} + \frac{y^{2}}{9} = 1 .$

Page No 26.23:

Question 19:

Find the equation of the set of all points whose distances from (0, 4) are $\frac{2}{3}$ of their distances from the line y = 9.

Answer:

We have
$PQ = \frac{2}{3} PL \Rightarrow \sqrt{{(x - 0)}^{2} + {(y - 4)}^{2}} = \frac{2}{3} (y - 9) \Rightarrow 3^{2} [x^{2} + {(y - 4)}^{2}] = 2^{2} {(y - 9)}^{2} \Rightarrow 9 x^{2} + 9 y^{2} - 72 y + 144 = 4 y^{2} - 72 y + 324 \Rightarrow 9 x^{2} + 5 y^{2} = 180 \Rightarrow \frac{x^{2}}{20} + \frac{y^{2}}{36} = 1$

Page No 26.27:

Question 1:

For the ellipse 12x² + 4y² + 24x − 16y + 25 = 0

(a) centre is (−1, 2)

(b) lengths of the axes are $\sqrt{3}$ and 1

(c) eccentricity = $\sqrt{\frac{2}{3}}$

(d) all of these

Answer:

$Disclaimer : The equation should be 12 x^{2} + 4 y^{2} + 24 x - 16 y + 24 = 0 instead of 12 x^{2} + 4 y^{2} + 24 x - 16 y + 25 = 0 . (d) all of these 12 x^{2} + 4 y^{2} + 24 x - 16 y + 24 = 0 \Rightarrow 12 (x^{2} + 2 x) + 4 (y^{2} - 4 y) = - 24 \Rightarrow 12 (x^{2} + 2 x + 1) + 4 (y^{2} - 4 y + 4) = - 24 + 12 + 16 \Rightarrow 12 {(x + 1)}^{2} + 4 {(y - 2)}^{2} = 4 \Rightarrow \frac{{(x + 1)}^{2}}{3} + \frac{{(y - 2)}^{2}}{1} = 1 So, the centre is (- 1, 2) . Here, a = \sqrt{3} and b = 1 The l engths of the axes are \sqrt{3} and 1. Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} e = \sqrt{1 - \frac{1}{3}} \Rightarrow e = \sqrt{\frac{2}{3}}$

Page No 26.27:

Question 2:

The equation of the ellipse with focus (−1, 1), directrix x − y + 3 = 0 and eccentricity 1/2 is
(a) 7x² + 2xy + 7y² + 10x + 10y + 7 = 0
(b) 7x² + 2xy + 7y² + 10x − 10y + 7 = 0
(c) 7x² + 2xy + 7y² + 10x − 10y − 7 = 0
(d) none of these

Answer:

$(b) 7 x^{2} + 7 y^{2} + 2 x y - 10 y + 10 x + 7 = 0$

$Let P(x,y) be any point on the ellipse whose focus and eccentricit y are S (- 1, 1) and e= \frac{1}{2}, respectively. Let PM be the perpendicular from P on the directrix. Then S P = e \times P M \Rightarrow S P = \frac{1}{2} \times P M \Rightarrow 2 S P = P M \Rightarrow 4 {(S P)}^{2} = P M^{2} \Rightarrow 4 [{(x + 1)}^{2} + {(y - 1)}^{2}] = {(\frac{x - y + 3}{\sqrt{1^{2} + {(- 1)}^{2}}})}^{2} \Rightarrow 4 [x^{2} + 1 + 2 x + y^{2} + 1 - 2 y] = \frac{x^{2} + y^{2} + 9 - 2 x y - 6 y + 6 x}{2} \Rightarrow 8 x^{2} + 8 + 16 x + 8 y^{2} + 8 - 16 y = x^{2} + y^{2} + 9 - 2 x y - 6 y + 6 x ∴ 7 x^{2} + 7 y^{2} + 2 x y - 10 y + 10 x + 7 = 0 This is the required equation of the ellipse.$

Page No 26.27:

Question 3:

The equation of the circle drawn with the two foci of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ as the end-points of a diameter is
(a) x²+ y² = a² + b²
(b) x²+ y² = a²
(c) x² + y² = 2a²
(d) x² + y² = a² − b²

Answer:

$(d) x^{2} + y^{2} = a^{2} - b^{2} We have r = a e Let the equation of the circle be x^{2} + y^{2} = r^{2} . Now, x^{2} + y^{2} = a^{2} e^{2} (∵ r = a e) \Rightarrow x^{2} + y^{2} = a^{2} (1 - \frac{b^{2}}{a^{2}}) (∵ e = \sqrt{1 - \frac{b^{2}}{a^{2}}}) \Rightarrow x^{2} + y^{2} = a^{2} - b^{2} ∴ The required equation of the circle is x^{2} + y^{2} = a^{2} - b^{2} .$

Page No 26.27:

Question 4:

The eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ if its latus rectum is equal to one half of its minor axis, is
(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{\sqrt{3}}{2}$

(c) $\frac{1}{2}$

(d) none of these

Answer:

$(b) e = \frac{\sqrt{3}}{2} According to the question, the latus rectum is half its minor axis. i . e . \frac{2 b^{2}}{a} = \frac{1}{2} \times 2 b \Rightarrow 2 b^{2} = a b \Rightarrow a = 2 b Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{b^{2}}{4 b^{2}}} \Rightarrow e = \sqrt{1 - \frac{1}{4}} \Rightarrow e = \frac{\sqrt{3}}{2}$

Page No 26.27:

Question 5:

The eccentricity of the ellipse, if the distance between the foci is equal to the length of the latus-rectum, is
(a) $\frac{\sqrt{5} - 1}{2}$

(b) $\frac{\sqrt{5} + 1}{2}$

(c) $\frac{\sqrt{5} - 1}{4}$

(d) none of these

Answer:

$(a) e = \frac{\sqrt{5} - 1}{2} According to the question, the distance between the foci is equal to the length of the latus rectum. \frac{2 b^{2}}{a} = 2 a e \Rightarrow b^{2} = a^{2} e Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{a^{2} e}{a^{2}}} \Rightarrow e = \sqrt{1 - e} On squaring both sides, we get: e^{2} + e - 1 = 0 \Rightarrow e = \frac{- 1 \pm \sqrt{1 + 4}}{2} \Rightarrow e = \frac{\sqrt{5} - 1}{2} (∵ e cannot be negative)$

