Let the foci of a hyperbola coincide with the foci of the ellipse x 2 36 + y 2 16 = 1 . If the eccentricity of the hyperbola is 5 then the length of its latus rectum is: [2026]

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Solution

Q.
Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1 .$ If the eccentricity of the hyperbola is 5 then the length of its latus rectum is: [2026]

1 $24 \sqrt{5}$

2 $\frac{96}{\sqrt{5}}$

3 $12$

4 16

Ans.
(2)

$Let e_{1} be eccentricity of ellipse$

$\Rightarrow e_{1} = \sqrt{1 - \frac{16}{36}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$

$So a e_{1} = 6 \cdot \frac{\sqrt{5}}{3} = 2 \sqrt{5}$

$Now H : \frac{x^{2}}{p^{2}} - \frac{y^{2}}{q^{2}} = 1$

$p . e = a e_{1}$

$p \cdot 5 = 2 \sqrt{5}$

$p = \frac{2}{\sqrt{5}} \Rightarrow e^{2} = 1 + \frac{q^{2}}{p^{2}} \Rightarrow 25 = 1 + \frac{5 q^{2}}{4} \Rightarrow q^{2} = \frac{96}{5}$

$So length of LR = \frac{2 q^{2}}{p} = \frac{96}{\sqrt{5}}$

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