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:

• A. $\sqrt{30}$
• B. $\frac{4 \sqrt{5}}{3}$
• C. $2 \sqrt{6}$
• D. $\frac{2 \sqrt{5}}{3}$

Answer 1

Answer: (B)

Given $\theta \in\left(0, \frac{\pi}{2}\right)$
equation of hyperbola $\Rightarrow x^{2}-y^{2} \sec ^{2} \theta=10$
$\Rightarrow \frac{x^{2}}{10}-\frac{y^{2}}{10 \cos ^{2} \theta}=1$
Hence eccentricity of hyperbola
$\left( e _{ H }\right)=\sqrt{1+\frac{10 \cos ^{2} \theta}{10}} \ldots(1) \quad\left\{ e =\sqrt{1+\frac{ b ^{2}}{ a ^{2}}}\right\}$
Now equation of ellipse $\Rightarrow x^{2} \sec ^{2} \theta+y^{2}=5$
$\Rightarrow \frac{ x ^{2}}{5 \cos ^{2} \theta}+\frac{ y ^{2}}{5}=1 \,\left\{ e =\sqrt{1-\frac{ a ^{2}}{ b ^{2}}}\right\}$
Hence eccenticity of ellipse
$\left(e_{E}\right)=\sqrt{1-\frac{5 \cos ^{2} \theta}{5}}$
$\left( e _{ E }\right)=\sqrt{1-\cos ^{2} \theta}=|\sin \theta|=\sin \theta \quad \ldots(2)$
$\left\{\because \theta \in\left(0, \frac{\pi}{2}\right)\right\}$
given $\Rightarrow e _{ H }=\sqrt{5} e$
Hence $1+\cos ^{2} \theta=5 \sin ^{2} \theta$
$1+\cos ^{2} \theta=5\left(1-\cos ^{2} \theta\right)$
$1+\cos ^{2} \theta=5-5 \cos ^{2} \theta$
$6 \cos ^{2} \theta=4$
$\cos ^{2} \theta=\frac{2}{3} \dots$(3)
Now length of latus rectum of ellipse
$=\frac{2 a^{2}}{b}=\frac{10 \cos ^{2} \theta}{\sqrt{5}}=\frac{20}{3 \sqrt{5}}=\frac{4 \sqrt{5}}{3}$