Page No 26.27:

Question 6:

The eccentricity of the ellipse, if the minor axis is equal to the distance between the foci, is
(a) $\frac{\sqrt{3}}{2}$

(b) $\frac{2}{\sqrt{3}}$

(c) $\frac{1}{\sqrt{2}}$

(d) $\frac{\sqrt{2}}{3}$

Answer:

$(c) \frac{1}{\sqrt{2}} According to the question, the minor axis is equal to the distance between the foci. i . e . 2 b = 2 a e \Rightarrow e = \frac{b}{a} . . . (1) Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - e^{2}} [From (1)] On squaring both sides, we get: e^{2} = 1 - e^{2} \Rightarrow 2 e^{2} = 1 \Rightarrow e^{2} = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$

Page No 26.27:

Question 7:

The difference between the lengths of the major axis and the latus-rectum of an ellipse is
(a) ae
(b) 2ae
(c) ae²
(d) 2ae²

Answer:

$(d) 2 a e^{2} Length of the latus rectum = \frac{2 b^{2}}{a} and e = \sqrt{1 - \frac{b^{2}}{a^{2}}} a^{2} e^{2} = a^{2} - b^{2} \Rightarrow b^{2} = a^{2} - a^{2} e^{2} \Rightarrow b^{2} = a^{2} (1 - e^{2}) ∴ Length of the latus rectum = \frac{2 a^{2} (1 - e^{2})}{a} = {2a(1-e}^{2}) Length of the major axis=2a Difference between length of latus rectum and length of major axis = 2 a - {2a(1-e}^{2}) = 2 a - 2 a + 2 a e^{2} = 2 a e^{2}$

Page No 26.27:

Question 8:

The eccentricity of the conic 9x² + 25y² = 225 is
(a) 2/5
(b) 4/5
(c) 1/3
(d) 1/5
(e) 3/5

Answer:

$(b) \frac{4}{5} 9 x^{2} + 25 y^{2} = 225 \Rightarrow \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 Comparing it with \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get : a = 5 and b = 3 Here, a > b, so the major and t h e minor ax e s of the ellipse are along t h e x - a x i s and y - axis, respectively . N o w, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{9}{25}} \Rightarrow e = \sqrt{\frac{16}{25}} \Rightarrow e = \frac{4}{5}$

Page No 26.27:

Question 9:

The latus-rectum of the conic 3x² + 4y² − 6x + 8y − 5 = 0 is

(a) 3

(b) $\frac{\sqrt{3}}{2}$

(c) $\frac{2}{\sqrt{3}}$

(d) none of these

Answer:

$(a) 3 3 x^{2} + 4 y^{2} - 6 x + 8 y - 5 = 0 \Rightarrow 3 (x^{2} - 2 x) + 4 (y^{2} + 2 y) = 5 \Rightarrow 3 (x^{2} - 2 x + 1) + 4 (y^{2} + 2 y + 1) = 5 + 3 + 4 \Rightarrow 3 (x - 1)^{2} + 4 (y + 1)^{2} = 12 \Rightarrow \frac{(x - 1)^{2}}{4} + \frac{(y + 1)^{2}}{3} = 1 So, a = 2 and b = \sqrt{3} ∴ Latus rectum = \frac{2 b^{2}}{a} = 2 \frac{{[\sqrt{3}]}^{2}}{2} = 3$

Page No 26.28:

Question 10:

The equations of the tangents to the ellipse 9x² + 16y² = 144 from the point (2, 3) are
(a) y = 3, x = 5
(b) x = 2, y = 3
(c) x = 3, y = 2
(d) x + y = 5, y = 3

Answer:

$(d) x + y = 5, y = 3 9 x^{2} + 16 y^{2} = 144 \Rightarrow \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 Equation of the tangent in case of an ellipse is given by y = m x + \sqrt{a^{2} m^{2} + b^{2}} \Rightarrow y = m x + \sqrt{16 m^{2} + 9} . . . (1) Substituting x =2 and y =3, we get: 3 = 2 m \pm \sqrt{16 m^{2} + 9} \Rightarrow 3 - 2 m = \sqrt{16 m^{2} + 9} On squaring both sides, we get: {(3 - 2 m)}^{2} = (16 m^{2} + 9) \Rightarrow 9 + 4 m^{2} - 12 m = (16 m^{2} + 9) \Rightarrow 12 m^{2} + 12 m = 0 \Rightarrow 12 m (m + 1) = 0 \Rightarrow m = 0, - 1 Substituting values of m in eq. (1), we get: For m = 0, y = 3 For m = - 1, y = - x + 5 or x + y = 5$

Page No 26.28:

Question 11:

The eccentricity of the ellipse 4x² + 9y² + 8x + 36y + 4 = 0 is

(a) $\frac{5}{6}$

(b) $\frac{3}{5}$

(c) $\frac{\sqrt{2}}{3}$

(d) $\frac{\sqrt{5}}{3}$

Answer:

$(d) \frac{\sqrt{5}}{3} 4 x^{2} + 9 y^{2} + 8 x + 36 y + 4 = 0 \Rightarrow 4 (x^{2} + 2 x) + 9 (y^{2} + 4 y) = - 4 \Rightarrow 4 (x^{2} + 2 x + 1) + 9 (y^{2} + 4 y + 4) = - 4 + 4 + 36 \Rightarrow 4 (x + 1)^{2} + 9 (y + 2)^{2} = 36 \Rightarrow \frac{4 (x + 1)^{2}}{36} + \frac{9 (y + 2)^{2}}{36} = 1 \Rightarrow \frac{(x + 1)^{2}}{9} + \frac{(y + 2)^{2}}{4} = 1 Comparing it with \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get : a = 3 a n d b = 2 S o, the major and t h e minor ax e s of the ellipse are along t h e x - a x i s and y - axis, respectively . N o w, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{4}{9}} \Rightarrow e = \sqrt{\frac{5}{9}} \Rightarrow e = \frac{\sqrt{5}}{3}$

Page No 26.28:

Question 12:

