JEE Main 2023 — Ellipse Question with Solution

Given:

$\frac{x^2}{36}+\frac{y^2}{16}=1$

Let any point on the ellipse be $P\left(6\cos \theta , 4\sin \theta \right)$.

Equation of normal of ellipse at any point $P$ is

$3\sec \theta x–2\mathrm{cosec}\theta y=10$

This normal is also the normal of the circle passing through the point $\left(2,0\right)$.

So, $6\sec \theta =10$ or $\sin \theta =0$ (Not possible)

$\cos \theta =\frac{3}{5}$ and $\sin \theta =\frac{4}{5}$ 

So, point $P\equiv \left(\frac{18}{5},\frac{16}{5}\right)$.

So, the largest radius of circle $r=\frac{\sqrt{320}}{5}=\sqrt{\frac{64}{5}}$

So the equation of circle ${\left(x-2\right)}^2+y^2=\frac{64}{5}$

This circle is passing through $\left(1,\alpha \right)$, hence

${\left(1-2\right)}^2+{\alpha }^2=\frac{64}{5}$

$\Rightarrow {\alpha }^2=\frac{59}{5}$

$\Rightarrow 10{\alpha }^2=118$.