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Hyperbola

120750 If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 2 \sqrt{3})$ is $5 x = 4 \sqrt{5}$ and its eccentricity is e, then

1

4 e^{4} - 12 e^{2} - 27 = 0

2

4 e^{4} - 24 e^{2} + 27 = 0

3

4 e^{4} + 8 e^{2} - 35 = 0

4

4 e^{4} - 24 e^{2} + 35 = 0

Explanation:

D The given directrix of hyperbola is -
$5 x = 4 \sqrt{5}$
$x = \frac{4 \sqrt{5}}{5}$
$∴ \frac{a}{e} = \frac{4}{\sqrt{5}}$
Now, the hyperbola
Given,
below passes through $(4, - 2 \sqrt{3})$
Substituting the value of $x$ and $y$
$\frac{4^{2}}{a^{2}} - \frac{(2 \sqrt{3})^{2}}{b^{2}} = 1$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$[∵ e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow a^{2} (e^{2} - 1) = b^{2}]$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$\frac{16 (e^{2} - 1) - 12}{a^{2} (e^{2} - 1)} = 1$
$16 e^{2} - 16 - 12 = a^{2} (e^{2} - 1)$
$16 e^{2} - 28 = a^{2} (e^{2} - 1)$
On substituting $a = \frac{4 e}{\sqrt{5}}$ in the above ${eq}^{n}$ , we get-
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$
On dividing the above equation with $4$
$\frac{16 e^{2}}{4} - \frac{96 e^{2}}{4} + \frac{140}{4} = 0$
$4 e^{2} - 24 e^{2} + 35 = 0$
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$ On dividing the above equation with 4

JEE Main 10.04.2019

Hyperbola

120751 If $5 x + 9 = 0$ is the directirx of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is

1

(- \frac{5}{3}, 0)

2

(- 5, 0)

3

(\frac{5}{3}, 0)

4

(5, 0)

Explanation:

B Given,
The equation of the hyperbola
$16 x^{2} - 9 y^{2} = 144$
On dividing by 144 both sides
$\frac{16 x^{2}}{144} - \frac{9 y^{2}}{144} = \frac{144}{144}$
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
For eccentricity (e) in hyperbola is -
$b^{2} = a^{2} (e^{2} - 1)$
$16 = 9 (e^{2} - 1)$
$\frac{16}{9} + 1 = e^{2}$
$\frac{16 + 9}{9} = e^{2}$
$\frac{25}{9} = e^{2}$
$e = \frac{5}{3}$
And, directirx $5 x + 9 = 0$
$5 x + 9 = 0$
$5 x = - 9$
$\frac{- 9}{5} = \frac{- a}{e} is$
$a = \frac{9}{5} \times \frac{5}{3}$
$a = 3$
Focus of hyperbola (-ae, 0 )
$= (- 3 \times \frac{5}{3}, 0)$
$= (- 5, 0)$

JEE Main 10.04.2019

Hyperbola

120752 For some $θ \in (0, \frac{π}{2})$ . If the eccentricity of the hyperbola, $x^{2} - y^{2} \sec^{2} θ = 10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is

1

2 \sqrt{6}

2

\sqrt{30}

3

\frac{2 \sqrt{5}}{3}

4

\frac{4 \sqrt{5}}{3}

Explanation:

D Given,
$θ \in (0, \frac{π}{2})$
Equation of hyperbola,
$x^{2} - y^{2} \cdot \sec^{2} θ = 10$
$\frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} θ} = 1$
Hence, eccentricity of hyperbola
$(e_{H}) = \sqrt{1 + \frac{10 \cos^{2} θ}{10}}$
$e_{H} = \sqrt{1 + \cos^{2} θ} \dots \dots (i)$
${∵ e = \sqrt{1 + \frac{b^{2}}{a^{2}}}}$
Now, equation of ellipse $\Rightarrow x^{2} \sec^{2} θ + y^{2} = 5$
$\frac{x^{2}}{5 \cos^{2} θ} + \frac{y^{2}}{5} = 1$
${∵ e = \sqrt{1 - \frac{a^{2}}{b^{2}}}}$
Hence, eccentricity of ellipse,
$(e_{E}) = \sqrt{1 - \frac{5 \cos^{2} θ}{5}}$
$(e_{E}) = \sqrt{1 - \cos^{2} θ} = | \sin θ | = \sin θ \dots . . (ii)$
$[∵ θ \in (0, \frac{π}{2})]$
Given,
$e_{H} = \sqrt{5} e_{e}$
Hence,
$1 + \cos^{2} θ = 5 \sin^{2} θ$
$1 + \cos^{2} θ = 5 (1 - \cos^{2} θ)$
$1 + \cos^{2} θ = 5 - 5 \cos^{2} θ$
$6 \cos^{2} θ = 4$
$\cos^{2} θ = \frac{4}{6}$
$\cos^{2} θ = \frac{2}{3}$
Now, length of latusrectum of ellipse,
$\frac{2 a^{2}}{b} = \frac{10 \cos^{2} θ}{\sqrt{5}}$
$= \frac{4 \sqrt{5}}{3}$

