# For some

Question:

For some $\theta \in\left(0, \frac{\pi}{2}\right)$, if the eccentricity of the hyperbola, $x^{2}-y^{2} \sec ^{2} \theta=10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec ^{2} \theta+y^{2}=5$, then the length of the latus rectum of the ellipse, is :

1. (1) $2 \sqrt{6}$

2. (2) $\sqrt{30}$

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

4. (4) $\frac{4 \sqrt{5}}{3}$

Correct Option: , 4

Solution:

Hyperbola : $\frac{x^{2}}{10}-\frac{y^{2}}{10 \cos ^{2} \theta}=1 \Rightarrow e_{1}=\sqrt{1+\cos ^{2} \theta}$

and Ellipse : $\frac{x^{2}}{5 \cos ^{2} \theta}+\frac{y^{2}}{5}=1$

$\Rightarrow e_{2}=\sqrt{1-\cos ^{2} \theta}=\sin \theta$

According to the question, $e_{1}=\sqrt{5} e_{2}$

$\Rightarrow 1+\cos ^{2} \theta=5 \sin ^{2} \theta \Rightarrow \cos ^{2} \theta=\frac{2}{3}$

Now length of latus rectum of ellipse

$=\frac{2 a^{2}}{b}=\frac{10 \cos ^{2} \theta}{\sqrt{5}}=\frac{20}{3 \sqrt{5}}=\frac{4 \sqrt{5}}{3}$