Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 27 - Hyperbola

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Page No 27.13:

Question 1:

The equation of the directrix of a hyperbola is x − y + 3 = 0. Its focus is (−1, 1) and eccentricity 3. Find the equation of the hyperbola.

Answer:

Let S be the focus and $P (x, y)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

$\Rightarrow \sqrt{{(x - (- 1))}^{2} + {(y - 1)}^{2}} = 3 \times (\frac{x - y + 3}{\sqrt{2}})$
Squaring both the sides, we get:
${(x + 1)}^{2} + {(y - 1)}^{2} = \frac{9}{2} {(x - y + 3)}^{2} \Rightarrow x^{2} + 2 x + 1 + y^{2} - 2 y + 1 = \frac{9}{2} (x^{2} + y^{2} + 9 - 2 x y - 6 y + 6 x) \Rightarrow 2 x^{2} + 4 x + 2 + 2 y^{2} - 4 y + 2 = 9 x^{2} + 9 y^{2} + 81 - 18 x y - 54 y + 54 x \Rightarrow 7 x^{2} + 7 y^{2} + 50 x - 50 y - 18 x y + 77 = 0$
Equation of the hyperbola:
$7 x^{2} + 7 y^{2} + 50 x - 50 y - 18 x y + 77 = 0$

Page No 27.13:

Question 2:

Find the equation of the hyperbola whose
(i) focus is (0, 3), directrix is x + y − 1 = 0 and eccentricity = 2
(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2
(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity = $\sqrt{3}$
(iv) focus is (2, −1), directrix is 2x + 3y = 1 and eccentricity = 2
(v) focus is (a, 0), directrix is 2x − y + a = 0 and eccentricity = $\frac{4}{3}$
(vi) focus is (2, 2), directrix is x + y = 9 and eccentricity = 2.

Answer:

(i) Let S be the focus and $P (x, y)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.

By definition:
SP = ePM
$\Rightarrow$ $\sqrt{(x - 0)^{2} + (y - 3)^{2}} = 2 (\frac{x + y - 1}{\sqrt{2}})$
Squaring both the sides:
$(x - 0)^{2} + (y - 3)^{2} = 4 {(\frac{x + y - 1}{\sqrt{2}})}^{2} \Rightarrow x^{2} + y^{2} + 9 - 6 y = 2 (x^{2} + y^{2} + 1 + 2 x y - 2 y - 2 x) \Rightarrow x^{2} + y^{2} + 4 x y + 2 y - 4 x - 7 = 0$
∴ Equation of the hyperbola = $x^{2} + y^{2} + 4 x y + 2 y - 4 x - 7 = 0$

(ii) Let S be the focus and $P (x, y)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

$\Rightarrow$ $\sqrt{(x - 1)^{2} + (y - 1)^{2}} = 2 (\frac{3 x + 4 y + 8}{5})$
Squaring both the sides:
$(x - 1)^{2} + (y - 1)^{2} = 4 {(\frac{3 x + 4 y + 8}{5})}^{2} \Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 - 2 y = \frac{4}{25} (9 x^{2} + 16 y^{2} + 64 + 24 x y + 64 y + 48 x) \Rightarrow 25 x^{2} + 25 - 50 x + 25 y^{2} + 25 - 50 y = 36 x^{2} + 64 y^{2} + 256 + 96 x y + 256 y + 192 x \Rightarrow 11 x^{2} + 39 y^{2} + 96 x y + 306 y + 242 x + 206 = 0$
∴ Equation of the hyperbola = $11 x^{2} + 39 y^{2} + 96 x y + 306 y + 242 x + 206 = 0$

(iii) Let S be the focus and $P (x, y)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

$\Rightarrow$ $\sqrt{(x - 1)^{2} + (y - 1)^{2}} = \sqrt{3} (\frac{2 x + y - 1}{\sqrt{5}})$
Squaring both the sides:
$(x - 1)^{2} + (y - 1)^{2} = 3 {(\frac{2 x + y - 1}{5})}^{2} \Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 - 2 y = \frac{3}{5} (4 x^{2} + y^{2} + 1 + 4 x y - 2 y - 4 x) \Rightarrow 5 x^{2} + 5 - 10 x + 5 y^{2} + 5 - 10 y = 12 x^{2} + 3 y^{2} + 3 + 12 x y - 6 y - 12 x \Rightarrow 7 x^{2} - 2 y^{2} + 12 x y + 4 y - 2 x - 7 = 0$
∴ Equation of the hyperbola = $7 x^{2} - 2 y^{2} + 12 x y + 4 y - 2 x - 7 = 0$

(iv) Let S be the focus and $P (x, y)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.

By definition:
SP = ePM
= ePM
$\Rightarrow$ $\sqrt{(x - 2)^{2} + (y + 1)^{2}} = 2 (\frac{2 x + 3 y - 1}{\sqrt{13}})$
Squaring both the sides:
$(x - 2)^{2} + (y + 1)^{2} = 4 {(\frac{2 x + 3 y - 1}{13})}^{2} \Rightarrow x^{2} + 4 - 4 x + y^{2} + 1 + 2 y = \frac{4}{13} (4 x^{2} + 9 y^{2} + 1 + 12 x y - 6 y - 4 x) \Rightarrow 13 x^{2} + 52 - 52 x + 13 y^{2} + 13 + 26 y = 16 x^{2} + 36 y^{2} + 4 + 48 x y - 24 y - 16 x \Rightarrow 3 x^{2} + 23 y^{2} + 48 x y - 50 y + 36 x - 61 = 0$
∴ Equation of the hyperbola = $3 x^{2} + 23 y^{2} + 48 x y - 50 y + 36 x - 61 = 0$

(v) Let S be the focus and $P (x, y)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.

By definition:
SP = ePM
$\Rightarrow$ $\sqrt{(x - a)^{2} + (y - 0)^{2}} = \frac{4}{3} (\frac{2 x - y + a}{\sqrt{5}})$
Squaring both the sides:
$(x - a)^{2} + (y)^{2} = \frac{16}{9} {(\frac{2 x - y + a}{5})}^{2} \Rightarrow x^{2} - 2 a x + a^{2} + y^{2} = \frac{16}{45} (4 x^{2} + y^{2} + a^{2} - 4 x y - 2 y a + 4 x a) \Rightarrow 45 x^{2} - 90 a x + 45 a^{2} + 45 y^{2} = 64 x^{2} + 16 y^{2} + 16 a^{2} - 64 x y - 32 a y + 64 a x \Rightarrow 19 x^{2} - 29 y^{2} - 64 x y - 32 a y + 154 a x - 29 a^{2} = 0$
∴ Equation of the hyperbola = $19 x^{2} - 29 y^{2} - 64 x y - 32 a y + 154 a x - 29 a^{2} = 0$

(vi) Let S be the focus and $P (x, y)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

$\Rightarrow$ $\sqrt{(x - 2)^{2} + (y - 2)^{2}} = 2 (\frac{x + y - 9}{\sqrt{2}})$
Squaring both the sides:
$(x - 2)^{2} + (y - 2)^{2} = 4 {(\frac{x + y - 9}{2})}^{2} \Rightarrow x^{2} - 4 x + 4 + y^{2} - 4 y + 4 = 2 (x^{2} + y^{2} + 81 + 2 x y - 18 y - 18 x) \Rightarrow x^{2} - 4 x + 4 + y^{2} - 4 y + 4 = 2 x^{2} + 2 y^{2} + 162 + 4 x y - 36 y - 36 x \Rightarrow x^{2} + y^{2} + 4 x y - 32 y - 32 x + 154 = 0$
∴ Equation of the hyperbola = $x^{2} + y^{2} + 4 x y - 32 y - 32 x + 154 = 0$

Page No 27.13:

Question 3:

Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola
(i) 9x² − 16y² = 144
(ii) 16x² − 9y² = −144
(iii) 4x² − 3y² = 36
(iv) 3x² − y² = 4
(v) 2x² − 3y² = 5.

Answer:

(i) Equation of the hyperbola:
$9 x^{2} - 16 y^{2} = 144$
This can be rewritten in the following way:
$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
This is the standard equation of a hyperbola, where $a^{2} = 16 and b^{2} = 9$ .

$\Rightarrow b^{2} = a^{2} (e^{2} - 1) \Rightarrow 9 = 16 (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{9}{16} \Rightarrow e^{2} = \frac{25}{16} \Rightarrow e = \frac{5}{4}$
Coordinates of the foci are given by $(\pm a e, 0)$ , i.e. $(\pm 5, 0)$ .
Equation of directrices:
$x = \pm \frac{a}{e}$
$\Rightarrow x = \pm \frac{4}{\frac{5}{4}} \Rightarrow 5 x \pm 16 = 0$
Length of the latus rectum of the hyperbola is $\frac{2 b^{2}}{a}$ .
Length of the latus rectum = $\frac{2 \times 9}{4} = \frac{9}{2}$

(ii) Equation of the hyperbola:
$16 x^{2} - 9 y^{2} = - 144$
This can be rewritten in the following way:
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = - 1$
This is the standard equation of a hyperbola, where $a^{2} = 9 and b^{2} = 16$ .

$\Rightarrow a^{2} = b^{2} (e^{2} - 1) \Rightarrow 9 = 16 (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{9}{16} \Rightarrow e^{2} = \frac{25}{16} \Rightarrow e = \frac{5}{4}$
Coordinates of foci are given by $(0, \pm a e)$ , i.e. $(0, \pm 5)$ .
Equation of the directrices:
$y = \pm \frac{a}{e}$
$\Rightarrow y = \pm \frac{4}{\frac{5}{4}} \Rightarrow 5 y \pm 16 = 0$
Length of the latus rectum of the hyperbola = $\frac{2 a^{2}}{b}$
Length of the latus rectum = $\frac{2 \times 9}{4} = \frac{9}{2}$

(iii) Equation of the hyperbola:
$4 x^{2} - 3 y^{2} = 36$
This can be rewritten in the following way:
$\frac{4 x^{2}}{36} - \frac{3 y^{2}}{36} = 1 \frac{x^{2}}{9} - \frac{y^{2}}{12} = 1$
This is the standard equation of a hyperbola, where $a^{2} = 9 and b^{2} = 12$ .