The eccentricity of the ellipse 4x² + 9y² = 36 is

(a) $\frac{1}{2 \sqrt{3}}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\frac{\sqrt{5}}{3}$

(d) $\frac{\sqrt{5}}{6}$

Answer:

$(c) \frac{\sqrt{5}}{3} 4 x^{2} + 9 y^{2} = 36 \Rightarrow \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 Comparing it with \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get : a = 3 and b = 2 Here, a > b, so the major and t h e minor ax e s of the ellipse are along t h e x - a x i s and y - axis, respectively . Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{4}{9}} \Rightarrow e = \sqrt{\frac{5}{9}} \Rightarrow e = \frac{\sqrt{5}}{3}$

Page No 26.28:

Question 13:

The eccentricity of the ellipse 5x² + 9y² = 1 is
(a) 2/3
(b) 3/4
(c) 4/5
(d) 1/2

Answer:

$(a) \frac{2}{3} 5 x^{2} + 9 y^{2} = 1 \Rightarrow \frac{x^{2}}{5} + \frac{y^{2}}{9} = 1 ∴ a = \sqrt{5} and b = 3 Here, b > a, so the major and the minor ax e s of the ellipse are along the x - a x i s and t h e y - axis, respectively . Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{5}{9}} \Rightarrow e = \sqrt{\frac{4}{9}} \Rightarrow e = \frac{2}{3}$

Page No 26.28:

Question 14:

For the ellipse x² + 4y² = 9
(a) the eccentricity is 1/2
(b) the latus-rectum is 3/2
(c) a focus is $(3 \sqrt{3}, 0)$
(d) a directrix is x = $- 2 \sqrt{3}$

Answer:

$(b) the latus rectum is \frac{3}{2} x^{2} + 4 y^{2} = 9 \Rightarrow \frac{x^{2}}{9} + \frac{y^{2}}{\frac{9}{4}} = 1 Here, a > b ∴ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{9}{\frac{9}{4}}} \Rightarrow e = \sqrt{1 - \frac{1}{4}} \Rightarrow e = \frac{\sqrt{3}}{2} Latus rectum= \frac{2 b^{2}}{a} = \frac{2 \times \frac{9}{4}}{3} = \frac{3}{2} Focus = (\pm a e, 0) = (\frac{3 \sqrt{3}}{2}, 0) Directrix x=± \frac{a}{e} = \pm \frac{3}{\frac{\sqrt{3}}{2}} = 2 \sqrt{3}$

Page No 26.28:

Question 15:

If the latus rectum of an ellipse is one half of its minor axis, then its eccentricity is

(a) $\frac{1}{2}$

(b) $\frac{1}{\sqrt{2}}$

(c) $\frac{\sqrt{3}}{2}$

(d) $\frac{\sqrt{3}}{4}$

Answer:

$(c) \frac{\sqrt{3}}{2} According to the question, the latus rectum of an ellipse is half its minor axis. i . e . \frac{2 b^{2}}{a} = \frac{1}{2} \times 2 b \Rightarrow 2 b^{2} = a b \Rightarrow a = 2 b Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{b^{2}}{4 b^{2}}} \Rightarrow e = \sqrt{1 - \frac{1}{4}} \Rightarrow e = \sqrt{\frac{3}{4}} \Rightarrow e = \frac{\sqrt{3}}{2}$

Page No 26.28:

Question 16:

An ellipse has its centre at (1, −1) and semi-major axis = 8 and it passes through the point
(1, 3). The equation of the ellipse is

(a) $\frac{{(x + 1)}^{2}}{64} + \frac{{(y + 1)}^{2}}{16} = 1$

(b) $\frac{{(x - 1)}^{2}}{64} + \frac{{(y + 1)}^{2}}{16} = 1$

(c) $\frac{{(x - 1)}^{2}}{16} + \frac{{(y + 1)}^{2}}{64} = 1$

(d) $\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 1)}^{2}}{16} = 1$

Answer:

$(b) \frac{{(x - 1)}^{2}}{64} + \frac{{(y + 1)}^{2}}{16} = 1 According to the question, the centre is at (1, - 1) . a = 8 (Given) \Rightarrow \frac{{(x - 1)}^{2}}{a^{2}} + \frac{{(y + 1)}^{2}}{b^{2}} = 1 . . . (1) It passes through point (1, 3) . i . e . x = 1 and y = 3 Putting these values in eq . (1), we get : \frac{{(1 - 1)}^{2}}{a^{2}} + \frac{{(3 + 1)}^{2}}{b^{2}} = 1 \Rightarrow \frac{16}{b^{2}} = 1 \Rightarrow b^{2} = 16 or b = 4 Substituting the values of a and b in eq. (1), we get: \frac{{(x - 1)}^{2}}{64} + \frac{{(y + 1)}^{2}}{16} = 1$

Page No 26.28:

Question 17:

The sum of the focal distances of any point on the ellipse 9x² + 16y² = 144 is
(a) 32
(b) 18
(c) 16
(d) 8

Answer:

$(d) 8 {9x}^{2} {+ 16y}^{2} = 144 \Rightarrow \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 Here, a =4 and b =3 The sum of the focal distances of any point on an ellipse is constant and equal to the length of the major axis of the ellipse. i . e . sum of the focal distances of any point on an ellips e = 2 a = 2 \times 4 = 8$

Page No 26.28:

Question 18:

If (2, 4) and (10, 10) are the ends of a latus-rectum of an ellipse with eccentricity 1/2, then the length of semi-major axis is
(a) 20/3
(b) 15/3
(c) 40/3
(d) none of these

Answer:

