JEE Main 2023 — Ellipse Question with Solution

Given,

$P\left(\frac{2\sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q, R$ and $S$ be four points on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$

Now, $OP=r_1=\sqrt{{\left(\frac{2\sqrt{3}}{\sqrt{7}}\right)}^2+{\left(\frac{6}{\sqrt{7}}\right)}^2}=\sqrt{\frac{48}{7}}$ where $O$ is origin,

Let $P$ be $\left(r_1\cos \theta , r_1\sin \theta \right)$

$P$ lies on ellipse, so we get,

$\frac{{r_1}^2{\cos }^2\theta }{4}+\frac{{\displaystyle {r_1}^2{\sin }^2\theta }}{{\displaystyle 9}}=1$

$\Rightarrow \frac{{\cos }^2\theta }{4}+\frac{{\sin }^2\theta }{9}=\frac{7}{48} ... \left(\mathrm{i}\right)$

Let $R$ be $\left(-r_2\sin \theta , r_2\cos \theta \right)$ as $PQ\hspace{0.17em}\& RS$ are perpendicular and pass through origin,

So, $\frac{{r_2}^2{\sin }^2\theta }{4}+\frac{{r_2}^2{\cos }^2\theta }{9}=1$

$\Rightarrow \frac{{\sin }^2\theta }{4}+\frac{{\cos }^2\theta }{9}=\frac{1}{{r_2}^2} ...\left(\mathrm{ii}\right)$

Now adding equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$ we get,

$\frac{1}{{r_2}^2}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}$

Now solving,

$\frac{1}{P{Q}^2}+\frac{1}{R{S}^2}=\frac{1}{4}\left(\frac{1}{O{P}^2}+\frac{1}{O{R}^2}\right)$

$\Rightarrow \frac{1}{P{Q}^2}+\frac{1}{R{S}^2}=\frac{1}{4}\left(\frac{1}{{r_1}^2}+\frac{1}{{r_2}^2}\right)$

$\Rightarrow \frac{1}{P{Q}^2}+\frac{1}{R{S}^2}=\frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)=\frac{13}{144}=\frac{p}{m}$

$\therefore p+m=157$