JEE Main 2023 — Ellipse Question with Solution

Given equation of ellipse is $\frac{x^2}{36}+\frac{y^2}{9}=1$

It is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

With $a=6, b=3$.

The centre of the circle is $C\left(2,0\right)$.

Let $P\left(a\cos \theta ,b\sin \theta \right)$ be the common point for ellipse and circle.

$P\equiv \left(6\cos \theta ,3\sin \theta \right)$

Equation of normal to the ellipse will be $\frac{ax}{\cos \theta }-\frac{by}{\sin {\displaystyle \theta }}=a^2-b^2$.

$\Rightarrow 6x\mathrm{sec\theta }-3y\mathrm{cosec\theta }=36-9$

But the normal through $P$ passes through the centre of the circle $C\left(2,0\right)$.

$\Rightarrow 6\left(2\right)\mathrm{sec\theta }-3\left(0\right)\mathrm{cosec\theta }=27$

$\Rightarrow \mathrm{sec\theta }=\frac{27}{12}$

$\Rightarrow \cos \theta =\frac{4}{9}$ and $\sin \theta =\frac{\sqrt{65}}{9}$

$P\equiv \left(6\times \frac{4}{9},3\times \frac{\sqrt{65}}{9}\right)$

$P\equiv \left(\frac{8}{3},\frac{\sqrt{65}}{3}\right)$

Now $CP=r=\sqrt{{\left(2-\frac{8}{3}\right)}^2+{\left(0-\frac{\sqrt{65}}{3}\right)}^2}$

$\Rightarrow r^2=\frac{69}{9}$

$\Rightarrow 12r^2=12\times \frac{69}{9}=92$

Therefore, the required answer is $92$