$\Rightarrow b^{2} = a^{2} (e^{2} - 1) \Rightarrow 12 = 9 (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{4}{3} \Rightarrow e^{2} = \frac{7}{3} \Rightarrow e = \sqrt{\frac{7}{3}}$
Coordinates of the foci are given by $(\pm a e, 0)$ , i.e. $(\pm \sqrt{21}, 0)$ .
Equation of the directrices:
$x = \pm \frac{a}{e}$
$\Rightarrow x = \pm \frac{3}{\sqrt{\frac{7}{3}}} \Rightarrow \sqrt{7} x \pm 3 \sqrt{3} = 0$
Length of the latus rectum of the hyperbola is $\frac{2 b^{2}}{a}$ .
$\Rightarrow \frac{2 \times 12}{3} = 8$

(iv) Equation of the hyperbola:
$3 x^{2} - y^{2} = 4$
This can be rewritten in the following way:
$\frac{3 x^{2}}{4} - \frac{y^{2}}{4} = 1 \Rightarrow \frac{x^{2}}{\frac{4}{3}} - \frac{y^{2}}{4} = 1$
This is the standard equation of a hyperbola, where $a^{2} = \frac{4}{3} and b^{2} = 4$ .

$\Rightarrow b^{2} = a^{2} (e^{2} - 1) \Rightarrow 4 = \frac{4}{3} (e^{2} - 1) \Rightarrow e^{2} - 1 = 3 \Rightarrow e^{2} = 4 \Rightarrow e = 2$
Coordinates of the foci are given by $(\pm a e, 0)$ , i.e. $(\pm \frac{4 \sqrt{3}}{3}, 0)$ .
Equation of the directrices:
$x = \pm \frac{a}{e}$
$x = \pm \frac{\sqrt{\frac{4}{3}}}{2} \Rightarrow \sqrt{3} x \pm 1 = 0$
Length of the latus rectum of the hyperbola = $\frac{2 b^{2}}{a}$
$\Rightarrow \frac{2 \times 4}{\sqrt{\frac{4}{3}}} = 4 \sqrt{3}$

(v) Equation of the hyperbola:
$2 x^{2} - 3 y^{2} = 5$
This can be rewritten in the following manner:
$\frac{2 x^{2}}{5} - \frac{3 y^{2}}{5} = 1 \Rightarrow \frac{x^{2}}{\frac{5}{2}} - \frac{y^{2}}{\frac{5}{3}} = 1$
This is the standard equation of a hyperbola, where $a^{2} = \frac{5}{2} and b^{2} = \frac{5}{3}$ .

$\Rightarrow b^{2} = a^{2} (e^{2} - 1) \Rightarrow \frac{5}{3} = \frac{5}{2} (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{2}{3} \Rightarrow e^{2} = \frac{5}{3} \Rightarrow e = \sqrt{\frac{5}{3}}$
Coordinates of the foci are given by $(\pm a e, 0)$ , i.e. $(\pm \frac{5 \sqrt{6}}{6}, 0)$ .
Equation of the directrices:
$x = \pm \frac{a}{e}$
$x = \pm \frac{\sqrt{\frac{5}{2}}}{\sqrt{\frac{5}{3}}} \Rightarrow x = \pm \frac{\sqrt{3}}{\sqrt{2}} \Rightarrow \sqrt{2} x \pm \sqrt{3} = 0$
Length of the latus rectum of the hyperbola is $\frac{2 b^{2}}{a}$ .

$\Rightarrow \frac{2 \times (\frac{5}{3})}{\sqrt{\frac{5}{2}}} = \frac{10}{3} \sqrt{\frac{2}{5}}$

Page No 27.13:

Question 4:

Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x² − 36y² = 225.

Answer:

Equation of the hyperbola:
$25 x^{2} - 36 y^{2} = 225$
This equation can be rewritten in the following way:
$\frac{25 x^{2}}{225} - \frac{36 y^{2}}{225} = 1 \Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{\frac{225}{36}} = 1$
This is the standard equation of the hyperbola, where $a^{2} = 9 and b^{2} = \frac{225}{36}$ .
Length of the transverse = $2 a = 2 \times 3 = 6$
Length of the conjugate axis = $2 b = 2 \times \frac{15}{6} = 5$

Eccentricity of the hyperbola is calculated using $b^{2} = a^{2} (e^{2} - 1)$ .

$\Rightarrow \frac{225}{36} = 9 (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{25}{36} \Rightarrow e^{2} = \frac{61}{36} \Rightarrow e = \frac{\sqrt{61}}{6}$
Length of the latus rectum = $\frac{2 b^{2}}{a} = \frac{2 \times (\frac{225}{36})}{3} = \frac{25}{6}$
The coordinates of the foci are given by $(\pm a e, 0)$ .
$\Rightarrow (\pm \frac{\sqrt{61}}{2}, 0)$

Page No 27.13:

Question 5:

Find the centre, eccentricity, foci and directrices of the hyperbola
(i) 16x² − 9y² + 32x + 36y − 164 = 0
(ii) x² − y² + 4x = 0
(iii) x² − 3y² − 2x = 8.

Answer:

(i) The equation $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ can be simplified in the following way:
$16 (x^{2} + 2 x) - 9 (y^{2} - 4 y) = 164 \Rightarrow 16 (x^{2} + 2 x + 1) - 9 (y^{2} - 4 y + 4) = 164 + 16 - 36 \Rightarrow 16 (x + 1)^{2} - 9 {(y - 2)}^{2} = 144 \Rightarrow \frac{(x + 1)^{2}}{9} - \frac{(y - 2)^{2}}{16} = 1$
Thus, the centre is $(- 1, 2)$ .
Eccentricity of the hyperbola = $\frac{\sqrt{a^{2} + b^{2}}}{a} = \frac{\sqrt{9 + 16}}{3} = \frac{5}{3}$
Foci = $(- 1 \pm 5, 2) = (- 6, 2), (4, 2)$

Equation of the directrices:
$x + 1 = \pm \frac{a}{e} \Rightarrow x = \pm \frac{3 \times 3}{5} - 1 \Rightarrow x = \pm \frac{9}{5} - 1 \Rightarrow 5 x - 4 = 0 or 5 x + 14 = 0$

(ii) The equation $x^{2} - y^{2} + 4 x = 0$ can be simplified in the following manner:
$(x^{2} + 4 x + 4) - y^{2} = 4 \Rightarrow {(x + 2)}^{2} - y^{2} = 4 \Rightarrow \frac{(x + 2)^{2}}{4} - \frac{y^{2}}{1} = 1$
Thus, the centre is $(- 2, 0)$ .
Eccentricity of the hyperbola = $\frac{\sqrt{a^{2} + b^{2}}}{a} = \frac{\sqrt{4 + 1}}{2} = \frac{\sqrt{5}}{2}$
Foci = $(- 2 \pm \sqrt{5}, 0)$
Equation of the directrices:
$x + 2 = \pm \frac{a}{e} \Rightarrow x + 2 = \pm \sqrt{5}$

(iii) The equation $x^{2} - 3 y^{2} - 2 x = 8$ can be simplified in the following manner:
$(x^{2} - 2 x + 1) - 3 y^{2} = 8 + 1 \Rightarrow {(x - 1)}^{2} - 3 y^{2} = 9 \Rightarrow \frac{(x - 1)^{2}}{9} - \frac{y^{2}}{3} = 1$
Thus, the centre is $(1, 0)$ .
Eccentricity of the hyperbola = $\frac{\sqrt{a^{2} + b^{2}}}{a} = \frac{\sqrt{9 + 3}}{3} = \frac{2 \sqrt{3}}{3}$
Foci = $(1 \pm 2 \sqrt{3}, 0)$
Equation of the directrices:
$x - 1 = \pm \frac{a}{e} \Rightarrow x = 1 \pm \frac{2 \sqrt{3}}{3}$

Page No 27.13:

Question 6:

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
(i) the distance between the foci = 16 and eccentricity = $\sqrt{2}$
(ii) conjugate axis is 5 and the distance between foci = 13
(iii) conjugate axis is 7 and passes through the point (3, −2).

Answer:

(i) The distance between the foci is $2 a e$ .

$∴ 2 a e = 16 \Rightarrow a e = 8$

$e = \sqrt{2}$
$∴ a \sqrt{2} = 8 \Rightarrow a = 4 \sqrt{2}$

$Also, b^{2} = a^{2} (e^{2} - 1) \Rightarrow b^{2} = 32 (2 - 1) \Rightarrow b^{2} = 32$
Therefore, the standard form of the hyperbola is given below:
$\frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 x^{2} - y^{2} = 32$

(ii) The distance between the foci is $2 a e$ .

$∴ 2 a e = 13 \Rightarrow a e = \frac{13}{2}$
Length of the conjugate axis, $2 b = 5$
$\Rightarrow b = \frac{5}{2}$

$Also, b^{2} = a^{2} (e^{2} - 1) \Rightarrow {(\frac{5}{2})}^{2} = {(\frac{13}{2})}^{2} - a^{2} \Rightarrow a^{2} = \frac{169 - 25}{4} \Rightarrow a^{2} = \frac{144}{4} = 36 \Rightarrow a = 6$
Therefore, the standard form of the hyperbola is $\frac{x^{2}}{36} - \frac{4 y^{2}}{25} = 1$ .
$or 25 x^{2} - 144 y^{2} = 900$

(iii) Length of the conjugate axis, $2 b = 7$
$\Rightarrow b = \frac{7}{2}$
Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .
It passes through $(3, - 2)$ .

$∴ \frac{3^{2}}{a^{2}} - \frac{(- 2)^{2}}{{(\frac{7}{2})}^{2}} = 1 \Rightarrow \frac{3^{2}}{a^{2}} - \frac{16}{49} = 1 \Rightarrow \frac{9}{a^{2}} = \frac{16}{49} + 1 \Rightarrow \frac{9}{a^{2}} = \frac{65}{49} \Rightarrow a^{2} = \frac{441}{65}$

Therefore, the standard form of the hyperbola is $\frac{65 x^{2}}{441} - \frac{4 y^{2}}{49} = 1$ .
$or 65 x^{2} - 36 y^{2} = 441$

Page No 27.14:

Question 7:

Find the equation of the hyperbola whose
(i) foci are (6, 4) and (−4, 4) and eccentricity is 2.
(ii) vertices are (−8, −1) and (16, −1) and focus is (17, −1)
(iii) foci are (4, 2) and (8, 2) and eccentricity is 2.
(iv) vertices are at (0 ± 7) and foci at $(0, \pm \frac{28}{3})$ .
(v) vertices are at (± 6, 0) and one of the directrices is x = 4. [NCERT EXEMPLAR]
(vi) foci at (± 2, 0) and eccentricity is 3/2. [NCERT EXEMPLAR]

Answer:

(i) The centre of the hyperbola is the midpoint of the line joining the two focii.
So, the coordinates of the centre are $(\frac{6 - 4}{2}, \frac{4 + 4}{2}), i . e . (1, 4)$ .
Let 2a and 2b be the length of the transverse and conjugate axis and let e be the eccentricity.