$(a) \frac{20}{3} e = \frac{1}{2} (Given) Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow \frac{1}{2} = \sqrt{1 - \frac{b^{2}}{a^{2}}} On squaring both sides, we get: \frac{1}{4} = 1 - \frac{b^{2}}{a^{2}} \Rightarrow \frac{1}{4} = \frac{a^{2} - b^{2}}{a^{2}} \Rightarrow 4 a^{2} - 4 b^{2} = a^{2} \Rightarrow 3 a^{2} = 4 b^{2} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{3}{4} \Rightarrow a^{2} = \frac{4 b^{2}}{3} or a = \frac{2 b}{\sqrt{3}} . . . (1) Latus rectum= \frac{2 b^{2}}{a^{2}} If (2, 4) and (10, 10) are the ends points of a latus rectum . i . e . \sqrt{{(10 - 2)}^{2} + {(10 - 4)}^{2}} = 2 b^{2} \times \frac{\sqrt{3}}{2 b} \Rightarrow \sqrt{64 + 36} = \sqrt{3} b On squaring both sides, we get: 100 = 3 b^{2} \Rightarrow b = \frac{10}{\sqrt{3}} Now, a = \frac{2 b}{\sqrt{3}} \Rightarrow a = \frac{20}{\sqrt{3}} \times \frac{1}{\sqrt{3}} \Rightarrow a = \frac{20}{3} So, the length of the semi major axis is \frac{20}{3} .$

Page No 26.28:

Question 19:

The equation $\frac{x^{2}}{2 - λ} + \frac{y^{2}}{λ - 5} + 1 = 0$ represents an ellipse, if
(a) λ < 5
(b) λ < 2
(c) 2 < λ < 5
(d) λ < 2 or λ > 5

Answer:

$(c) 2 < λ < 5 \frac{x^{2}}{2 - λ} + \frac{y^{2}}{λ - 5} + 1 = 0 \Rightarrow \frac{x^{2}}{λ - 2} + \frac{y^{2}}{5 - λ} = 1 To represent the equation of ellipse, we have: λ - 2 > 0 \Rightarrow λ > 2 and 5 - λ > 0 \Rightarrow 5 < λ ∴ 2 < λ < 5$

Page No 26.28:

Question 20:

The eccentricity of the ellipse 9x² + 25y² − 18x − 100y − 116 = 0, is
(a) 25/16
(b) 4/5
(c) 16/25
(d) 5/4

Answer:

$(b) \frac{4}{5} 9 x^{2} - 18 x + 25 y^{2} - 100 y - 116 = 0 \Rightarrow 9 (x^{2} - 2 x) + 25 (y^{2} - 4 y) = 116 \Rightarrow 9 (x^{2} - 2 x + 1) + 25 (y^{2} - 4 y + 4) = 116 + 100 + 9 \Rightarrow 9 (x - 1)^{2} + 25 (y - 2)^{2} = 225 \Rightarrow \frac{9 (x - 1)^{2}}{225} + \frac{25 (y - 2)^{2}}{225} = 1 \Rightarrow \frac{(x - 1)^{2}}{25} + \frac{(y - 2)^{2}}{9} = 1 Comparing it with \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get : a = 5 and b = 3 Here, a > b, so the major and the minor ax e s of the ellipse are along t h e x - a x i s and y - axis, respectively . N o w, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{9}{25}} \Rightarrow e = \sqrt{\frac{16}{25}} \Rightarrow e = \frac{4}{5}$

Page No 26.28:

Question 21:

If the major axis of an ellipse is three times the minor axis, then its eccentricity is equal to
(a) $\frac{1}{3}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\frac{1}{\sqrt{2}}$

(d) $\frac{2 \sqrt{2}}{3}$

(e) $\frac{2}{3 \sqrt{2}}$

Answer:

$(d) \frac{2 \sqrt{2}}{3} Length of the major axis = 2 b Length of the minor axis = 2 a According to question, the major axis of the ellipse is three times the minor axis . i . e . 2 b = 3 (2 a) \Rightarrow 2 b = 6 a \Rightarrow a = 2 b / 6 \Rightarrow a = b / 3, b = 3 a Here, a < b, s o the major and the minor axes of the ellipse are along the x - axis and the y - axis, respectively . Now, e = \sqrt{1 - \frac{a^{^{2}}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{b^{2}}{9}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{1}{9}} \Rightarrow e = \sqrt{\frac{8}{9}} \Rightarrow e = \frac{2 \sqrt{2}}{3}$

Page No 26.28:

Question 22:

The eccentricity of the ellipse 25x² + 16y² = 400 is
(a) 3/5
(b) 1/3
(c) 2/5
(d) 1/5

Answer:

$(a) \frac{3}{5} 25 x^{2} + 16 y^{2} = 400 \Rightarrow \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 . . . (1) Comparing equation (1) with \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get : a^{2} = 16 and b^{2} = 25 Here, a < b, so the major and the minor axes of the given ellipse are along the y - a x i s and the x - axis, respectively . Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{16}{25}} \Rightarrow e = \sqrt{\frac{9}{25}} \Rightarrow e = \frac{3}{5}$

Page No 26.28:

Question 23:

The eccentricity of the ellipse 5x² + 9y² = 1 is
(a) 2/3
(b) 3/4
(c) 4/5
(d) 1/2

Answer:

$(a) \frac{2}{3} 5 x^{2} + 9 y^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{5}} + \frac{y^{2}}{\frac{1}{9}} = 1 Comparing with the standard equation of the ellipse, we get : a^{2} = \frac{1}{5} and b^{2} = \frac{1}{9}, i . e . a = \frac{1}{\sqrt{5}} and b = \frac{1}{3} Here, a > b Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{1}{9}}{\frac{1}{5}}} \Rightarrow e = \sqrt{1 - \frac{5}{9}} \Rightarrow e = \sqrt{\frac{4}{9}} \Rightarrow e = \frac{2}{3}$

Page No 26.29:

Question 24:

The eccentricity of the ellipse 4x² + 9y² = 36 is

(a) $\frac{1}{2 \sqrt{3}}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\frac{\sqrt{5}}{3}$

(d) $\frac{\sqrt{5}}{6}$

Answer:

$(c) \frac{\sqrt{5}}{3}$
$4 x^{2} + 9 y^{2} = 36 \Rightarrow \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 . . (1) Compairing equation (1) with \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get a^{2} = 9 and b^{2} = 4 Here, a > b, so the major and the minor axes of the given ellipse are along the x - axis and the y - axis, respectively . Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{4}{9}} \Rightarrow e = \sqrt{\frac{5}{9}} \Rightarrow e = \frac{\sqrt{5}}{3}$

Page No 26.29:

Question 25:

The length of the latusrectum of the ellipse 3x² + y² = 12 is
(a) 4
(b) 3
(c) 8
(d) $\frac{4}{\sqrt{3}}$

Answer:

Given equation of ellipse is
$3 x^{2} + y^{2} = 12 i . e \frac{3 x^{2}}{12} + \frac{y^{2}}{12} = 1 i . e \frac{x^{2}}{4} + \frac{y^{2}}{{(\sqrt{12})}^{2}} = 1 i . e \frac{x^{2}}{{(2)}^{2}} + \frac{y^{2}}{{(\sqrt{12})}^{2}} = 1 since a = 2, b = \sqrt{12} i . e b > a \Rightarrow length of latus rectum \frac{2 \times 4}{\sqrt{12}} = \frac{2 \times 4}{2 \sqrt{3}} = \frac{4}{\sqrt{3}}$
Hence, the correct answer is option D.

