JEE Main 2023 — Hyperbola Question with Solution

Given:

$\frac{x^2}{4}+\frac{y^2}{b^2}=1$

Equation of normal to the ellipse is

$2x\sec \theta -by\mathrm{cosec}\theta =4-b^2$

Distance of normal from origin is

$d=\left|\frac{4-b^2}{\sqrt{4{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta }}\right|$

For maximum distance, denominator must be minimum.

$\sqrt{4{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta }$

$=\sqrt{4+4{\tan }^2\theta +b^2+b^2{\cot }^2\theta }$

Now,

$\frac{4{\tan }^2\theta +b^2{\cot }^2\theta }{2}\ge \sqrt{4{\tan }^2\theta \times b^2{\cot }^2\theta }$

$\Rightarrow 4{\tan }^2\theta +b^2{\cot }^2\theta \ge 4b$

$\Rightarrow 4+b^2+4{\tan }^2\theta +b^2{\cot }^2\theta \ge 4+b^2+4b$

$\Rightarrow 4+b^2+4{\tan }^2\theta +b^2{\cot }^2\theta \ge {\left(2+b\right)}^2$

So, minimum value of $\sqrt{4{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta }$ is $\sqrt{{\left(2+b\right)}^2}=b+2$

So,

$d=\left|\frac{4-b^2}{2+b}\right|=1$

$\Rightarrow \left|2-b\right|=1$

$\Rightarrow 2-b=\pm 1$

$\Rightarrow b=1, 3$

But $b<2$, so $b=1$, hence

$e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$