$\Rightarrow \frac{{(x - 1)}^{2}}{a^{2}} - \frac{{(y - 4)}^{2}}{b^{2}} = 1$
Distance between the two focii = 2ae

$2 a e = \sqrt{{(6 + 4)}^{2} + {(4 - 4)}^{2}} \Rightarrow 2 a e = 10 \Rightarrow a e = 5 \Rightarrow a = \frac{5}{2}$

$Also, b^{2} = {(a e)}^{2} - {(a)}^{2} \Rightarrow b^{2} = 25 - (\frac{25}{4}) \Rightarrow b^{2} = \frac{75}{4}$

Equation of the hyperbola is given below:
$\frac{4 {(x - 1)}^{2}}{25} - \frac{4 {(y - 4)}^{2}}{75} = 1 \Rightarrow \frac{4 (x^{2} - 2 x + 1)}{25} - \frac{(y^{2} - 8 y + 16)}{75} = 1 \Rightarrow \frac{(4 x^{2} - 8 x + 4)}{25} - \frac{(4 y^{2} - 32 y + 64)}{75} = 1 \Rightarrow 3 (4 x^{2} - 8 x + 4) - (4 y^{2} - 32 y + 64) = 75 \Rightarrow 12 x^{2} - 24 x + 12 - 4 y^{2} + 32 y - 64 = 75 \Rightarrow 12 x^{2} - 24 x - 4 y^{2} + 32 y - 127 = 0$

(ii) The centre of the hyperbola is given below:
$(\frac{- 8 + 16}{2}, \frac{- 1 - 1}{2}) = (4, - 1)$
If the other focus is $S' (m, n)$ , then it is calculated in the following way:
$4 = \frac{17 + m}{2} \Rightarrow m = - 9 A n d - 1 = \frac{- 1 + n}{2} \Rightarrow n = - 1$
Thus, the other focus is $(- 9, - 1)$ .

$Distance between the vertices : 2 a = \sqrt{{(16 + 8)}^{2} + {(- 1 + 1)}^{2}} \Rightarrow 2 a = 24 \Rightarrow a = 12$
Distance between the foci:
$2 c = \sqrt{{(17 + 9)}^{2} + {(- 1 + 1)}^{2}} \Rightarrow 2 c = 26 \Rightarrow c = 13$

$Also, c^{2} = a^{2} + b^{2} \Rightarrow b^{2} = 169 - 144 \Rightarrow b^{2} = 25$

Equation of the hyperbola is given below:
$\frac{{(x - 4)}^{2}}{144} - \frac{{(y + 1)}^{2}}{25} = 1 \Rightarrow \frac{x^{2} - 8 x + 16}{144} - \frac{(y^{2} + 2 y + 1)}{25} = 1 \Rightarrow 25 x^{2} - 200 x + 400 - (144 y^{2} + 288 y + 144) = 3600 \Rightarrow 25 x^{2} - 200 x - 144 y^{2} - 288 y - 3344 = 0$

(iii) The centre of the hyperbola is the midpoint of the line joining the two focii.
So, the coordinates of the centre are $(\frac{4 + 8}{2}, \frac{2 + 2}{2}), i . e . (6, 2)$ .
Let 2a and 2b be the length of the transverse and conjugate axis. Let e be the eccentricity.

$\Rightarrow \frac{{(x - 6)}^{2}}{a^{2}} - \frac{{(y - 2)}^{2}}{b^{2}} = 1$
Distance between the two focii = 2ae

$2 a e = \sqrt{{(4 - 8)}^{2} + {(2 - 2)}^{2}} \Rightarrow 2 a e = 4 \Rightarrow a e = 2 \Rightarrow a = 1$

$Also, b^{2} = {(a e)}^{2} - {(a)}^{2} \Rightarrow b^{2} = 4 - 1 \Rightarrow b^{2} = 3$

Equation of the hyperbola:
$\frac{{(x - 6)}^{2}}{1} - \frac{{(y - 2)}^{2}}{3} = 1 \Rightarrow \frac{(x^{2} - 12 x + 36)}{1} - \frac{(y^{2} - 4 y + 4)}{3} = 1 \Rightarrow 3 (x^{2} - 12 x + 36) - (y^{2} - 4 y + 4) = 3 \Rightarrow 3 x^{2} - 36 x + 108 - y^{2} + 4 y - 4 = 3 \Rightarrow 3 x^{2} - y^{2} - 36 x + 4 y + 101 = 0$

(iv) The Vertices of the hyperbola are $(0, \pm 7)$ .
∴ $b = 7$
The foci is $(0, \pm \frac{28}{3})$ .
∴ $b e = \frac{28}{3}$

$Also, a^{2} = b^{2} (e^{2} - 1) \Rightarrow a^{2} = {(\frac{28}{3})}^{2} - 49 \Rightarrow a^{2} = \frac{343}{9}$
Therefore, the equation of the hyperbola is $- \frac{9 x^{2}}{343} + \frac{y^{2}}{49} = 1$ .

(v) The Vertices of the hyperbola are $(\pm 6, 0)$ .
∴ $a = 6$
⇒ a² = 36
Now, x = 4
$\frac{a}{e} = 4 \Rightarrow e = \frac{3}{2} [∵ a = 6]$
Now,
${(a e)}^{2} = a^{2} + b^{2} \Rightarrow {(6 \times \frac{3}{2})}^{2} = 6^{2} + b^{2} \Rightarrow 81 - 36 = b^{2} \Rightarrow b^{2} = 45$

Therefore, the equation of the hyperbola is $\frac{x^{2}}{36} - \frac{y^{2}}{45} = 1$ .

(vi) The foci of the hyperbola are $(\pm 2, 0)$ .
∴ $a e = 2 \Rightarrow a = 2 \times \frac{2}{3} = \frac{4}{3} \Rightarrow a^{2} = \frac{16}{9}$
Now,
${(a e)}^{2} = a^{2} + b^{2} \Rightarrow {(2)}^{2} = {(\frac{4}{3})}^{2} + b^{2} \Rightarrow 4 - \frac{16}{9} = b^{2} \Rightarrow b^{2} = \frac{20}{9}$

Therefore, the equation of the hyperbola is given by
$\frac{9 x^{2}}{16} - \frac{9 y^{2}}{20} = 1 \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9}$

Page No 27.14:

Question 8:

Find the eccentricity of the hyperbola, the length of whose conjugate axis is $\frac{3}{4}$ of the length of transverse axis.

Answer:

The lengths of the conjugate axis and the transverse axis are $2 b$ and $2 a$ , respectively.
We have:
$\Rightarrow 2 b = \frac{3}{4} \times 2 a \Rightarrow b = \frac{3}{4} a$
Using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:

$\Rightarrow {(\frac{3}{4})}^{2} a^{2} = a^{2} (e^{2} - 1) \Rightarrow \frac{9}{16} = e^{2} - 1 \Rightarrow e^{2} = \frac{25}{16} \Rightarrow e = \frac{5}{4}$
Therefore, the eccentricity of the hyperbola is $\frac{5}{4}$ .

Page No 27.14:

Question 9:

Find the equation of the hyperboala whose
(i) focus is at (5, 2), vertex at (4, 2) and centre at (3, 2)
(ii) focus is at (4, 2), centre at (6, 2) and e = 2.

Answer:

(i) The equation of the hyperbola with centre (x₀,y₀) is given by
$\frac{{(x - x_{0})}^{2}}{a^{2}} - \frac{{(y - y_{0})}^{2}}{b^{2}} = 1 Focus = (a e + x_{0}, y_{0})$

Vertex = (a+x₀, y₀)

$∴ a e = 2 and a = 1 b^{2} = {(2)}^{2} - a^{2} \Rightarrow b^{2} = (2^{2}) - {(1)}^{2} \Rightarrow b^{2} = 3$

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

(ii) The equation of the hyperbola with centre (x₀,y₀) is given by
$\frac{{(x - x_{0})}^{2}}{a^{2}} - \frac{{(y - y_{0})}^{2}}{b^{2}} = 1 Focus = (a e + x_{0}, y_{0})$

$∴ a e = - 2 \Rightarrow a = - 1 b^{2} = {(2)}^{2} - a^{2} \Rightarrow b^{2} = {(- 2)}^{2} - {(- 1)}^{2} \Rightarrow b^{2} = 3$

$\Rightarrow \frac{{(x - 6)}^{2}}{1} - \frac{{(y - 2)}^{2}}{3} = 1 \Rightarrow 3 {(x - 6)}^{2} - {(y - 2)}^{2} = 3$

Page No 27.14:

Question 10:

If P is any point on the hyperbola whose axis are equal, prove that SP. S'P = CP².

Answer:

Equation of the hyperbola:
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

If the axes of the hyperbola are equal, then $a = b$ .
Then, equation of the hyperbola becomes $x^{2} - y^{2} = a^{2}$ .