JEE Main 02.09.2020

Hyperbola

120753 A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

1

\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

3

\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1

4

x^{2} - y^{2} = 9

Explanation:

A Given,
The ellipse, $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ ,
Comparing with equation,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$a^{2} = 25, b^{2} = 16$
Let, $e_{1}$ is eccentricity of hyperbola-
$e_{1} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e_{1} = \sqrt{1 - \frac{16}{25}}$
$e_{1} = \frac{3}{5}$
Foci $= (\pm 3, 0) [∵$ Foci $= (ae, 0)]$
II- Let the equation of hyperbola be
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 [∴ passes through (\pm 3, 0)]$
Let, $e_{2}$ is eccentricity of hyperbola
$a^{2} = 9$
$a = 3$
$∴ e_{2} = \frac{5}{3}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} - 1 = \frac{b^{2}}{9}$
$\frac{16}{9} = \frac{b^{2}}{9}$
$b^{2} = 16$
$∴$ The equation of the hyperbola is
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Put, the value of $a$ and $b$ in ${eq}^{n}$ . (i) we get,
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

JEE Main 25.02.2021

Hyperbola

120754 If $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse, $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the
hyperbola, $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ respectively and $(e_{1}, e_{2})$ is a point on the ellipse, $15 x^{2} + 3 y^{2} = k$ , then $k$ is equal to

1 14

2 15

3 17

4 16

Explanation:

D Given,
Case I: $- \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$
$a^{2} = 18, b^{2} = 4$
Eccentricity $(e_{1}) = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$= \sqrt{1 - \frac{4}{18}}$
$= \sqrt{\frac{14}{18}}$
$e_{1} = \frac{\sqrt{7}}{3}$
Case II:- $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
$a^{2} = 9, b^{2} = 4$
Eccentricity $(e_{2}) = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e_{2} = \sqrt{1 + \frac{4}{9}}$
$e_{2} = \frac{\sqrt{13}}{3}$
The value of $e_{1}, e_{2}$ lies on ellipse,
$∴ 15 x^{2} + 3 y^{2} = k [\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}]$
$15 {(\frac{\sqrt{7}}{3})}^{2} + 3 {(\frac{\sqrt{13}}{3})}^{2} = k$
$15 \times \frac{7}{9} + 3 \times \frac{13}{9} = k$
$\frac{105}{9} + \frac{39}{9} = k$
$\frac{105 + 39}{9} = k$
$\frac{144}{9} = k$
$k = 16$

JEE Main 09.01.2020

Hyperbola

120750 If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 2 \sqrt{3})$ is $5 x = 4 \sqrt{5}$ and its eccentricity is e, then

1

4 e^{4} - 12 e^{2} - 27 = 0

2

4 e^{4} - 24 e^{2} + 27 = 0

3

4 e^{4} + 8 e^{2} - 35 = 0

4

4 e^{4} - 24 e^{2} + 35 = 0

Explanation:

D The given directrix of hyperbola is -
$5 x = 4 \sqrt{5}$
$x = \frac{4 \sqrt{5}}{5}$
$∴ \frac{a}{e} = \frac{4}{\sqrt{5}}$
Now, the hyperbola
Given,
below passes through $(4, - 2 \sqrt{3})$
Substituting the value of $x$ and $y$
$\frac{4^{2}}{a^{2}} - \frac{(2 \sqrt{3})^{2}}{b^{2}} = 1$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$[∵ e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow a^{2} (e^{2} - 1) = b^{2}]$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$\frac{16 (e^{2} - 1) - 12}{a^{2} (e^{2} - 1)} = 1$
$16 e^{2} - 16 - 12 = a^{2} (e^{2} - 1)$
$16 e^{2} - 28 = a^{2} (e^{2} - 1)$
On substituting $a = \frac{4 e}{\sqrt{5}}$ in the above ${eq}^{n}$ , we get-
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$
On dividing the above equation with $4$
$\frac{16 e^{2}}{4} - \frac{96 e^{2}}{4} + \frac{140}{4} = 0$
$4 e^{2} - 24 e^{2} + 35 = 0$
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$ On dividing the above equation with 4

JEE Main 10.04.2019

Hyperbola

120751 If $5 x + 9 = 0$ is the directirx of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is

1

(- \frac{5}{3}, 0)

2

(- 5, 0)

3

(\frac{5}{3}, 0)

4

(5, 0)

Explanation:

