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Let $P \left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q , R$ and $S$ be four points on the ellipse $9 x^2+4 y^2=36$. Let $P Q$ and $RS$ be mutually perpendicular and pass through the origin. If $\frac{1}{( PQ )^2}+\frac{1}{( RS )^2}=\frac{ p }{ q }$, where $p$ and $q$ are coprime, then $p+q$ is equal to $.........$.
$143$
$137$
$157$
$147$
Solution
Let $R (2 \cos \theta, 3 \sin \theta)$
as $OP \perp OR$
$\text { so } \frac{3 \sin \theta}{2 \cos \theta} \times \frac{\frac{6}{\sqrt{7}}}{\frac{2 \sqrt{3}}{\sqrt{7}}}=-1$
$\Rightarrow \tan \theta=\frac{-2}{3 \sqrt{3}}$
$\Rightarrow R\left(\frac{-6 \sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}}\right) \text { or } R \left(\frac{6 \sqrt{3}}{\sqrt{31}}, \frac{-6}{\sqrt{31}}\right)$
$\text { Now }=\frac{1}{( PQ )^2}+\frac{1}{( RS )^2}=\frac{1}{4}\left(\frac{1}{( OP )^2}+\frac{1}{( OR )^2}\right)$
$=\frac{1}{4}\left(\frac{1}{\frac{48}{7}}+\frac{1}{\frac{144}{31}}\right)=\frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)$
$=\frac{13}{144}$
$\Rightarrow p+q=157$