Page No 26.29:

Question 26:

If e is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a < b),$ then
(a) b² = a²(1 – e²)
(b) a² = b²(1 – e²)
(c) a² = b²(e² – 1)
(d) b² = a²(e² – 1)

Answer:

For a < b,
Given equation of ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Eccentricity $e = \sqrt{1 - \frac{a^{2}}{b^{2}}}$
$i . e e^{2} = \frac{b^{2} - a^{2}}{b^{2}} i . e b^{2} e^{2} = b^{2} - a^{2} i . e a^{2} = b^{2} (1 - e^{2})$

Hence, the correct answer is option B.

Page No 26.29:

Question 27:

The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which asses through the points (–3, 1) and (2, –2) is
(a) 5x² + 3y² = 32
(b) 3x² + 5y² = 32
(c) 5x² – 3y² = 32
(d) 3x² + 5y² + 32 = 0

Answer:

for an ellipse, centre is given as (0, 0) major axis as x axis
i.e equation is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 where a > b$
Since ellipses passes through (–3, 1) and (2, –2)
We get,
$\frac{9}{a^{2}} + \frac{1}{b^{2}} = 1 and \frac{4}{a^{2}} + \frac{4}{b^{2}} = 1 i . e - \frac{9}{a^{2}} + 1 = \frac{1}{b^{2}} and \frac{4}{a^{2}} + 4 (\frac{1}{b^{2}}) = 1 i . e \frac{4}{a^{2}} + 4 (1 - \frac{9}{a^{2}}) = 1 i . e \frac{4}{a^{2}} + 4 - \frac{36}{a^{2}} = 1 i . e \frac{4 - 36}{a^{2}} = 1 - 4 i . e \frac{- 32}{a^{2}} = - 3 i . e \frac{1}{a^{2}} = \frac{3}{32}$
$∴ \frac{1}{b^{2}} = 1 - \frac{9}{a^{2}} = 1 - \frac{9 \times 3}{32} \frac{1}{b^{2}} = \frac{32 - 27}{32} i . e \frac{1}{b^{2}} = \frac{5}{32} Therefore equation of ellipse is \frac{3 x^{2}}{32} + \frac{5 y^{2}}{32} = 1 i . e 3 x^{2} + 5 y^{2} = 32$

Hence, the correct answer is option C.

Page No 26.29:

Question 1:

If the latusrectum of an ellipse be equal to half of its minor axis, then its eccentricity is ___________.

Answer:

For ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where a > b
Major axis is given by 2a and minor axis is 2b
$e^{2} = \frac{a^{2} - b^{2}}{a^{2}} . . . (1)$
Since latus rectum is $\frac{2 b^{2}}{a} . . . (2)$
Given latus rectum is half of its minor axis
i.e $\frac{2 b^{2}}{a} = \frac{1}{2} \times 2 b$
i.e a = 2b
$i . e e^{2} = \frac{(a^{2} - b^{2})}{a^{2}} = \frac{4 b^{2} - b^{2}}{{(2 b)}^{2}} e^{2} = \frac{3 b^{2}}{4 b^{2}} i . e e = \frac{\sqrt{3}}{2}$

Page No 26.29:

Question 2:

The eccentricity of the ellipse 16x² + 7y² = 112 is ___________.

Answer:

For given ellipse 16x²+ 7y² =112
$i . e \frac{16}{112} x^{2} + \frac{7}{112} y^{2} = 1 i . e \frac{x^{2}}{7} + \frac{y^{2}}{16} = 1 i . e \frac{x^{2}}{{(\sqrt{7})}^{2}} + \frac{y^{2}}{{(4)}^{2}} = 1 i . e a = \sqrt{7} and b = 4 i . e b > a ∴ eccentricity is \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{16 - 7}{16}} = \sqrt{\frac{9}{16}} i . e eccentricity = \frac{3}{4}$

Page No 26.29:

Question 3:

If the distance between the directrices of an ellipse be three times the distance between its foci, then the eccentricity is ___________.

Answer:

for an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where a > b
Distance between directrix is $\frac{2 a}{e}$ where e is eccentricity and Distance between foci is 2ae
According to given condition,
distance between directrix = 3(distance between foci)
$i . e \frac{2 a}{e} = 3 \times 2 a e i . e e^{2} = \frac{1}{3} i . e e = \frac{1}{\sqrt{3}}$

Page No 26.29:

Question 4:

If the eccentricity of an ellipse is $\frac{5}{8}$ and the distance between its foci is 10, then the length of the latusrectum is ___________.

Answer:

For an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
eccentricity = $\frac{5}{8}$
Distance between foci is 10 (given)
i.e 2ae = 10
i.e  ae = 5
i.e  a $\times \frac{5}{8}$ = 5
i.e  a = 8
Length of latus rectum is $\frac{2 b^{2}}{a} . . . (1)$
Since $e^{2} = 1 - {(\frac{b}{a})}^{2}$
i.e $b^{2} = a^{2} (1 - e^{2})$
$∴ Length of latus rectum is 2 (a^{2} (1 - e^{2})) \frac{1}{a} = 2 a (1 - e^{2}) = 2 \times 8 (1 - \frac{25}{64}) = 16 (\frac{64 - 25}{64}) Hence, Length of latus rectum = \frac{39}{4}$

Page No 26.29:

Question 5:

If the coordinates of foci and vertices of an ellipse are (±1, 0) and (±2, 0) respectively, then the minor axis is of length ___________.