B Given,
The equation of the hyperbola
$16 x^{2} - 9 y^{2} = 144$
On dividing by 144 both sides
$\frac{16 x^{2}}{144} - \frac{9 y^{2}}{144} = \frac{144}{144}$
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
For eccentricity (e) in hyperbola is -
$b^{2} = a^{2} (e^{2} - 1)$
$16 = 9 (e^{2} - 1)$
$\frac{16}{9} + 1 = e^{2}$
$\frac{16 + 9}{9} = e^{2}$
$\frac{25}{9} = e^{2}$
$e = \frac{5}{3}$
And, directirx $5 x + 9 = 0$
$5 x + 9 = 0$
$5 x = - 9$
$\frac{- 9}{5} = \frac{- a}{e} is$
$a = \frac{9}{5} \times \frac{5}{3}$
$a = 3$
Focus of hyperbola (-ae, 0 )
$= (- 3 \times \frac{5}{3}, 0)$
$= (- 5, 0)$

JEE Main 10.04.2019

Hyperbola

120752 For some $θ \in (0, \frac{π}{2})$ . If the eccentricity of the hyperbola, $x^{2} - y^{2} \sec^{2} θ = 10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is

1

2 \sqrt{6}

2

\sqrt{30}

3

\frac{2 \sqrt{5}}{3}

4

\frac{4 \sqrt{5}}{3}

Explanation:

D Given,
$θ \in (0, \frac{π}{2})$
Equation of hyperbola,
$x^{2} - y^{2} \cdot \sec^{2} θ = 10$
$\frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} θ} = 1$
Hence, eccentricity of hyperbola
$(e_{H}) = \sqrt{1 + \frac{10 \cos^{2} θ}{10}}$
$e_{H} = \sqrt{1 + \cos^{2} θ} \dots \dots (i)$
${∵ e = \sqrt{1 + \frac{b^{2}}{a^{2}}}}$
Now, equation of ellipse $\Rightarrow x^{2} \sec^{2} θ + y^{2} = 5$
$\frac{x^{2}}{5 \cos^{2} θ} + \frac{y^{2}}{5} = 1$
${∵ e = \sqrt{1 - \frac{a^{2}}{b^{2}}}}$
Hence, eccentricity of ellipse,
$(e_{E}) = \sqrt{1 - \frac{5 \cos^{2} θ}{5}}$
$(e_{E}) = \sqrt{1 - \cos^{2} θ} = | \sin θ | = \sin θ \dots . . (ii)$
$[∵ θ \in (0, \frac{π}{2})]$
Given,
$e_{H} = \sqrt{5} e_{e}$
Hence,
$1 + \cos^{2} θ = 5 \sin^{2} θ$
$1 + \cos^{2} θ = 5 (1 - \cos^{2} θ)$
$1 + \cos^{2} θ = 5 - 5 \cos^{2} θ$
$6 \cos^{2} θ = 4$
$\cos^{2} θ = \frac{4}{6}$
$\cos^{2} θ = \frac{2}{3}$
Now, length of latusrectum of ellipse,
$\frac{2 a^{2}}{b} = \frac{10 \cos^{2} θ}{\sqrt{5}}$
$= \frac{4 \sqrt{5}}{3}$

JEE Main 02.09.2020

Hyperbola

120753 A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

1

\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

3

\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1

4

x^{2} - y^{2} = 9

Explanation:

A Given,
The ellipse, $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ ,
Comparing with equation,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$a^{2} = 25, b^{2} = 16$
Let, $e_{1}$ is eccentricity of hyperbola-
$e_{1} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e_{1} = \sqrt{1 - \frac{16}{25}}$
$e_{1} = \frac{3}{5}$
Foci $= (\pm 3, 0) [∵$ Foci $= (ae, 0)]$
II- Let the equation of hyperbola be
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 [∴ passes through (\pm 3, 0)]$
Let, $e_{2}$ is eccentricity of hyperbola
$a^{2} = 9$
$a = 3$
$∴ e_{2} = \frac{5}{3}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} - 1 = \frac{b^{2}}{9}$
$\frac{16}{9} = \frac{b^{2}}{9}$
$b^{2} = 16$
$∴$ The equation of the hyperbola is
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Put, the value of $a$ and $b$ in ${eq}^{n}$ . (i) we get,
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

JEE Main 25.02.2021

Hyperbola

120754 If $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse, $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the
hyperbola, $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ respectively and $(e_{1}, e_{2})$ is a point on the ellipse, $15 x^{2} + 3 y^{2} = k$ , then $k$ is equal to

1 14

2 15

3 17

4 16

Explanation:

