Question

For some $\theta \in (0,\frac{\pi }{2})$, 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. $\frac{4\sqrt{5}}{3}$
• B. $\frac{2\sqrt{5}}{3}$
• C. $2\sqrt{6}$
• D. $\sqrt{30}$

Answer 1

The correct option is A $\frac{4\sqrt{5}}{3}$

Equation of the Hyperbola is $x^2-y^2{\sec }^2\theta =10$

Equation of the Ellipse is $x^2{\sec }^2\theta +y^2=5$

Given that $\sqrt{1+\frac{10{\cos }^2\theta }{10}}=\sqrt{5}\sqrt{1-\frac{5{\cos }^2\theta }{5}}$
$\Rightarrow 1+{\cos }^2\theta =5-5{\cos }^2\theta$
$\Rightarrow 6{\cos }^2\theta =4$
$\Rightarrow \cos \theta =\pm \sqrt{\frac{2}{3}}$

Length of the Latus rectum of ellipse = $\frac{2\cdot 5{\cos }^2\theta }{\sqrt{5}}$
$=2\sqrt{5}\cdot \frac{2}{3}=\frac{4\sqrt{5}}{3}$