$∴ b^{2} = a^{2} (e^{2} - 1) \Rightarrow a^{2} = a^{2} (e^{2} - 1) \Rightarrow 1 = (e^{2} - 1) \Rightarrow e^{2} = 2 \Rightarrow e = \sqrt{2}$
Thus, the centre $C (0, 0)$ and the focus are given by $S (\sqrt{2} a, 0)$ and $S' (- \sqrt{2} a, 0)$ , respectively.
Let $P (α, β)$ be any point on the parabola.
So, it will satisfy the equation.
$α^{2} - β^{2} = a^{2}$

$∴ S P^{2} = {(\sqrt{2} a - α)}^{2} + β^{2} = 2 a^{2} + α^{2} - 2 \sqrt{2} a α + β^{2}$

$S' P^{2} = {(- \sqrt{2} a - α)}^{2} + β^{2} = 2 a^{2} + α^{2} + 2 \sqrt{2} a α + β^{2}$

$N o w, S P^{2} . S' P^{2} = (2 a^{2} + α^{2} - 2 \sqrt{2} a α + β^{2}) (2 a^{2} + α^{2} + 2 \sqrt{2} a α + β^{2}) = 4 a^{4} + 4 a^{2} (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} - 8 a^{2} α^{2} = 4 a^{2} (a^{2} - 2 α^{2}) + 4 a^{2} (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} = 4 a^{2} (α^{2} - β^{2} - 2 α^{2}) + 4 a^{2} (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} = - 4 a^{2} (α^{2} + β^{2}) + 4 a^{2} (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} = {(α^{2} + β^{2})}^{2} = C P^{4}$
∴ $S P . S' P = C P^{2}$

Page No 27.14:

Question 11:

In each of the following find the equations of the hyperbola satisfying the given conditions:
(i) vertices (± 2, 0), foci (± 3, 0)
(ii) vertices (0, ± 5), foci (0, ± 8)
(iii) vertices (0, ± 3), foci (0, ± 5)
(iv) foci (± 5, 0), transverse axis = 8
(v) foci (0, ± 13), conjugate axis = 24
(vi) foci (± $3 \sqrt{5}$ , 0), the latus-rectum = 8
(vii) foci (± 4, 0), the latus-rectum = 12
(viii) vertices (± 7, 0), $e = \frac{4}{3}$
(ix) foci (0, ± $\sqrt{10}$ ), passing through (2, 3)
(x) foci (0, ± 12), latus-rectum = 36

Answer:

(i) The vertices of the hyperbola are $(\pm 2, 0)$ and the foci are $(\pm 3, 0)$ .
Thus, the value of $a = 2$ and $a e = 3$ .
Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow b^{2} = 9 - 4 \Rightarrow b^{2} = 5$
Thus, the equation of the hyperbola is $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$ .

(ii) The vertices of the hyperbola are $(0, \pm 5)$ and the foci are $(0, \pm 8)$
Thus, the value of $a = 5$ and $a e = 8$ .
Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow b^{2} = 64 - 25 \Rightarrow b^{2} = 39$
Thus, the equation of the hyperbola is $- \frac{x^{2}}{39} + \frac{y^{2}}{25} = 1$ .

(iii) The vertices of the hyperbola are $(0, \pm 3)$ and the foci are $(0, \pm 5)$ .
Thus, the value of $a = 3$ and $a e = 5$ .
Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow b^{2} = 25 - 9 \Rightarrow b^{2} = 16$
Thus, the equation of the hyperbola is   $- \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ .

(iv) The foci of the hyperbola are $(\pm 5, 0)$ and the transverse axis is 8.
Thus, the value of   $a e = 5$ and 2a = 8.
$\Rightarrow a = 4$
Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow b^{2} = 25 - 16 \Rightarrow b^{2} = 9$
Thus, the equation of the hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ .

(v) The foci of the hyperbola are $(0, \pm 13)$ and the conjugate axis is 24.
Thus, the value of   $a e = 13$ and 2b = 24.
⇒ b = 12

Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow a^{2} = 169 - 144 \Rightarrow a^{2} = 25$
Thus, the equation of the hyperbola is $- \frac{x^{2}}{144} + \frac{y^{2}}{25} = 1$ .

(vi) The foci of the hyperbola are $(\pm 3 \sqrt{5}, 0)$ and the latus rectum is 8.
Thus, the value of   $a e = 3 \sqrt{5}$
$and \frac{2 b^{2}}{a} = 8 \Rightarrow b^{2} = 4 a$
Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow 4 a = 45 - a^{2} \Rightarrow a^{2} + 4 a - 45 = 0 \Rightarrow (a - 5) (a + 9) = 0 \Rightarrow a = - 9, 5$
$b^{2} = - 36 or 20$
Since negative value is not possible, it is equal to 20.
Thus, the equation of the hyperbola is $\frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$ .

(vii) The foci of the hyperbola are $(\pm 4, 0)$ and the latus rectum is 12.
Thus, the value of   $a e = 4$
$and \frac{2 b^{2}}{a} = 12 \Rightarrow b^{2} = 6 a$
Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow 6 a = 16 - a^{2} \Rightarrow a^{2} + 6 a - 16 = 0 \Rightarrow (a - 2) (a + 8) = 0 \Rightarrow a = 2, or - 8$
$b^{2} = 12 or - 48$
Since negative value is not possible, its value is 12.
Thus, the equation of the hyperbola is $\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$ .

(viii) The vertices of hyperbola are $(\pm 7, 0)$ and eccentricity is $\frac{4}{3}$
Thus, the value of $a = 7$ .
Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow b^{2} = 49 (\frac{16}{9} - 1) \Rightarrow b^{2} = 49 \times \frac{7}{9} = \frac{343}{9}$
Thus, the equation of the hyperbola is $\frac{x^{2}}{49} - \frac{9 y^{2}}{343} = 1$ .

(ix) The foci of hyperbola are $(0, \pm \sqrt{10})$ that pass through $(2, 3)$ .

$Thus, the value of a e = \sqrt{10} . By squaring both the sides, we get : {(a e)}^{2} = 10 \Rightarrow a^{2} + b^{2} = 10 \Rightarrow b^{2} = 10 - a^{2}$
Let the equation of the hyperbola be $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$ .
It passes through $(2, 3)$ .
$\Rightarrow \frac{3^{2}}{a^{2}} - \frac{2^{2}}{10 - a^{2}} = 1 \Rightarrow 90 - 9 a^{2} - 4 a^{2} = 10 a^{2} - a^{4} \Rightarrow a^{4} - 23 a^{2} + 90 = 0 \Rightarrow (a^{2} - 18) (a^{2} - 5) = 0 \Rightarrow a^{2} = 18, 5$
Now, $b^{2} = - 8 or 5$
If we neglect the negative value, then b²= 5.
Thus, the equation of the hyperbola is $\frac{y^{2}}{5} - \frac{x^{2}}{5} = 1$ .

(x) The foci of the hyperbola are $(0, \pm 12)$ and the latus rectum is 36.
Thus, the value of   $a e = 12$ .
$and \frac{2 b^{2}}{a} = 36 \Rightarrow b^{2} = 18 a$

Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$\Rightarrow 18 a = 144 - a^{2} \Rightarrow a^{2} + 18 a - 144 = 0 \Rightarrow (a + 24) (a - 6) = 0 \Rightarrow a = - 24, or 6$
$b^{2} = - 432 or 108 (but negative value is not possible)$

Thus, the equation of the hyperbola is $\frac{y^{2}}{36} - \frac{x^{2}}{108} = 1$ .

Page No 27.14:

Question 12:

If the distance between the foci of a hyperbola is 16 and its ecentricity is $\sqrt{2}$ , then obtain its equation.

Answer:

We have
$2 a e = 16 \Rightarrow a e = 8 \Rightarrow a = \frac{8}{\sqrt{2}} = 4 \sqrt{2} \Rightarrow a^{2} = 32$
Now,
${(a e)}^{2} = a^{2} + b^{2} \Rightarrow {(8)}^{2} = 32 + b^{2} \Rightarrow 64 - 32 = b^{2} \Rightarrow b^{2} = 32$

Therefore, the equation of the hyperbola is given by
$\frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 \Rightarrow x^{2} - y^{2} = 32$

Page No 27.14:

Question 13:

Show that the set of all points such that the difference of their distances from (4, 0) and (− 4,0) is always equal to 2 represents a hyperbola.

Answer:

Let the point be P(x, y)
$∴ [\sqrt{{(x - 4)}^{2} + {(y - 0)}^{2}}] - [\sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}}] = 2 \Rightarrow {[\sqrt{{(x - 4)}^{2} + {(y - 0)}^{2}}]}^{2} = {[2 + \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}}]}^{2} \Rightarrow {(x - 4)}^{2} + y^{2} = 4 + {(x + 4)}^{2} + y^{2} + 4 \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}} \Rightarrow {(x - 4)}^{2} - {(x + 4)}^{2} = 4 + 4 \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}}$
$\Rightarrow - 16 x = 4 + 4 \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}} \Rightarrow - 16 x - 4 = 4 \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}} \Rightarrow - 4 (4 x + 1) = 4 \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}} \Rightarrow - (4 x + 1) = \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}} \Rightarrow 16 x^{2} + 8 x + 1 = x^{2} + 8 x + 16 + y^{2} \Rightarrow 15 x^{2} - y^{2} = 15 \Rightarrow \frac{x^{2}}{1} - \frac{y^{2}}{15} = 1$
Which is the equation of a hyperbola.

Page No 27.18:

Question 1:

Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
(a) 16x² − 9y² = 144
(b) 9x² − 16y² = 144
(c) 25x² − 9y² = 225
(d) 9x² − 25y² = 81

Answer:

(a) 16x² − 9y² = 144

The vertices of the hyperbola are $(\pm 3, 0)$ and foci are $(\pm 5, 0)$ .
Thus, the values of a and ae are 3 and 5, respectively.

Now, using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:
$b^{2} = 25 - 9 \Rightarrow b^{2} = 16$
Equation of the hyperbola is given below:
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 \Rightarrow 16 x^{2} - 9 y^{2} = 144$

Page No 27.18:

Question 2:

If e₁ and e₂ are respectively the eccentricities of the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ , then the relation between e₁ and e₂ is
(a) 3 e₁² + e₂² = 2
(b) e₁² + 2 e₂² = 3
(c) 2 e₁² + e₂² = 3
(d) e₁² + 3 e₂² = 2

Answer:

(c) 2 e₁² + e₂² = 3
The standard form of the ellipse is $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ , where $a^{2} = 18 and b^{2} = 4$ .
So, the eccentricity is calculated in the following way:
$b^{2} = a^{2} (1 - {e_{1}}^{2}) \Rightarrow 4 = 18 (1 - {e_{1}}^{2}) \Rightarrow \frac{2}{9} = 1 - {e_{1}}^{2} \Rightarrow {e_{1}}^{2} = \frac{7}{9}$
The standard form of the hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ , where $a^{2} = 9 and b^{2} = 4$ .
So, the eccentricity is calculated in the following way:
$b^{2} = a^{2} ({e_{2}}^{2} - 1) \Rightarrow 4 = 9 ({e_{2}}^{2} - 1) \Rightarrow \frac{4}{9} = {e_{2}}^{2} - 1 \Rightarrow {e_{2}}^{2} = \frac{13}{9}$

$∴ 2 {e_{1}}^{2} + {e_{2}}^{2} = \frac{2 \times 7}{9} + \frac{13}{9} = \frac{27}{9} = 3$

Page No 27.18:

Question 3:

The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is
(a) $8 \sqrt{2}$
(b) $16 \sqrt{2}$
(c) $4 \sqrt{2}$
(d) $6 \sqrt{2}$

Answer:

(a) $8 \sqrt{2}$

We have:
$x = 8 \sec θ, y = 8 \tan θ$

$On squaring and subtracting : x^{2} - y^{2} = 8 \sec^{2} θ - 8 \tan^{2} θ \Rightarrow x^{2} - y^{2} = 8 \Rightarrow \frac{x^{2}}{8} - \frac{y^{2}}{8} = 1$

∴ a = b = 8
Distance between the directrices of the hyperbola = $\frac{2 a^{2}}{\sqrt{a^{2} + b^{2}}}$
$Distance between the directrices = \frac{2 \times 64}{\sqrt{64 + 64}} = \frac{128}{8 \sqrt{2}} = \frac{16}{\sqrt{2}} = 8 \sqrt{2}$

Page No 27.18:

Question 4:

The equation of the conic with focus at (1, −1) directrix along x − y + 1 = 0 and eccentricity $\sqrt{2}$ is
(a) xy = 1
(b) 2xy + 4x − 4y − 1= 0
(c) x² − y² = 1
(d) 2xy − 4x + 4y + 1 = 0

Answer:

(d) 2xy − 4x + 4y + 1 = 0

Let $P (x, y)$ be any point on the hyperbola.
Then, the distance of any point from the focus is eccentricity times the distance from the directrix.