D Given,
Case I: $- \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$
$a^{2} = 18, b^{2} = 4$
Eccentricity $(e_{1}) = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$= \sqrt{1 - \frac{4}{18}}$
$= \sqrt{\frac{14}{18}}$
$e_{1} = \frac{\sqrt{7}}{3}$
Case II:- $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
$a^{2} = 9, b^{2} = 4$
Eccentricity $(e_{2}) = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e_{2} = \sqrt{1 + \frac{4}{9}}$
$e_{2} = \frac{\sqrt{13}}{3}$
The value of $e_{1}, e_{2}$ lies on ellipse,
$∴ 15 x^{2} + 3 y^{2} = k [\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}]$
$15 {(\frac{\sqrt{7}}{3})}^{2} + 3 {(\frac{\sqrt{13}}{3})}^{2} = k$
$15 \times \frac{7}{9} + 3 \times \frac{13}{9} = k$
$\frac{105}{9} + \frac{39}{9} = k$
$\frac{105 + 39}{9} = k$
$\frac{144}{9} = k$
$k = 16$

JEE Main 09.01.2020

Hyperbola

120750 If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 2 \sqrt{3})$ is $5 x = 4 \sqrt{5}$ and its eccentricity is e, then

1

4 e^{4} - 12 e^{2} - 27 = 0

2

4 e^{4} - 24 e^{2} + 27 = 0

3

4 e^{4} + 8 e^{2} - 35 = 0

4

4 e^{4} - 24 e^{2} + 35 = 0

Explanation:

D The given directrix of hyperbola is -
$5 x = 4 \sqrt{5}$
$x = \frac{4 \sqrt{5}}{5}$
$∴ \frac{a}{e} = \frac{4}{\sqrt{5}}$
Now, the hyperbola
Given,
below passes through $(4, - 2 \sqrt{3})$
Substituting the value of $x$ and $y$
$\frac{4^{2}}{a^{2}} - \frac{(2 \sqrt{3})^{2}}{b^{2}} = 1$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$[∵ e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow a^{2} (e^{2} - 1) = b^{2}]$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$\frac{16 (e^{2} - 1) - 12}{a^{2} (e^{2} - 1)} = 1$
$16 e^{2} - 16 - 12 = a^{2} (e^{2} - 1)$
$16 e^{2} - 28 = a^{2} (e^{2} - 1)$
On substituting $a = \frac{4 e}{\sqrt{5}}$ in the above ${eq}^{n}$ , we get-
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$
On dividing the above equation with $4$
$\frac{16 e^{2}}{4} - \frac{96 e^{2}}{4} + \frac{140}{4} = 0$
$4 e^{2} - 24 e^{2} + 35 = 0$
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$ On dividing the above equation with 4

JEE Main 10.04.2019

Hyperbola

120751 If $5 x + 9 = 0$ is the directirx of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is

1

(- \frac{5}{3}, 0)

2

(- 5, 0)

3

(\frac{5}{3}, 0)

4

(5, 0)

Explanation:

B Given,
The equation of the hyperbola
$16 x^{2} - 9 y^{2} = 144$
On dividing by 144 both sides
$\frac{16 x^{2}}{144} - \frac{9 y^{2}}{144} = \frac{144}{144}$
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
For eccentricity (e) in hyperbola is -
$b^{2} = a^{2} (e^{2} - 1)$
$16 = 9 (e^{2} - 1)$
$\frac{16}{9} + 1 = e^{2}$
$\frac{16 + 9}{9} = e^{2}$
$\frac{25}{9} = e^{2}$
$e = \frac{5}{3}$
And, directirx $5 x + 9 = 0$
$5 x + 9 = 0$
$5 x = - 9$
$\frac{- 9}{5} = \frac{- a}{e} is$
$a = \frac{9}{5} \times \frac{5}{3}$
$a = 3$
Focus of hyperbola (-ae, 0 )
$= (- 3 \times \frac{5}{3}, 0)$
$= (- 5, 0)$

JEE Main 10.04.2019

Hyperbola

120752 For some $θ \in (0, \frac{π}{2})$ . If the eccentricity of the hyperbola, $x^{2} - y^{2} \sec^{2} θ = 10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is

1

2 \sqrt{6}

2

\sqrt{30}

3

\frac{2 \sqrt{5}}{3}

4

\frac{4 \sqrt{5}}{3}

Explanation:

D Given,
$θ \in (0, \frac{π}{2})$
Equation of hyperbola,
$x^{2} - y^{2} \cdot \sec^{2} θ = 10$
$\frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} θ} = 1$
Hence, eccentricity of hyperbola
$(e_{H}) = \sqrt{1 + \frac{10 \cos^{2} θ}{10}}$
$e_{H} = \sqrt{1 + \cos^{2} θ} \dots \dots (i)$
${∵ e = \sqrt{1 + \frac{b^{2}}{a^{2}}}}$
Now, equation of ellipse $\Rightarrow x^{2} \sec^{2} θ + y^{2} = 5$
$\frac{x^{2}}{5 \cos^{2} θ} + \frac{y^{2}}{5} = 1$
${∵ e = \sqrt{1 - \frac{a^{2}}{b^{2}}}}$
Hence, eccentricity of ellipse,
$(e_{E}) = \sqrt{1 - \frac{5 \cos^{2} θ}{5}}$
$(e_{E}) = \sqrt{1 - \cos^{2} θ} = | \sin θ | = \sin θ \dots . . (ii)$
$[∵ θ \in (0, \frac{π}{2})]$
Given,
$e_{H} = \sqrt{5} e_{e}$
Hence,
$1 + \cos^{2} θ = 5 \sin^{2} θ$
$1 + \cos^{2} θ = 5 (1 - \cos^{2} θ)$
$1 + \cos^{2} θ = 5 - 5 \cos^{2} θ$
$6 \cos^{2} θ = 4$
$\cos^{2} θ = \frac{4}{6}$
$\cos^{2} θ = \frac{2}{3}$
Now, length of latusrectum of ellipse,
$\frac{2 a^{2}}{b} = \frac{10 \cos^{2} θ}{\sqrt{5}}$
$= \frac{4 \sqrt{5}}{3}$