Answer:

For an ellipse,
co-ordinates of foci is (±1, 0)
co-ordinates of vertices is (± 2, 0)
since y-co-ordinate is zero,
∴ equation of ellipse is of form
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; a > b$
Since vertex in general is given by (± a, 0)
and foci is given by (± ae, 0) ; where e is eccentricity.
∴ a = 2 and ae = 1
$\Rightarrow e = \frac{1}{2} \Rightarrow e^{2} = 1 - \frac{b^{2}}{a^{2}} i . e {(\frac{1}{2})}^{2} = 1 - \frac{b^{2}}{4} i . e \frac{b^{2}}{y} = 1 - \frac{1}{4} = \frac{3}{4} i . e b^{2} = 3 i . e b = \sqrt{3} ∴ length of minor axis is 2 \sqrt{3}$

Page No 26.29:

Question 6:

The distance between the directrices of the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$ is ___________.

Answer:

Given parabola is $\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$
i.e a²= 36 and b²= 20
$∴ e = \sqrt{1 - \frac{20}{36}} = \sqrt{\frac{36 - 20}{36}} = \sqrt{\frac{16}{36}} e = \frac{4}{6} = \frac{2}{3}$
∴ equation of directrices is $x = \frac{a}{e}$ and $x = - \frac{a}{e}$
i.e $x = \frac{6}{2} \times 3 and x = - 9$
i.e x = 9 and x = $-$ 9 are equations of directrices
$∴$ Distance between directrices is 18

Page No 26.29:

Question 7:

The distance between the foci of the ellipse 3x² + 4y² = 48 is ___________.

Answer:

Given ellipse is 3x² + 4y² = 48
$i . e \frac{3 x^{2}}{48} + \frac{4 y^{2}}{48} = 1 i . e \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1 i . e a^{2} = 16, b^{2} = 12 ∴ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{12}{16}} = \sqrt{\frac{16 - 12}{16}} = \sqrt{\frac{4}{16}} i . e e = \frac{1}{2} ∴ Coordinate of foci are (\pm a e, 0) i . e (\pm 4 \times \frac{1}{2}, 0) i . e (\pm 2, 0)$
∴ Distance of foci is 4

Page No 26.29:

Question 8:

The eccentricity of the ellipse whose latusrectum is equal to the distance between the foci, is ___________.

Answer:

For an ellipse, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Given, latusrectum = Distance between foci
i.e $\frac{2 b^{2}}{a}$ = 2ae
i.e 2b² = 2a²e
i.e $\frac{b^{2}}{a^{2}}$ = e
$Since e = \sqrt{1 - \frac{b^{2}}{a^{2}}} i . e e = \sqrt{1 - e} i . e e^{2} = 1 - e i . e e^{2} + e - 1 = 0 i . e e = \frac{- 1 \pm \sqrt{1 + 4}}{2} e = \frac{\pm \sqrt{5} - 1}{2} ∴ eccentricity is \frac{\sqrt{5} - 1}{2}$

Page No 26.29:

Question 9:

The equation $\frac{x^{2}}{2 - r} + \frac{y^{2}}{r - 5} + 1 = 0$ represents an ellipse, if ___________.

Answer:

For given equation $\frac{x^{2}}{2 - r} + \frac{y^{2}}{r - 5} + 1 = 0$
Represents an ellipse if
2 $-$ r < 0 and r $-$ 5 < 0
i.e r > 2 and r < 5
i.e 2 < r < 5

Page No 26.29:

Question 10:

An ellipse is described by using an endless string which passed over two points. If the axes are 6 cm and 4 cm, the length of the string and distance between the points are ___________.

Answer:

Let equation of ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Length of major axis i.e 2a = 6
i.e a = 3
and Length of minor axis is 4 = 2b
i.e b = 2
$∴$ eccentricity $e = 1 - \frac{b^{2}}{a^{2}}$
i.e $e = 1 - \frac{4}{9} = \frac{5}{9}$
Let S and S' represents the foci of ellipse and P be any point an ellipse
$∴$ SP + S'P = 2a
$∴$ Length of the endless string SP + S'P + SS'
= 2a + 2ae
$= 2 \times 3 + 2 \times 3 \times \frac{\sqrt{5}}{3}$
i.e Length of the endless string = 6 + $2 \sqrt{5}$ and distance between the points is SS' = $2 \sqrt{5}$

Page No 26.29:

Question 11:

The distance between the directrices of the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$ is ___________.

Answer:

Given ellipse is of the form,
$\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$
i.e a² = 36 and b² = 20
$∴ e = \sqrt{1 - \frac{20}{36}} = \sqrt{\frac{36 - 20}{36}} = \sqrt{\frac{16}{36}} e = \frac{4}{6} = \frac{2}{3}$
∴ equation of directrices is $x = \frac{a}{e}$ and $x = - \frac{a}{e}$
i.e $x = \frac{6}{2} \times 3 and x = - 9$
i.e x = 9 and x = $-$ 9 are equations of directrices
$∴$ Distance between directrices is 18

Page No 26.30:

Question 12:

The equation of the ellipse having foci (0, 1),(0, –1) and minor axis of length 1 is ___________.

Answer:

For an ellipse, length of minor axis is given = l and foci is given by = (0, $\pm$ 1)
i.e ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; a < b$
i.e be = $\pm$ 1
Also, 2a = 1 i.e $\frac{1}{a}$ = 2
i.e a = $\frac{1}{2}$
and e = $\pm \frac{1}{b}$
$Since eccentricity e = \sqrt{1 - \frac{a^{2}}{b^{2}}} i . e e^{2} = 1 - \frac{a^{2}}{b^{2}} i . e \frac{1}{b^{2}} = 1 - \frac{1}{4 b^{2}} \frac{1}{b^{2}} + \frac{1}{4 b^{2}} = 1 i . e \frac{4 + 1}{4 b^{2}} = 1 i . e \frac{1}{b^{2}} = \frac{4}{5} ∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 i . e \frac{x^{2}}{\frac{1}{4}} + \frac{4 y^{2}}{5} = 1 i . e 4 x^{2} + \frac{4 y^{2}}{5} = 1 i . e 20 x^{2} + 4 y^{2} = 5$

Hence, equation of ellipse is 20x² + 4y² = 5.

