JEE Main 2023 — Hyperbola Question with Solution

Given,

The eccentricity of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is reciprocal to that of the hyperbola $2x^2-2y^2=1$,

Now eccentricity of rectangular hyperbola $2x^2-2y^2=1$ is $e_H=\sqrt{2}$

So, $e_e=\frac{1}{\sqrt{2}}$

Now, focus of hyperbola $=\left(\pm 1, 0\right)$

Now given that both curve intersect orthogonally, so ellipse and hyperbola are confocal

So, for ellipse $a{e}_e=1\Rightarrow a=\sqrt{2}$

Now length of latusrectum $\mathrm{L}.\mathrm{R}.=\frac{2b^2}{a}=2a\left(1-{e_e}^2\right)$

$=2\sqrt{2}.\frac{1}{2}=\sqrt{2}$