JEE Main 02.09.2020

Hyperbola

120753 A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

1

\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

3

\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1

4

x^{2} - y^{2} = 9

Explanation:

A Given,
The ellipse, $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ ,
Comparing with equation,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$a^{2} = 25, b^{2} = 16$
Let, $e_{1}$ is eccentricity of hyperbola-
$e_{1} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e_{1} = \sqrt{1 - \frac{16}{25}}$
$e_{1} = \frac{3}{5}$
Foci $= (\pm 3, 0) [∵$ Foci $= (ae, 0)]$
II- Let the equation of hyperbola be
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 [∴ passes through (\pm 3, 0)]$
Let, $e_{2}$ is eccentricity of hyperbola
$a^{2} = 9$
$a = 3$
$∴ e_{2} = \frac{5}{3}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} - 1 = \frac{b^{2}}{9}$
$\frac{16}{9} = \frac{b^{2}}{9}$
$b^{2} = 16$
$∴$ The equation of the hyperbola is
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Put, the value of $a$ and $b$ in ${eq}^{n}$ . (i) we get,
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

JEE Main 25.02.2021

Hyperbola

120754 If $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse, $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the
hyperbola, $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ respectively and $(e_{1}, e_{2})$ is a point on the ellipse, $15 x^{2} + 3 y^{2} = k$ , then $k$ is equal to

1 14

2 15

3 17

4 16

Explanation:

D Given,
Case I: $- \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$
$a^{2} = 18, b^{2} = 4$
Eccentricity $(e_{1}) = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$= \sqrt{1 - \frac{4}{18}}$
$= \sqrt{\frac{14}{18}}$
$e_{1} = \frac{\sqrt{7}}{3}$
Case II:- $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
$a^{2} = 9, b^{2} = 4$
Eccentricity $(e_{2}) = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e_{2} = \sqrt{1 + \frac{4}{9}}$
$e_{2} = \frac{\sqrt{13}}{3}$
The value of $e_{1}, e_{2}$ lies on ellipse,
$∴ 15 x^{2} + 3 y^{2} = k [\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}]$
$15 {(\frac{\sqrt{7}}{3})}^{2} + 3 {(\frac{\sqrt{13}}{3})}^{2} = k$
$15 \times \frac{7}{9} + 3 \times \frac{13}{9} = k$
$\frac{105}{9} + \frac{39}{9} = k$
$\frac{105 + 39}{9} = k$
$\frac{144}{9} = k$
$k = 16$

JEE Main 09.01.2020

Hyperbola

120750 If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 2 \sqrt{3})$ is $5 x = 4 \sqrt{5}$ and its eccentricity is e, then

1

4 e^{4} - 12 e^{2} - 27 = 0

2

4 e^{4} - 24 e^{2} + 27 = 0

3

4 e^{4} + 8 e^{2} - 35 = 0

4

4 e^{4} - 24 e^{2} + 35 = 0

Explanation:

D The given directrix of hyperbola is -
$5 x = 4 \sqrt{5}$
$x = \frac{4 \sqrt{5}}{5}$
$∴ \frac{a}{e} = \frac{4}{\sqrt{5}}$
Now, the hyperbola
Given,
below passes through $(4, - 2 \sqrt{3})$
Substituting the value of $x$ and $y$
$\frac{4^{2}}{a^{2}} - \frac{(2 \sqrt{3})^{2}}{b^{2}} = 1$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$[∵ e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow a^{2} (e^{2} - 1) = b^{2}]$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$\frac{16 (e^{2} - 1) - 12}{a^{2} (e^{2} - 1)} = 1$
$16 e^{2} - 16 - 12 = a^{2} (e^{2} - 1)$
$16 e^{2} - 28 = a^{2} (e^{2} - 1)$
On substituting $a = \frac{4 e}{\sqrt{5}}$ in the above ${eq}^{n}$ , we get-
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$
On dividing the above equation with $4$
$\frac{16 e^{2}}{4} - \frac{96 e^{2}}{4} + \frac{140}{4} = 0$
$4 e^{2} - 24 e^{2} + 35 = 0$
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$ On dividing the above equation with 4

