JEE Main 2020 (Online) 2nd September Evening ShiftQuestion 1
For some $\theta\in\left(0,\frac\pi2\right)$, if the eccentricity of the hyperbola, $x^2-y^2\sec^2\theta=10$ is $\sqrt5$ 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 :
Ellipse
Solution
Let $c=\cos^2\theta$. Hyperbola $\frac{x^2}{10}-\frac{y^2}{10c}=1$: $e_h^2=1+c$. Ellipse $\frac{x^2}{5c}+\frac{y^2}{5}=1$ (since $c<1$, major axis along $y$): $e_e^2=1-c$. $e_h^2=5e_e^2\Rightarrow1+c=5-5c\Rightarrow c=\tfrac23$. Latus rectum $=\frac{2b^2}{a}=\frac{2\cdot5c}{\sqrt5}=\frac{20/3}{\sqrt5}=\frac{4\sqrt5}{3}$. Option (C).