$∴ \sqrt{{(x - 1)}^{2} + {(y + 1)}^{2}} = \sqrt{2} |\frac{x - y + 1}{\sqrt{2}}|$

Squaring both the sides, we get:
$(x - 1)^{2} + (y + 1)^{2} = {(x - y + 1)}^{2} x^{2} - 2 x + 1 + y^{2} + 1 + 2 y = x^{2} + y^{2} + 1 - 2 x y - 2 y + 2 x 2 x y - 4 x + 4 y + 1 = 0$

Page No 27.18:

Question 5:

The eccentricity of the conic 9x² − 16y² = 144 is
(a) $\frac{5}{4}$

(b) $\frac{4}{3}$

(c) $\frac{4}{5}$

(d) $\sqrt{7}$

Answer:

(a) $\frac{5}{4}$

Standard form of a hyperbola = $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
Here, $a^{2} = 16 and y^{2} = 9$

The eccentricity is calculated in the following way:
$b^{2} = a^{2} (e^{2} - 1) \Rightarrow 9 = 16 (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{9}{16} \Rightarrow e^{2} = \frac{25}{16} \Rightarrow e = \frac{5}{4}$

Page No 27.18:

Question 6:

A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PA − PB = k (k ≠ 0), then the locus of P is
(a) a hyperbola
(b) a branch of the hyperbola
(c) a parabola
(d) an ellipse

Answer:

(a) a hyperbola
$Let P (x, y) be any point on the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . B y definition, we have : P A = e (x - \frac{a}{e}) = e x - a and P B = e (x + \frac{a}{e}) = e x + a ∴ P B - P A = (e x + a) - (e x - a) = 2 a = k$

Page No 27.18:

Question 7:

The eccentricity of the hyperbola whose latus-rectum is half of its transverse axis, is
(a) $\frac{1}{\sqrt{2}}$

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

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

(d) none of these.

Answer:

(c) $\sqrt{\frac{3}{2}}$
The lengths of the latus rectum and the transverse axis are $\frac{2 b^{2}}{a}$ and $2 a$ , respectively.
According to the given statement, length of the latus rectum is half of its transverse axis.
$∴ \frac{2 b^{2}}{a} = \frac{1}{2} \times 2 a \Rightarrow \frac{2 b^{2}}{a} = a \Rightarrow 2 b^{2} = a^{2}$

Eccentricity, $e = \frac{\sqrt{a^{2} + b^{2}}}{a}$
Substituting the value $b^{2} = \frac{a^{2}}{2}$ , we get:
$e = \frac{\sqrt{a^{2} + \frac{a^{2}}{2}}}{a} = \frac{a \sqrt{\frac{3}{2}}}{a} = \sqrt{\frac{3}{2}}$
∴ Eccentricity is $\sqrt{\frac{3}{2}}$

Page No 27.18:

Question 8:

The eccentricity of the hyperbola x² − 4y² = 1 is

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

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

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

(d) $\frac{2}{\sqrt{5}}$

Answer:

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

The equation of the hyperbola is $x^{2} - 4 y^{2} = 1$ .
This can be rewritten in the following way:
$\frac{x^{2}}{1} - \frac{y^{2}}{\frac{1}{4}} = 1$
This is the standard form of a hyperbola, where $a^{2} = 1 and b^{2} = \frac{1}{4}$ .
The value of eccentricity is calculated in the following way:
$b^{2} = a^{2} (e^{2} - 1) \Rightarrow \frac{1}{4} = (e^{2} - 1) \Rightarrow e^{2} = \frac{5}{4} \Rightarrow e = \frac{\sqrt{5}}{2}$

Page No 27.18:

Question 9:

The difference of the focal distances of any point on the hyperbola is equal to
(a) length of the conjugate axis
(b) eccentricity
(c) length of the transverse axis
(d) Latus-rectum

Answer:

(c) length of the transverse axis

Let $P (x, y)$ be any point on the hyperbola, and $S, S'$ be the focus with coordinates $(\pm a e, 0)$ .

⇒ $S' P - S P = 2 a$
Thus, the difference of the focal distances of any point on the hyperbola is equal to the length of the transverse axis.

Page No 27.19:

Question 10:

The foci of the hyperbola 9x² − 16y² = 144 are
(a) (± 4, 0)
(b) (0, ± 4)
(c) (± 5, 0)
(d) (0, ± 5)

Answer:

(c) (± 5, 0)

The equation of the hyperbola is given below:
$9 x^{2} - 16 y^{2} = 144$
This equation can be rewritten in the following way:
$\frac{9 x^{2}}{144} - \frac{16 y^{2}}{144} = 1 \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
This is the standard equation of a hyperbola, where $a^{2} = 16 and b^{2} = 9$ .

The eccentricity is calculated in the following way:
$b^{2} = a^{2} (e^{2} - 1) \Rightarrow 9 = 16 (e^{2} - 1) \Rightarrow \frac{9}{16} = e^{2} - 1 \Rightarrow e = \frac{5}{4}$
Foci = $(\pm a e, 0) = (\pm 5, 0)$

Page No 27.19:

Question 11:

The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$ , then equation of the hyperbola is
(a) x² + y² = 32
(b) x² − y² = 16
(c) x² + y² = 16
(d) x² − y²= 32

Answer:

(d) x² − y²= 32

The distance between the foci is $2 a e$ .

$∴ 2 a e = 16 \Rightarrow a e = 8$

$e = \sqrt{2}$
$∴ a \sqrt{2} = 8 \Rightarrow a = 4 \sqrt{2}$

$Also, b^{2} = a^{2} (e^{2} - 1) \Rightarrow b^{2} = 32 (2 - 1) \Rightarrow b^{2} = 32$
Standard form of the hyperbola is given below:
$\frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 x^{2} - y^{2} = 32$

Page No 27.19:

Question 12:

If e₁ is the eccentricity of the conic 9x² + 4y² = 36 and e₂ is the eccentricity of the conic 9x² − 4y² = 36, then
(a) e₁² − e₂² = 2
(b) 2 < e₂² − e₁² < 3
(c) e₂² − e₁² = 2
(d) e₂² − e₁² > 3

Answer:

(b) 2 < e₂² − e₁² < 3
The conic $9 x^{2} + 4 y^{2} = 36$ can rewritten in the following way:
$\frac{9 x^{2}}{36} + \frac{4 y^{2}}{36} = 1 \Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
This is the standard equation of an ellipse.
$∴ b^{2} = a^{2} {(1 - e_{1})}^{2} \Rightarrow 9 = 4 {(1 - e_{1})}^{2} \Rightarrow {(e_{1})}^{2} = \frac{- 5}{4}$
The conic $9 x^{2} - 4 y^{2} = 36$ can rewritten in the following way:
$\frac{9 x^{2}}{36} - \frac{4 y^{2}}{36} = 1 \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$
This is the standard equation of a hyperbola.
$∴ b^{2} = a^{2} ({e_{2}}^{2} - 1) \Rightarrow 9 = 4 ({e_{2}}^{2} - 1) \Rightarrow {(e_{2})}^{2} = \frac{13}{4}$

$∴ {e_{2}}^{2} - {e_{1}}^{2} = \frac{13}{4} + \frac{5}{4} = 2.5$

Page No 27.19:

Question 13:

If the eccentricity of the hyperbola x² − y² sec²α = 5 is $\sqrt{3}$ times the eccentricity of the ellipse x² sec² α + y² = 25, then α =

(a) $\frac{π}{6}$

(b) $\frac{π}{4}$

(c) $\frac{π}{3}$

(d) $\frac{π}{2}$

Answer:

(b) $\frac{π}{4}$
The hyperbola $x^{2} - y^{2} s e c^{2} α = 5$ can be rewritten in the following way:
$\frac{x^{2}}{5} - \frac{y^{2}}{5 \cos^{2} α} = 1$

This is the standard form of a hyperbola, where $a^{2} = 5 and b^{2} = 5 \cos^{2} α$ .

$\Rightarrow b^{2} = a^{2} ({e_{1}}^{2} - 1) \Rightarrow 5 \cos^{2} α = 5 ({e_{1}}^{2} - 1) \Rightarrow {e_{1}}^{2} = \cos^{2} α + 1 . . . . . (1)$

The ellipse $x^{2} s e c^{2} α + y^{2} = 25$ can be rewritten in the following way:
$\frac{x^{2}}{25 \cos^{2} α} + \frac{y^{2}}{25} = 1$

This is the standard form of an ellipse, where $a^{2} = 25 and b^{2} = 25 \cos^{2} α$ .