JEE Main 10.04.2019

Hyperbola

120751 If $5 x + 9 = 0$ is the directirx of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is

1

(- \frac{5}{3}, 0)

2

(- 5, 0)

3

(\frac{5}{3}, 0)

4

(5, 0)

Explanation:

B Given,
The equation of the hyperbola
$16 x^{2} - 9 y^{2} = 144$
On dividing by 144 both sides
$\frac{16 x^{2}}{144} - \frac{9 y^{2}}{144} = \frac{144}{144}$
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
For eccentricity (e) in hyperbola is -
$b^{2} = a^{2} (e^{2} - 1)$
$16 = 9 (e^{2} - 1)$
$\frac{16}{9} + 1 = e^{2}$
$\frac{16 + 9}{9} = e^{2}$
$\frac{25}{9} = e^{2}$
$e = \frac{5}{3}$
And, directirx $5 x + 9 = 0$
$5 x + 9 = 0$
$5 x = - 9$
$\frac{- 9}{5} = \frac{- a}{e} is$
$a = \frac{9}{5} \times \frac{5}{3}$
$a = 3$
Focus of hyperbola (-ae, 0 )
$= (- 3 \times \frac{5}{3}, 0)$
$= (- 5, 0)$

JEE Main 10.04.2019

Hyperbola

120752 For some $θ \in (0, \frac{π}{2})$ . If the eccentricity of the hyperbola, $x^{2} - y^{2} \sec^{2} θ = 10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is

1

2 \sqrt{6}

2

\sqrt{30}

3

\frac{2 \sqrt{5}}{3}

4

\frac{4 \sqrt{5}}{3}

Explanation:

D Given,
$θ \in (0, \frac{π}{2})$
Equation of hyperbola,
$x^{2} - y^{2} \cdot \sec^{2} θ = 10$
$\frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} θ} = 1$
Hence, eccentricity of hyperbola
$(e_{H}) = \sqrt{1 + \frac{10 \cos^{2} θ}{10}}$
$e_{H} = \sqrt{1 + \cos^{2} θ} \dots \dots (i)$
${∵ e = \sqrt{1 + \frac{b^{2}}{a^{2}}}}$
Now, equation of ellipse $\Rightarrow x^{2} \sec^{2} θ + y^{2} = 5$
$\frac{x^{2}}{5 \cos^{2} θ} + \frac{y^{2}}{5} = 1$
${∵ e = \sqrt{1 - \frac{a^{2}}{b^{2}}}}$
Hence, eccentricity of ellipse,
$(e_{E}) = \sqrt{1 - \frac{5 \cos^{2} θ}{5}}$
$(e_{E}) = \sqrt{1 - \cos^{2} θ} = | \sin θ | = \sin θ \dots . . (ii)$
$[∵ θ \in (0, \frac{π}{2})]$
Given,
$e_{H} = \sqrt{5} e_{e}$
Hence,
$1 + \cos^{2} θ = 5 \sin^{2} θ$
$1 + \cos^{2} θ = 5 (1 - \cos^{2} θ)$
$1 + \cos^{2} θ = 5 - 5 \cos^{2} θ$
$6 \cos^{2} θ = 4$
$\cos^{2} θ = \frac{4}{6}$
$\cos^{2} θ = \frac{2}{3}$
Now, length of latusrectum of ellipse,
$\frac{2 a^{2}}{b} = \frac{10 \cos^{2} θ}{\sqrt{5}}$
$= \frac{4 \sqrt{5}}{3}$

JEE Main 02.09.2020

Hyperbola

120753 A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

1

\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

3

\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1

4

x^{2} - y^{2} = 9

Explanation:

A Given,
The ellipse, $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ ,
Comparing with equation,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$a^{2} = 25, b^{2} = 16$
Let, $e_{1}$ is eccentricity of hyperbola-
$e_{1} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e_{1} = \sqrt{1 - \frac{16}{25}}$
$e_{1} = \frac{3}{5}$
Foci $= (\pm 3, 0) [∵$ Foci $= (ae, 0)]$
II- Let the equation of hyperbola be
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 [∴ passes through (\pm 3, 0)]$
Let, $e_{2}$ is eccentricity of hyperbola
$a^{2} = 9$
$a = 3$
$∴ e_{2} = \frac{5}{3}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} - 1 = \frac{b^{2}}{9}$
$\frac{16}{9} = \frac{b^{2}}{9}$
$b^{2} = 16$
$∴$ The equation of the hyperbola is
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Put, the value of $a$ and $b$ in ${eq}^{n}$ . (i) we get,
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

