JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Ellipse

The ellipse is $\frac{x^2}{4}+\frac{y^2}{9}=1$. Point $P$ lies on it. Let $R(2\cos \theta ,3\sin \theta )$ be another point.

Since $OP\perp OR$, the product of their slopes is $-1$:
$\frac{3\sin \theta }{2\cos \theta }\cdot \frac{6/\sqrt{7}}{2\sqrt{3}/\sqrt{7}}=-1$

This gives $\tan \theta =-\frac{2}{3\sqrt{3}}$, leading to $R\left(\frac{6\sqrt{3}}{\sqrt{31}},\frac{-6}{\sqrt{31}}\right)$.

For perpendicular chords through the origin:
$\frac{1}{(PQ{)}^2}+\frac{1}{(RS{)}^2}=\frac{1}{4}\left(\frac{1}{(OP{)}^2}+\frac{1}{(OR{)}^2}\right)$

Computing: $O{P}^2=\frac{48}{7}$, $O{R}^2=\frac{144}{31}$

$\frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)=\frac{1}{4}\cdot \frac{52}{144}=\frac{13}{144}$

So $p=13$, $q=144$, and $p+q=157$.