$b^{2} = a^{2} (1 - {e_{2}}^{2}) \Rightarrow {e_{2}}^{2} = 1 - \cos^{2} α \Rightarrow {e_{2}}^{2} = \sin^{2} α . . . . . . (2)$

According to the question,

$\cos^{2} α + 1 = 3 (\sin^{2} α) \Rightarrow 2 = 4 \sin^{2} α \Rightarrow \sin α = \frac{1}{\sqrt{2}} \Rightarrow α = \frac{π}{4}$

Page No 27.19:

Question 14:

The equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity 2, is

(a) $\frac{(x - 1)^{2}}{25 / 4} - \frac{(y - 4)^{2}}{75 / 4} = 1$

(b) $\frac{(x + 1)^{2}}{25 / 4} - \frac{(y + 4)^{2}}{75 / 4} = 1$

(c) $\frac{(x - 1)^{2}}{75 / 4} - \frac{(y - 4)^{2}}{25 / 4} = 1$

(d) none of these

Answer:

(a) $\frac{(x - 1)^{2}}{25 / 4} - \frac{(y - 4)^{2}}{75 / 4} = 1$

The centre of the hyperbola is the midpoint of the line joining the two foci.
So, the coordinates of the centre are $(\frac{6 - 4}{2}, \frac{4 + 4}{2}), i . e . (1, 4) .$
Let 2a and 2b be the length of the transverse and the conjugate axes, respectively. Also, let e be the eccentricity.

$\Rightarrow \frac{{(x - 1)}^{2}}{a^{2}} - \frac{{(y - 4)}^{2}}{b^{2}} = 1$

Now, distance between the two foci = 2ae

$2 a e = \sqrt{{(6 + 4)}^{2} + {(4 - 4)}^{2}} \Rightarrow 2 a e = 10 \Rightarrow a e = 5 \Rightarrow a = \frac{5}{2}$

$Also, b^{2} = {(a e)}^{2} - {(a)}^{2} \Rightarrow b^{2} = 25 - (\frac{25}{4}) \Rightarrow b^{2} = \frac{75}{4}$

Equation of the hyperbola is given below:
$\frac{{(x - 1)}^{2}}{25 / 4} - \frac{{(y - 4)}^{2}}{75 / 4} = 1$

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Question 15:

The length of the straight line x − 3y = 1 intercepted by the hyperbola x² − 4y² = 1 is

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

(b) $3 \sqrt{\frac{2}{5}}$

(c) $6 \sqrt{\frac{2}{5}}$

(d) none of these

Answer:

(c) $6 \sqrt{\frac{2}{5}}$

The point of intersection of $x - 3 y = 1$ and the hyperbola $x^{2} - 4 y^{2} = 1$ is calculated in the following way:
${(1 + 3 y)}^{2} - 4 y^{2} = 1 \Rightarrow 1 + 6 y + 9 y^{2} - 4 y^{2} = 1 \Rightarrow 5 y^{2} + 6 y = 0 \Rightarrow y = 0 or y = - \frac{6}{5}$
If $y = 0$ , then $x = 1$ .
If $y = - \frac{6}{5}$ , then $x = 1 + 3 \times (- \frac{6}{5}) = - \frac{13}{5}$ .

So, the points are $(1, 0)$ and $(- \frac{13}{5}, - \frac{6}{5})$ .

∴ Length = $\sqrt{{(1 + \frac{13}{5})}^{2} + {(0 + \frac{6}{5})}^{2}} = 6 \sqrt{\frac{2}{5}}$

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Question 16:

The latus-rectum of the hyperbola 16x² − 9y² = 144 is
(a) 16/3
(b) 32/3
(c) 8/3
(d) 4/3

Answer:

(b) 32/3

The standard form of the hyperbola $16 x^{2} - 9 y^{2} = 144$ is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ .
Here, $a^{2} = 9, b^{2} = 16$
Latus rectum of the hyperbola = $\frac{2 b^{2}}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$

Page No 27.19:

Question 17:

The foci of the hyperbola 2x² − 3y² = 5 are
(a) $(\pm 5 / \sqrt{6}, 0)$
(b) (± 5/6, 0)
(c) $(\pm \sqrt{5} / 6, 0)$
(d) none of these

Answer:

(a) $(\pm 5 / \sqrt{6}, 0)$

The given equation of hyperbola is $2 x^{2} - 3 y^{2} = 5$ . It can be rewritten in the following way:
$\frac{2 x^{2}}{5} - \frac{3 y^{2}}{5} = 1 \Rightarrow \frac{x^{2}}{\frac{5}{2}} - \frac{y^{2}}{\frac{5}{3}} = 1$

This is the standard equation of a parabola, where $a^{2} = \frac{5}{2} and b^{2} = \frac{5}{3}$ .
The eccentricity can be calculated in the following way:
$b^{2} = a^{2} (e^{2} - 1) \Rightarrow \frac{5}{3} = \frac{5}{2} (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{2}{3} \Rightarrow e^{2} = \frac{5}{3} \Rightarrow e = \sqrt{\frac{5}{3}}$
Coordinates of the foci = $(\pm a e, 0) = (\pm \frac{5}{\sqrt{6}}, 0)$

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Question 18:

The eccentricity the hyperbola $x = \frac{a}{2} (t + \frac{1}{t}), y = \frac{a}{2} (t - \frac{1}{t})$ is
(a) $\sqrt{2}$
(b) $\sqrt{3}$
(c) $2 \sqrt{3}$
(d) $3 \sqrt{2}$

Answer:

(a) $\sqrt{2}$
We eliminate t in $x = \frac{a}{2} (t + \frac{1}{t}), y = \frac{a}{2} (t - \frac{1}{t})$ .
On squaring both the sides in the equations $x = \frac{a}{2} (t + \frac{1}{t}), y = \frac{a}{2} (t - \frac{1}{t})$ , we get:
$\frac{4 x^{2}}{a^{2}} = t^{2} + \frac{1}{t^{2}} + 2 \Rightarrow \frac{4 x^{2}}{a^{2}} - 2 = t^{2} + \frac{1}{t^{2}} . . . (1)$

$Also, \frac{4 y^{2}}{a^{2}} = t^{2} + \frac{1}{t^{2}} - 2 \Rightarrow \frac{4 y^{2}}{a^{2}} + 2 = t^{2} + \frac{1}{t^{2}} . . . (2)$

From (1) and (2), we get:
$\frac{4 x^{2}}{a^{2}} - \frac{4 y^{2}}{a^{2}} = 4 \Rightarrow \frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}} = 1$

This is the standard equation of a hyperbola, where $a^{2} = b^{2}$ .

Eccentricity of the hyperbola, $e = \frac{\sqrt{a^{2} + a^{2}}}{a} = \sqrt{2}$

Page No 27.19:

Question 19:

The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is
(a) 3 (x − 6)² − (y −2)² = 3
(b) (x − 6)² − 3 (y − 2)² = 1
(c) (x − 6)² − 2 (y −2)² = 1
(d) 2 (x − 6)² − (y − 2)² = 1

Answer:

(a) 3 (x − 6)² − (y −2)² = 3

The equation of the hyperbola with centre (x₀,y₀) is given by
$\frac{{(x - x_{0})}^{2}}{a^{2}} - \frac{{(y - y_{0})}^{2}}{b^{2}} = 1 Focus = (a e + x_{0}, y_{0})$

$∴ a e = - 2 \Rightarrow a = - 1 b^{2} = {(2)}^{2} - a^{2} \Rightarrow b^{2} = {(- 2)}^{2} - {(- 1)}^{2} \Rightarrow b^{2} = 3$

$∴ \frac{{(x - 6)}^{2}}{1} - \frac{{(y - 2)}^{2}}{3} = 1 \Rightarrow 3 {(x - 6)}^{2} - {(y - 2)}^{2} = 3$

Page No 27.19:

Question 20:

The locus of the point of intersection of the lines $\sqrt{3} x - y - 4 \sqrt{3} λ = 0 and \sqrt{3} λ x + λ y - 4 \sqrt{3} = 0$ is a hyperbola of eccentricity
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(b) 2

The equations of lines $\sqrt{3} x - y - 4 \sqrt{3} λ = 0 and \sqrt{3} λ x + λ y - 4 \sqrt{3} = 0$ can be rewritten as $\sqrt{3} x - y = 4 \sqrt{3} λ and \sqrt{3} λ x + λ y = 4 \sqrt{3}$ , respectively.

Multiplying the equations:

$3 λ x^{2} - λ y^{2} = 48 λ \Rightarrow \frac{3 λ x^{2}}{48 λ} - \frac{λ y^{2}}{48 λ} = 1 \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{48} = 1$
This is the standard equation of a hyperbola, where $a^{2} = 16 and b^{2} = 48$ .

$Eccentricity, e = \sqrt{\frac{a^{2} + b^{2}}{a^{2}}} \Rightarrow e = \sqrt{\frac{16 + 48}{16}} \Rightarrow e = \frac{8}{4} \Rightarrow e = 2$

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Question 21:

Equation of the hyperbola with eccentricity $\frac{3}{2}$ and foci at (±2, 0) is

(a) $\frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9}$

(b) $\frac{x^{2}}{9} - \frac{y^{2}}{9} = \frac{4}{9}$

(c) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$

(d) none of these

Answer:

$For an hyperbola, eccentricity is given as \frac{3}{2} and foci at (\pm 2, 0) i . e . hyperbola has equation of the form \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 i . e . \pm a e = \pm 2 i . e . a e = 2 i . e . a \times \frac{3}{2} = 2 i . e . a = \frac{4}{3}, i . e . a^{2} = \frac{16}{9} ∴ Since b^{2} = a^{2} (e^{2} - 1) \Rightarrow b^{2} = \frac{16}{9} (\frac{9}{4} - 1) = \frac{16}{9} \times (\frac{5}{4}) b^{2} = \frac{20}{9} Hence, equation of hyperbola is \frac{x^{2}}{\frac{16}{9}} - \frac{y^{2}}{\frac{20}{9}} = 1 i . e . \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9}; Hence, the correct answer is option A .$

Page No 27.20:

Question 22:

The distance between the foci of a hyperbola is 16 and eccentricity is $\sqrt{2} .$ Its equation is

(a) x² – y² = 32

(b) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$

(c) 2x² – 3y² = 7

(d) none of these

Answer:

$For an hyperbola, eccentricity is \sqrt{2} (given) Distance between foci is 16 i . e . 2 a e = 16 i . e . 2 \times a \times \sqrt{2} = 16 i . e . a = \frac{8}{\sqrt{2}} \times \frac{\sqrt{2}}{2} = 4 \sqrt{2} i . e . a^{2} = 32 ∴ b^{2} = 32 (2 - 1) b^{2} = 32 ∴ equation of hyperbola is \frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 i . e . x^{2} - y^{2} = 32$
Hence, the correct answer option is A.

