JEE Main 2022 — Ellipse Question with Solution

Given ellipse is $\frac{x^2}{2}+\frac{y^2}{4}=1 ...........\left(1\right)$,

So its eccentricity will be $a^2=b^2\left(1-e^2\right)$

$\Rightarrow 2=4\left(1-e^2\right)$

$\Rightarrow \frac{1}{2}=1-e^2$

$\Rightarrow e=\frac{1}{\sqrt{2}}$

So, focus $S$ will be $S\equiv \left(0,-ae\right)\equiv \left(0,-\sqrt{2}\right)$

Now, equation of chord of contact will be $T=0\Rightarrow$ $\frac{x}{\sqrt{2}}+\frac{\left(2\sqrt{2}-2\right)y}{4}=1$

$\Rightarrow \frac{x}{\sqrt{2}}=1-\frac{\left(\sqrt{2}-1\right)y}{2} ..........\left(2\right)$

Now on solving equation $\left(1\right) \& \left(2\right)$ we get,

$\Rightarrow y=0,\sqrt{2} \& x=\sqrt{2},1$

So points $P \& Q$ is given by $P\equiv \left(1,\sqrt{2}\right) \&\hspace{0.17em}Q\equiv \left(\sqrt{2},0\right)$

Now using the distance we get,

${\left(SP\right)}^2+{\left(SQ\right)}^2=13$