JEE Main 25.02.2021

Hyperbola

120754 If $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse, $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the
hyperbola, $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ respectively and $(e_{1}, e_{2})$ is a point on the ellipse, $15 x^{2} + 3 y^{2} = k$ , then $k$ is equal to

1 14

2 15

3 17

4 16

Explanation:

D Given,
Case I: $- \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$
$a^{2} = 18, b^{2} = 4$
Eccentricity $(e_{1}) = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$= \sqrt{1 - \frac{4}{18}}$
$= \sqrt{\frac{14}{18}}$
$e_{1} = \frac{\sqrt{7}}{3}$
Case II:- $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
$a^{2} = 9, b^{2} = 4$
Eccentricity $(e_{2}) = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e_{2} = \sqrt{1 + \frac{4}{9}}$
$e_{2} = \frac{\sqrt{13}}{3}$
The value of $e_{1}, e_{2}$ lies on ellipse,
$∴ 15 x^{2} + 3 y^{2} = k [\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}]$
$15 {(\frac{\sqrt{7}}{3})}^{2} + 3 {(\frac{\sqrt{13}}{3})}^{2} = k$
$15 \times \frac{7}{9} + 3 \times \frac{13}{9} = k$
$\frac{105}{9} + \frac{39}{9} = k$
$\frac{105 + 39}{9} = k$
$\frac{144}{9} = k$
$k = 16$

JEE Main 09.01.2020

Hyperbola

120750 If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 2 \sqrt{3})$ is $5 x = 4 \sqrt{5}$ and its eccentricity is e, then

1

4 e^{4} - 12 e^{2} - 27 = 0

2

4 e^{4} - 24 e^{2} + 27 = 0

3

4 e^{4} + 8 e^{2} - 35 = 0

4

4 e^{4} - 24 e^{2} + 35 = 0

Explanation:

D The given directrix of hyperbola is -
$5 x = 4 \sqrt{5}$
$x = \frac{4 \sqrt{5}}{5}$
$∴ \frac{a}{e} = \frac{4}{\sqrt{5}}$
Now, the hyperbola
Given,
below passes through $(4, - 2 \sqrt{3})$
Substituting the value of $x$ and $y$
$\frac{4^{2}}{a^{2}} - \frac{(2 \sqrt{3})^{2}}{b^{2}} = 1$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$[∵ e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow a^{2} (e^{2} - 1) = b^{2}]$
$\frac{16}{a^{2}} - \frac{12}{a^{2} (e^{2} - 1)} = 1$
$\frac{16 (e^{2} - 1) - 12}{a^{2} (e^{2} - 1)} = 1$
$16 e^{2} - 16 - 12 = a^{2} (e^{2} - 1)$
$16 e^{2} - 28 = a^{2} (e^{2} - 1)$
On substituting $a = \frac{4 e}{\sqrt{5}}$ in the above ${eq}^{n}$ , we get-
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$
On dividing the above equation with $4$
$\frac{16 e^{2}}{4} - \frac{96 e^{2}}{4} + \frac{140}{4} = 0$
$4 e^{2} - 24 e^{2} + 35 = 0$
$\Rightarrow 16 e^{2} - 28 = {(\frac{4 e}{\sqrt{5}})}^{2} (e^{2} - 1)$
$\Rightarrow 16 e^{2} - 28 = \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5}$
$\Rightarrow \frac{16 e^{4}}{5} - \frac{16 e^{2}}{5} - 16 e^{2} + 28 = 0$
$\frac{16 e^{4} - 16 e^{2} - 80 e^{2} + 140}{5} = 0$
$16 e^{4} - 16 e^{2} - 80 e^{2} + 140 = 0$
$16 e^{4} - 96 e^{2} + 140 = 0$ On dividing the above equation with 4

JEE Main 10.04.2019

Hyperbola

120751 If $5 x + 9 = 0$ is the directirx of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is

1

(- \frac{5}{3}, 0)

2

(- 5, 0)

3

(\frac{5}{3}, 0)

4

(5, 0)

Explanation:

B Given,
The equation of the hyperbola
$16 x^{2} - 9 y^{2} = 144$
On dividing by 144 both sides
$\frac{16 x^{2}}{144} - \frac{9 y^{2}}{144} = \frac{144}{144}$
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
For eccentricity (e) in hyperbola is -
$b^{2} = a^{2} (e^{2} - 1)$
$16 = 9 (e^{2} - 1)$
$\frac{16}{9} + 1 = e^{2}$
$\frac{16 + 9}{9} = e^{2}$
$\frac{25}{9} = e^{2}$
$e = \frac{5}{3}$
And, directirx $5 x + 9 = 0$
$5 x + 9 = 0$
$5 x = - 9$
$\frac{- 9}{5} = \frac{- a}{e} is$
$a = \frac{9}{5} \times \frac{5}{3}$
$a = 3$
Focus of hyperbola (-ae, 0 )
$= (- 3 \times \frac{5}{3}, 0)$
$= (- 5, 0)$

