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
$\left(e_H\right)=\sqrt{1+\frac{10{\cos }^2\theta }{10}}\dots (1)\left\{e=\sqrt{1+\frac{b^2}{a^2}}\right\}$
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
$\left(e_E\right)=\sqrt{1-\frac{5{\cos }^2\theta }{5}}$
$\left(e_E\right)=\sqrt{1-{\cos }^2\theta }=|\sin \theta |=\sin \theta \dots (2)$
$\left\{\because \theta \in \left(0,\frac{\pi }{2}\right)\right\}$
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
$=\frac{2a^2}{b}=\frac{10{\cos }^2\theta }{\sqrt{5}}=\frac{20}{3\sqrt{5}}=\frac{4\sqrt{5}}{3}$