Page No 26.30:

Question 13:

The eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a < b,$ is given by ___________.

Answer:

For $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; a < b$
eccentricity, $e = \sqrt{1 - \frac{a^{2}}{b^{2}}}$
$i . e e^{2} = 1 - \frac{a^{2}}{b^{2}} i . e b^{2} e^{2} = b^{2} - a^{2} i . e a^{2} = b^{2} (1 - e^{2})$

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Question 14:

The length of the latusretum of the ellipse 3x² + y² = 12 is ___________.

Answer:

Given ellipse is 3x²+ y² = 12
$i . e \frac{x^{2}}{4} + \frac{y^{2}}{12} = 1$
i.e a² = 4 and b²= 12
$∴$ Length of latus rectum is $\frac{2 b^{2}}{a}$ = $\frac{2 \times 12}{2} = 12$

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Question 15:

If the latusrectum of an ellipse with axis along x-axis and centre at origin is 10, distance between foci = length of minor axis, then equation of the ellipse is ___________.

Answer:

Centre of ellipse is given (0, 0)
Let the equation of ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Latusrectum is given, which is 10
$i . e \frac{2 b^{2}}{a} = 10 i . e \frac{b^{2}}{a} = 5 . . . (1)$
Also given, distance between foci = Length of minor axis
i.e 2ae = 2b
i.e ae = b
i.e e = $\frac{b}{a}$
Since b² = a² (1 – e²)
i.e a²e² = a²(1 – e²)
i.e e² = 1 – e²
i.e 2e² = 1
i.e e = $\pm \frac{1}{\sqrt{2}}$
i.e e = $\frac{1}{\sqrt{2}}$
$\Rightarrow a = b \sqrt{2}$
Since $\frac{2 b^{2}}{a} = 10$
$\Rightarrow \frac{2 b^{2}}{b \sqrt{2}} = 10 i . e b = \frac{10}{\sqrt{2}} i . e b = 5 \sqrt{2} \Rightarrow a = 10$
$∴$ equation of ellipse is $\frac{x^{2}}{{(10)}^{2}} + \frac{b^{2}}{{(5 \sqrt{2})}^{2}} = 1$
i.e $\frac{x^{2}}{100} + \frac{b^{2}}{50} = 1$
i.e x² + 2b² – 100
Hence equation of ellipse in x² + 2b² = 100

Page No 26.30:

Question 1:

If the lengths of semi-major and semi-minor axes of an ellipse are 2 and $\sqrt{3}$ and their corresponding equations are y − 5 = 0 and x + 3 = 0, then write the equation of the ellipse.

Answer:

$The lengths of the semi major and the semi minor axes of the ellipse are 2 and \sqrt{3}, respectively. Their corresponding equations are y-5=0 and x+3 =0. S o, a =2 and b = \sqrt{3} The e quation of the ellipse is given by \frac{{(x + 3)}^{2}}{4} + \frac{{(y - 5)}^{2}}{3} = 1 \Rightarrow 3 {(x + 3)}^{2} + 4 {(y - 5)}^{2} = 12 \Rightarrow 3 (x^{2} + 9 + 6 x) + 4 (y^{2} + 25 - 10 y) = 12 \Rightarrow 3 x^{2} + 27 + 18 x + 4 y^{2} + 100 - 40 y = 12 \Rightarrow 3 x^{2} + 4 y^{2} + 18 x - 40 y + 115 = 0$

Page No 26.30:

Question 2:

Write the eccentricity of the ellipse 9x² + 5y² − 18x − 2y − 16 = 0.

Answer:

$9 x^{2} + 5 y^{2} - 18 x - 2 y - 16 = 0 \Rightarrow 9 (x^{2} - 2 x) + 5 (y^{2} - \frac{2 y}{5}) = 16 \Rightarrow 9 (x^{2} - 2 x + 1) + 5 (y^{2} - \frac{2 y}{5} + \frac{1}{25}) = 16 + 9 + \frac{1}{5} \Rightarrow 9 {(x - 1)}^{2} + 5 {(y - \frac{1}{5})}^{2} = \frac{126}{5} \Rightarrow \frac{{(x - 1)}^{2}}{\frac{126}{45}} + \frac{{(y - \frac{1}{5})}^{2}}{\frac{126}{25}} = 1 \Rightarrow a^{2} = \frac{126}{45} and b^{2} = \frac{126}{25} Clearly, a < b Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{126}{45}}{\frac{126}{25}}} \Rightarrow e = \sqrt{1 - \frac{5}{9}} \Rightarrow e = \frac{2}{3}$

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Question 3:

Write the centre and eccentricity of the ellipse 3x² + 4y² − 6x + 8y − 5 = 0.

Answer:

$3 x^{2} - 6 x + 4 y^{2} + 8 y - 5 = 0 \Rightarrow 3 (x^{2} - 2 x) + 4 (y^{2} + 2 y) = 5 \Rightarrow 3 (x^{2} - 2 x + 1) + 4 (y^{2} + 2 y + 1) = 5 + 3 + 4 \Rightarrow 3 (x - 1)^{2} + 4 (y + 1)^{2} = 12 \Rightarrow \frac{3 (x - 1)^{2}}{12} + \frac{4 (y + 1)^{2}}{12} = 1 \Rightarrow \frac{(x - 1)^{2}}{4} + \frac{(y + 1)^{2}}{3} = 1 Compairing it with \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get : a = 2 and b = \sqrt{3} Here, a > b, so the major and t h e minor ax e s of the ellipse are along the x - axis and y - ax i s, respectively . Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{3}{4}} \Rightarrow e = \sqrt{\frac{1}{4}} ∴ e = \frac{1}{2} and centre = (1, - 1)$

Page No 26.30:

Question 4:

PSQ is a focal chord of the ellipse 4x² + 9y² = 36 such that SP = 4. If S' is the another focus, write the value of S'Q.