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Question 23:

The eccentrcity of the hyperbola whose latusrectum is 8 and conjugate axis is equal to half of the distance between the foci is

(a) $\frac{4}{3}$

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

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

(d) none of these

Answer:

$For an hyperbola, Latus rectum is 8 and conjugate axis = hlaf of the distance between foci i . e . 2 b = \frac{1}{2} (2 a e) i . e . 2 b = a e i . e . e = \frac{2 b}{a} i . e . \frac{b}{a} = \frac{e}{2} Since \frac{2 b^{2}}{a} = 8 (i . e . length of latus rectum) i . e . \frac{b^{2}}{a} = 4 i . e . b^{2} = 4 a ∴ e = \sqrt{\frac{a^{2} + b^{2}}{a^{2}}} i . e . e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{e^{2}}{4} i . e . e^{2} - \frac{e^{2}}{4} = 1 i . e . \frac{3 e^{2}}{4} = 1 i . e . e = \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} Hence, the correct answer is option C .$

Page No 27.20:

Question 1:

The equation of the hyperbola with vertices at (0, ± 6) and eccentricity $\frac{5}{3}$ is __________________ and its foci are __________________.

Answer:

$For the hyperbola, Vertices is given at (0, \pm 6) and eccentricity is \frac{5}{3} . Since vertices lie on y ‐ a x i s \Rightarrow equation of hyperbola is of the form, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1 i . e . b = 6 Since e = \frac{5}{3} \Rightarrow a^{2} = b^{2} (e^{2} - 1) i . e . a^{2} = 36 (\frac{25}{9} - 1) = 36 (\frac{16}{9}) a^{2} = 4 \times 16 i . e . equation of hyperbola is \frac{x^{2}}{64} - \frac{y^{2}}{36} = - 1 i . e . \frac{y^{2}}{36} - \frac{x^{2}}{64} = 1 and foci is (0, \pm b e) = (0, \frac{\pm 5}{3} \times 6) = (0, \pm 10)$

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Question 2:

The locus of the point of intersection of the lines $\sqrt{3} x - y - 4 \sqrt{3} k = 0 and \sqrt{3} k x + k y - 4 \sqrt{3} = 0$ for different values of k is the hyperbola__________________________.

Answer:

$Since given lines are \sqrt{3} x - y - 4 \sqrt{3 k} = 0 i . e . k = \frac{\sqrt{3} x - y}{4 \sqrt{3}} and \sqrt{3} k x + k y - 4 \sqrt{3} = 0 i . e . k = \frac{4 \sqrt{3}}{\sqrt{3} x + y} equating, the value of k, we get, \frac{\sqrt{3} x - y}{4 \sqrt{3}} = \frac{4 \sqrt{3}}{\sqrt{3} x + y} i . e . (\sqrt{3} x - y) (\sqrt{3} x + y) = 16 \times 3 i . e . 3 x^{2} - y^{2} = 48 i . e . locus of point of intersection of the above line is the hyperbola given by, 3 x^{2} - y^{2} = 48$

Page No 27.20:

Question 3:

The eccentricity of the hyperbola passing through the points (3, 0) and ( $3 \sqrt{2}$ , 2) is _______________________.

Answer:

$Suppose the equation of hyperbola is \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 which passes through (3, 0) and (3 \sqrt{2}, 2) i . e . \frac{{(3)}^{2}}{a^{2}} - \frac{{(0)}^{2}}{b^{2}} = 1 i . e . a^{2} = 9 and \frac{{(3 \sqrt{2})}^{2}}{a^{2}} - \frac{{(2)}^{2}}{b^{2}} = 1 i . e . \frac{9 \times 2}{9} - \frac{4}{b^{2}} = 1 i . e . 2 - \frac{4}{b^{2}} = 1 i . e . \frac{4}{b^{2}} = 1 i . e . b^{2} = 4 ∴ eccentricity is \sqrt{\frac{a^{2} + b^{2}}{a^{2}}} = \sqrt{\frac{9 + 4}{9}} i . e . eccentricity = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} .$

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Question 4:

The equation of the hyperbola having its eccentricity 2 and the distance between foci 8, is ________________________.

Answer:

For the hyperbola,
Eccentricity is given i.e. 2 and distance between foci is 8
$i . e . 2 a e = 8 i . e . 2 a \times 2 = 8 i . e . a = 2 i . e . b^{2} = a^{2} (- 1 + e^{2}) = a^{2} (e^{2} - 1) i . e . b^{2} = 4 (- 1 + 4) = 4 (+ 3) b^{2} = + 12 ∴ equation of hyperbola is \frac{x^{2}}{4} - \frac{y^{2}}{12} = 1 i . e . 3 x^{2} - y^{2} = 12$

Page No 27.20:

Question 5:

The eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$ = –1 is given by __________________.

Answer:

For given hyperbola
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1 i . e . \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1 eccentricity e = \sqrt{\frac{b^{2} + a^{2}}{b^{2}}} i . e . e^{2} = \frac{b^{2} + a^{2}}{b^{2}} i . e . b^{2} e^{2} = b^{2} + a^{2} i . e . a^{2} = b^{2} (c^{2} - 1)$

Page No 27.20:

Question 6:

If t is parameter, then the equations x = a $(t + \frac{1}{t}), y = b (t - \frac{1}{t})$ represent __________________.

Answer:

Given equations,

$x = a (t + \frac{1}{t}) and y = b (t - \frac{1}{t}) can be written as, \frac{x}{a} = t + \frac{1}{t} a n d \frac{y}{b} = t - \frac{1}{t} ∴ {(\frac{x}{a})}^{2} - {(\frac{4}{b})}^{2} = {(t + \frac{1}{t})}^{2} - {(t - \frac{1}{t})}^{2} = t^{2} + \frac{1}{t^{2}} + 2 - (t^{2} + \frac{1}{t^{2}} - 2) i . e . {(\frac{x}{a})}^{2} - {(\frac{4}{b})}^{2} = 4 i . e . \frac{x^{2}}{4 a^{2}} - \frac{y^{2}}{4 b^{2}} = 1 which represents a hyperbola$

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Question 7:

If P is a point on the hyperbola 16x² – 9y² = 144 having foci at S and S , then S'P – SP = ___________________.

Answer:

For a given hyperbola,
Let P(x, y) be any point on hyperbola,

$16 x^{2} - 9 y^{2} = 144 i . e . \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 i . e . a = 3 and b = 4 Let S and S^{1} represent foci then |S P - S P^{1}| = 2 a i . e . S P - S P^{1} = 6$

Page No 27.20:

Question 8:

If the distance between the foci and the distance between the directries of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ are in the ratio 3:2, then its eccentricity is __________________.

Answer:

$For the hyperbola, \frac{Distance between foci}{Distance between directrix} = \frac{3}{2} i . e . \frac{2 a e}{\frac{2 a}{e}} = \frac{3}{2} i . e . e^{2} = \frac{3}{2} i . e . e = \sqrt{\frac{3}{2}}$

Page No 27.20:

Question 9:

The eccentricity of the hyperbola $x^{2} - y^{2} = a^{2}$ is ___________________.

Answer:

For the hyperbola,
$∴ eccentricity e = \sqrt{\frac{a^{2} + b^{2}}{a^{2}}} = \sqrt{\frac{a^{2} + a^{2}}{a^{2}}} i . e . eccentricity' e' = \sqrt{2}$

Page No 27.20:

Question 10:

The eccentricity of the hyperbola $5 x^{2} - 4 y^{2} - 20 x + 8 y + 4 = 0$ is ______________________________.

Answer:

Given hyperbola is,

$5 x^{2} - 4 y^{2} + 20 x + 8 y + 4 = 0 i . e . 5 (x^{2} + 4 x) - 4 (y^{2} - 2 y) + 4 = 0 i . e . 5 (x^{2} + 4 x + 4 - 4) - 4 (y^{2} - 2 y + 1 - 1) + 4 = 0 i . e . 5 [{(x + 2)}^{2} - 4] - 4 [{(y - 1)}^{2} - 1] + 4 = 0 i . e . 5 {(x + 2)}^{2} - 4 {(y - 1)}^{2} - 20 + 4 + 4 = 0 i . e . 5 {(x + 2)}^{2} - 4 {(y - 1)}^{2} - 12 = 0 i . e . \frac{5 {(x + 2)}^{2}}{12} - \frac{{(y - 1)}^{2}}{3} = 0 i . e . a^{2} = (\frac{12}{5}) and b^{2} = 3$

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Question 11:

The latusrectum of the hyperbola $9 x^{2} - 16 y^{2} + 72 x - 32 y - 16 = 0$ is _______________________.

Answer:

Given equation of hyperbola,

$9 x^{2} - 16 y^{2} + 72 x - 32 y - 16 = 0 i . e . 9 (x^{2} + 8 x) - 16 (y^{2} - 2 y) = 16 i . e . 9 (x^{2} + 2 (4) x + 16 - 16) - 16 (y^{2} - 2 y + 1 - 1) = 16 i . e . 9 [{(x + 4)}^{2} - 16] - 16 [{(y - 1)}^{2} - 1] = 16 i . e . 9 {(x + 4)}^{2} - 16 {(y - 1)}^{2} - 16 \times 9 + 16 = 16 i . e . 9 {(x + 4)}^{2} - 16 {(y - 1)}^{2} = 16 \times 9 i . e . \frac{{(x + 4)}^{2}}{16} - \frac{{(y - 1)}^{2}}{9} = 1 i . e . a^{2} = 16, b^{2} = 9 ∴ e^{2} = 1 + \frac{b^{2}}{a^{2}} i . e . e^{2} = 1 + \frac{9}{16} = \frac{25}{16} i . e . e = \frac{5}{4} then, latus rectum is \frac{2 b^{2}}{a} i . e . \frac{2 \times 9}{4} = \frac{9}{2}$

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Question 12:

The eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$ =1 which passes through the points(3, 0) and $(3 \sqrt{2}, 2)$ , is ________________.

Answer:

$Suppose th equation of hyperbola is \frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}} = 1 which passes through (3, 0) and (3 \sqrt{2}, 2) i . e . \frac{{(3)}^{2}}{a^{2}} - \frac{{(a)}^{2}}{b^{2}} = 1 i . e . a^{2} = 9 and \frac{{(3 \sqrt{2})}^{2}}{a^{2}} - \frac{{(4)}^{2}}{b^{2}} = 1 i . e . \frac{9 \times 2}{9} - \frac{4}{b^{2}} = 1 i . e . 2 - \frac{4}{b^{2}} = 1 i . e . b^{2} = 4 ∴ eccentricity is \sqrt{\frac{a^{2} + b^{2}}{a^{2}}} = \sqrt{\frac{9 + 4}{9}} i . e . eccentricity = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} .$

Page No 27.21:

Question 1:

Write the eccentricity of the hyperbola 9x² − 16y² = 144.