JEE Main 10.04.2019

Hyperbola

120752 For some $θ \in (0, \frac{π}{2})$ . If the eccentricity of the hyperbola, $x^{2} - y^{2} \sec^{2} θ = 10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is

1

2 \sqrt{6}

2

\sqrt{30}

3

\frac{2 \sqrt{5}}{3}

4

\frac{4 \sqrt{5}}{3}

Explanation:

D Given,
$θ \in (0, \frac{π}{2})$
Equation of hyperbola,
$x^{2} - y^{2} \cdot \sec^{2} θ = 10$
$\frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} θ} = 1$
Hence, eccentricity of hyperbola
$(e_{H}) = \sqrt{1 + \frac{10 \cos^{2} θ}{10}}$
$e_{H} = \sqrt{1 + \cos^{2} θ} \dots \dots (i)$
${∵ e = \sqrt{1 + \frac{b^{2}}{a^{2}}}}$
Now, equation of ellipse $\Rightarrow x^{2} \sec^{2} θ + y^{2} = 5$
$\frac{x^{2}}{5 \cos^{2} θ} + \frac{y^{2}}{5} = 1$
${∵ e = \sqrt{1 - \frac{a^{2}}{b^{2}}}}$
Hence, eccentricity of ellipse,
$(e_{E}) = \sqrt{1 - \frac{5 \cos^{2} θ}{5}}$
$(e_{E}) = \sqrt{1 - \cos^{2} θ} = | \sin θ | = \sin θ \dots . . (ii)$
$[∵ θ \in (0, \frac{π}{2})]$
Given,
$e_{H} = \sqrt{5} e_{e}$
Hence,
$1 + \cos^{2} θ = 5 \sin^{2} θ$
$1 + \cos^{2} θ = 5 (1 - \cos^{2} θ)$
$1 + \cos^{2} θ = 5 - 5 \cos^{2} θ$
$6 \cos^{2} θ = 4$
$\cos^{2} θ = \frac{4}{6}$
$\cos^{2} θ = \frac{2}{3}$
Now, length of latusrectum of ellipse,
$\frac{2 a^{2}}{b} = \frac{10 \cos^{2} θ}{\sqrt{5}}$
$= \frac{4 \sqrt{5}}{3}$

JEE Main 02.09.2020

Hyperbola

120753 A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

1

\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

3

\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1

4

x^{2} - y^{2} = 9

Explanation:

A Given,
The ellipse, $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ ,
Comparing with equation,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$a^{2} = 25, b^{2} = 16$
Let, $e_{1}$ is eccentricity of hyperbola-
$e_{1} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e_{1} = \sqrt{1 - \frac{16}{25}}$
$e_{1} = \frac{3}{5}$
Foci $= (\pm 3, 0) [∵$ Foci $= (ae, 0)]$
II- Let the equation of hyperbola be
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 [∴ passes through (\pm 3, 0)]$
Let, $e_{2}$ is eccentricity of hyperbola
$a^{2} = 9$
$a = 3$
$∴ e_{2} = \frac{5}{3}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} = 1 + \frac{b^{2}}{a^{2}}$
$\frac{25}{9} - 1 = \frac{b^{2}}{9}$
$\frac{16}{9} = \frac{b^{2}}{9}$
$b^{2} = 16$
$∴$ The equation of the hyperbola is
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Put, the value of $a$ and $b$ in ${eq}^{n}$ . (i) we get,
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

JEE Main 25.02.2021

Hyperbola

120754 If $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse, $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the
hyperbola, $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ respectively and $(e_{1}, e_{2})$ is a point on the ellipse, $15 x^{2} + 3 y^{2} = k$ , then $k$ is equal to

1 14

2 15

3 17

4 16

Explanation:

D Given,
Case I: $- \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$
$a^{2} = 18, b^{2} = 4$
Eccentricity $(e_{1}) = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$= \sqrt{1 - \frac{4}{18}}$
$= \sqrt{\frac{14}{18}}$
$e_{1} = \frac{\sqrt{7}}{3}$
Case II:- $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
$a^{2} = 9, b^{2} = 4$
Eccentricity $(e_{2}) = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e_{2} = \sqrt{1 + \frac{4}{9}}$
$e_{2} = \frac{\sqrt{13}}{3}$
The value of $e_{1}, e_{2}$ lies on ellipse,
$∴ 15 x^{2} + 3 y^{2} = k [\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}]$
$15 {(\frac{\sqrt{7}}{3})}^{2} + 3 {(\frac{\sqrt{13}}{3})}^{2} = k$
$15 \times \frac{7}{9} + 3 \times \frac{13}{9} = k$
$\frac{105}{9} + \frac{39}{9} = k$
$\frac{105 + 39}{9} = k$
$\frac{144}{9} = k$
$k = 16$

JEE Main 09.01.2020

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Question Bank Series

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