Answer:

$The given equation of the ellipse is 4 x^{2} + 9 y^{2} = 36 . . . (i) \Rightarrow \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = 9 and b^{2} = 4 i . e a = 3 and b = 2 Clearly a > b, therefore the major axis and minor axis of the ellipse are along x axis and y axis respectively . Let, e be the ecentricity of the ellipse . Then, ∴ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3} Therefore, coordinates of focus at S i . e (a e, 0) = (\sqrt{5}, 0) & coordinates of focus at S' = (- a e, 0) = (- \sqrt{5}, 0) It is given that P S Q is a focal chord As we know that, S P + S' P = 2 a \Rightarrow 4 + S' P = 6 [∵ S P = 4] \Rightarrow S' P = 2 Let coordinates of P be (m, n) A s S' P = 2 ∴ \sqrt{{(n - 0)}^{2} + {(m + \sqrt{5})}^{2}} = 2 \Rightarrow n^{2} + m^{2} + 5 + 2 \sqrt{5} m = 4 . . . (ii) & S P = 4 ∴ \sqrt{{(n - 0)}^{2} + {(m - \sqrt{5})}^{2}} = 4 \Rightarrow n^{2} + m^{2} + 5 - 2 \sqrt{5} m = 16 . . . (iii) Subtracting eq (iii) from eq (ii), we get 4 \sqrt{5} m = - 12 \Rightarrow m = \frac{- 3}{\sqrt{5}} & we get n = \frac{4}{\sqrt{5}} ∴ Coordinates of P are (\frac{- 3}{\sqrt{5}}, \frac{4}{\sqrt{5}}) and coordinates of S are (\sqrt{5}, 0) ∴ Equation of the line segment P S which is extended to P Q is given by x + 2 y = \sqrt{5} . . . (i v) Solving eq (i) & (iv) we get, x = \frac{- 3}{\sqrt{5}} and y = \frac{4}{\sqrt{5}} which are the coordinates of P & x = \frac{66 \sqrt{5}}{50} and y = \frac{- 8 \sqrt{5}}{50} which would be the coordinates of Q . ∴ S' Q = \sqrt{{(\frac{- 8 \sqrt{5}}{50} - 0)}^{2} + {(\frac{66 \sqrt{5}}{50} + \sqrt{5})}^{2}} = \frac{26}{5}$

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Question 5:

Write the eccentricity of an ellipse whose latus-rectum is one half of the minor axis.

Answer:

$According to the question, the latus rectum is half its minor axis. i . e . \frac{2 b^{2}}{a} = \frac{1}{2} \times 2 b \Rightarrow 2 b = a Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{b^{2}}{4 b^{2}}} \Rightarrow e = \sqrt{1 - \frac{1}{4}} \Rightarrow e = \frac{\sqrt{3}}{2}$

Page No 26.30:

Question 6:

If the distance between the foci of an ellipse is equal to the length of the latus-rectum, write the eccentricity of the ellipse.

Answer:

$According to the question, the distance between the foci of an ellipse is equal to the length of the latus rectum. i . e . \frac{2 b^{2}}{a} = 2 a e \Rightarrow e = \frac{b^{2}}{a^{2}} But e = \sqrt{1 - \frac{b^{2}}{a^{2}}} Then e = \sqrt{1 - e} Squaring both sides, we get: e^{2} + e - 1 = 0 \Rightarrow e = \frac{- 1 \pm \sqrt{1 + 4}}{2} (∵ Ecentricity cannot be negative) \Rightarrow e = \frac{\sqrt{5} - 1}{2}$

Page No 26.30:

Question 7:

If S and S' are two foci of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and B is an end of the minor axis such that ∆BSS' is equilateral, then write the eccentricity of the ellipse.

Answer:

$W e know that t h e focal distance of a point B (0, b) is a \pm e . 0 = a i . e . S B = S B' = a ∴ S B + S B^{'} = 2 a Since ∆ {BSS}^{'} is equilateral, we have: S B = S S^{'} = S^{'} B = 2 a e \Rightarrow 2 a e + 2 a e = 2 a \Rightarrow 4 a e = 2 a \Rightarrow e = \frac{2}{4} \Rightarrow e = \frac{1}{2}$

Page No 26.30:

Question 8:

If the minor axis of an ellipse subtends an equilateral triangle with vertex at one end of major axis, then write the eccentricity of the ellipse.

Answer:

$According to the question, the minor axis of the ellipse subtends an equilateral triangle with the vertex at one end of the major axis. AB= \sqrt{a^{2} + b^{2}} We know that ABC is an equilateral triangle . ∴ AB = BB' \Rightarrow \sqrt{a^{2} + b^{2}} = 2 b On squaring both sides, we have: a^{2} + b^{2} = 4 b^{2} \Rightarrow a^{2} = 3 b^{2} Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{b^{2}}{3 b^{2}}} \Rightarrow e = \sqrt{1 - \frac{1}{3}} \Rightarrow e = \sqrt{\frac{2}{3}}$

Page No 26.30:

Question 9:

If a latus rectum of an ellipse subtends a right angle at the centre of the ellipse, then write the eccentricity of the ellipse.

Answer:

$According to the Pythogoras theorem, we have: O A^{2} + O B^{2} = A B^{2} From the figure, we can see that O A = \sqrt{{(\frac{b^{2}}{a} - 0)}^{2} + {(a e - 0)}^{2}} = \sqrt{\frac{b^{4}}{a^{2}}} + a^{2} e^{2} = OB and AB = \frac{2 b^{2}}{a} Now, 2 [a^{2} e^{2} + \frac{b^{4}}{a^{2}}] = \frac{4 b^{4}}{a^{2}} \Rightarrow a^{2} e^{2} + \frac{b^{4}}{a^{2}} = \frac{2 b^{4}}{a^{2}} \Rightarrow a^{2} e^{2} = - \frac{b^{4}}{a^{2}} + \frac{2 b^{4}}{a^{2}} \Rightarrow a^{2} e^{2} = \frac{b^{4}}{a^{2}} \Rightarrow e^{2} = \frac{b^{4}}{a^{4}} \Rightarrow e = \frac{b^{2}}{a^{2}} We know that e = \sqrt{1 - \frac{b^{2}}{a^{2}}} e = \sqrt{1 - e} On squaring both sides, we get: e^{2} + e - 1 = 0 \Rightarrow e = \frac{- 1 \pm \sqrt{1 + 4}}{2} (∵ Ecentricity cannot be negative) \Rightarrow e = \frac{\sqrt{5} - 1}{2}$

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