Answer:

Equation of the hyperbola:
$9 x^{2} - 16 y^{2} = 144$ ..... (1)

Equation (1) can be rewritten in the following way:
$\frac{9 x^{2}}{144} - \frac{16 y^{2}}{144} = \frac{144}{144} \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$\Rightarrow \frac{x^{2}}{4^{2}} - \frac{y^{2}}{3^{2}} = 1$
This becomes the standard equation of the hyperbola with its major axis $a = 4$ and minor axis $b = 3$ .
Eccentricity, $e = \frac{\sqrt{a^{2} + b^{2}}}{a}$
Substituting the value of a and b, we get:
$e = \frac{\sqrt{4^{2} + 3^{2}}}{4} = \frac{\sqrt{16 + 9}}{4} = \frac{5}{4}$
Therefore, the eccentricity is $\frac{5}{4}$ .

Page No 27.21:

Question 2:

Write the eccentricity of the hyperbola whose latus-rectum is half of its transverse axis.

Answer:

The lengths of the latus rectum and the transverse axis are $\frac{2 b^{2}}{a}$ and $2 a$ , respectively.
According to the given statement, length of the latus rectum is half of its trasverse axis.

$∴ \frac{2 b^{2}}{a} = \frac{1}{2} \times 2 a \Rightarrow \frac{2 b^{2}}{a} = a \Rightarrow 2 b^{2} = a^{2}$

Eccentricity, $e = \frac{\sqrt{a^{2} + b^{2}}}{a}$
Substituting the value $b^{2} = \frac{a^{2}}{2}$ , we get:
$e = \frac{\sqrt{a^{2} + \frac{a^{2}}{2}}}{a} = \frac{a \sqrt{\frac{3}{2}}}{a} = \sqrt{\frac{3}{2}}$
Therefore, the eccentricity is $\sqrt{\frac{3}{2}}$ .

Page No 27.21:

Question 3:

Write the coordinates of the foci of the hyperbola 9x² − 16y² = 144.

Answer:

Equation of the hyperbola:
$9 x^{2} - 16 y^{2} = 144$ ..... (1)

This equation (1) can be rewritten in the following way:
$\Rightarrow \frac{9 x^{2}}{144} - \frac{16 y^{2}}{144} = 1 \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

This becomes a standard form of the hyperbola with transverse axis $a = 4$ and conjugate axis $b = 3$ .

$Eccentricity of the hyperbola, e = \frac{\sqrt{a^{2} + b^{2}}}{a} = \frac{\sqrt{4^{2} + 3^{2}}}{4} = \frac{\sqrt{16 + 9}}{4} = \frac{5}{4}$
The foci of the hyperbola are of the form $(a e, 0)$ and $(- a e, 0)$ .
Therefore, the foci are $(5, 0)$ and $(- 5, 0)$ .4

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Question 4:

Write the equation of the hyperbola of eccentricity $\sqrt{2}$ , if it is known that the distance between its foci is 16.

Answer:

The foci of the hyperbola are of the form $(a e, 0)$ and $(- a e, 0)$ .

$Distance between the foci = \sqrt{{(a e - (- a e)}^{2} + 0^{^{2}}} = \sqrt{{(2 a e)}^{^{2}}} = 2 a e$
Distance between the foci is 16 and eccentricity of the hyperbola is $\sqrt{2}$ .

$∴ 2 a e = 16 \Rightarrow 2 \sqrt{2} a = 16 \Rightarrow a = 4 \sqrt{2}$

$Now, b^{2} = a^{2} (e^{2} - 1) \Rightarrow b^{2} = {(4 \sqrt{2})}^{2} ((\sqrt{2})^{2} - 1) \Rightarrow b^{2} = 32$

Equation of the hyperbola is given below:
$\frac{x^{2}}{{(4 \sqrt{2})}^{2}} - \frac{y^{2}}{32} = 1 \Rightarrow \frac{x^{2}}{32} - \frac{y^{2}}{32} = 1$

Page No 27.21:

Question 5:

If the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$ coincide, write the value of b².

Answer:

Equation of the hyperbola:
$\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$

Simplifying the above equation, we get:
$\frac{25 x^{2}}{144} - \frac{25 y^{2}}{81} = 1 \Rightarrow \frac{x^{2}}{\frac{144}{25}} - \frac{y^{2}}{\frac{81}{25}} = 1 \Rightarrow \frac{x^{2}}{{(\frac{12}{5})}^{2}} - \frac{y^{2}}{{(\frac{9}{5})}^{2}} = 1$
This is the standard form of hyperbola with $a = \frac{12}{5}$ and $b = \frac{9}{5}$ .

$Eccentricity of the h yperbola, e = \frac{\sqrt{{(\frac{12}{5})}^{2} + {(\frac{9}{5})}^{2}}}{\frac{12}{5}} = \frac{15}{12}$
The foci of the hyperbola are of the form $(a e, 0)$ and $(- a e, 0)$ .
So, foci of the hyperbola are $(3, 0)$ and $(- 3, 0)$ .
The foci of the hyperbola coincide with the foci of the ellipse.

For ellipse,
$a = 4$

Using the relation $\sqrt{a^{2} - b^{2}} = a e$ , we get:
$\sqrt{4^{2} - b^{2}} = 3 \Rightarrow 16 - b^{2} = 9 \Rightarrow b^{2} = 7$
Therefore, the value of $b^{2}$ is 7.

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Question 6:

Write the length of the latus-rectum of the hyperbola 16x² − 9y² = 144.

Answer:

Equation of the hyperbola:
$16 x^{2} - 9 y^{2} = 144$
This equation can be rewritten in the following way:
$\frac{16 x^{2}}{144} - \frac{9 y^{2}}{144} = 1 \Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 \Rightarrow \frac{x^{2}}{3^{2}} - \frac{y^{2}}{4^{2}} = 1$
This is the standard form of a hyperbola with $a = 3$ and $b = 4$ .
Length of the latus rectum = $\frac{2 b^{2}}{a}$
Substituting the value of a and b, we get:
Length of the latus rectum $= \frac{2 \times 4^{2}}{3} = \frac{32}{3}$

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Question 7:

If the latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex, then write the eccentricity of the hyperbola.

Answer:

The standard equation of hyperbola is given below:
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

Let the latus rectum be QR, which passes through one of the focus. QR subtends a right angle at point P.

It passes through the focus $(a e, \pm y)$ .
$∴ \frac{{(a e)}^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{y^{2}}{b^{2}} = \frac{{(a e)}^{2}}{a^{2}} - 1 \Rightarrow \frac{y^{2}}{b^{2}} = e^{2} - 1 \Rightarrow y = \pm b \sqrt{e^{2} - 1}$
Now, the slope of PQ is $\frac{b \sqrt{e^{2} - 1}}{a e + a}$ and the slope of PR is $- \frac{b \sqrt{e^{2} - 1}}{a e + a}$ .
The lines PQ and PR are perpendicular, so the product of the slope is $- 1$ .

$\Rightarrow \frac{b \sqrt{e^{2} - 1}}{a e + a} \times \frac{- b \sqrt{e^{2} - 1}}{a e + a} = - 1 \Rightarrow b^{2} (e^{2} - 1) = {(a e + a)}^{2} \Rightarrow b^{2} (e^{2} - 1) = a^{2} {(e + 1)}^{2} \Rightarrow {(e^{2} - 1)}^{2} = {(e + 1)}^{2} \Rightarrow e^{4} - 2 e^{2} + 1 = e^{2} + 2 e + 1 \Rightarrow e^{4} - 3 e^{2} - 2 e = 0 \Rightarrow e (e^{3} - 3 e - 2) = 0 \Rightarrow e (e - 2) (e + 1) (e + 1) = 0$

∴ $e = 0, - 1, 2$

The value of e can neither be negative nor zero.

Therefore, the value of eccentricity is 2.

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Question 8:

Write the distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ.

Answer:

We have:
$x = 8 \sec θ, y = 8 \tan θ$

$On squaring and subtracting, we get : x^{2} - y^{2} = 64 \sec^{2} θ - 64 \tan^{2} θ \Rightarrow x^{2} - y^{2} = 64 \Rightarrow \frac{x^{2}}{64} - \frac{y^{2}}{64} = 1$

∴ a = b = 8
Distance between the directrices of hyperbola is $\frac{2 a^{2}}{\sqrt{a^{2} + b^{2}}}$ .

$\Rightarrow \frac{2 \times 64}{\sqrt{64 + 64}} = \frac{128}{8 \sqrt{2}} = \frac{16}{\sqrt{2}} = 8 \sqrt{2}$

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Question 9:

Write the equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0).

Answer:

The vertices of the hyperbola and the foci are $(\pm a, 0)$ and $(\pm a e, 0)$ , respectively.

$∴ a = 3 and a e = 5$

Using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we get:

$\Rightarrow b^{2} = {(a e)}^{2} - a^{2} \Rightarrow b^{2} = 25 - 9 \Rightarrow b^{2} = 16 \Rightarrow b = 4$
Therefore, the equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ .

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Question 10:

If e₁ and e₂ are respectively the eccentricities of the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ , then write the value of 2 e₁² + e₂².

Answer:

The standard form of the ellipse is $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ , where $a^{2} = 18 and b^{2} = 4$ .
So, the eccentricity is calculated in the following way:
$\Rightarrow b^{2} = a^{2} (1 - {e_{1}}^{2}) \Rightarrow 4 = 18 (1 - {e_{1}}^{2}) \Rightarrow \frac{2}{9} = 1 - {e_{1}}^{2} \Rightarrow {e_{1}}^{2} = \frac{7}{9}$
The standard form of the hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ , where $a^{2} = 9 and b^{2} = 4$ .
So, the eccentricity is calculated in the following way:
$b^{2} = a^{2} ({e_{2}}^{2} - 1) \Rightarrow 4 = 9 ({e_{2}}^{2} - 1) \Rightarrow \frac{4}{9} = {e_{2}}^{2} - 1 \Rightarrow {e_{2}}^{2} = \frac{13}{9}$

$∴ 2 {e_{1}}^{2} + {e_{2}}^{2} = \frac{2 \times 7}{9} + \frac{13}{9} = \frac{27}{9} = 3$

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