Let P left(frac{2 √{3}}{√{7}}, frac{6}{√{7}}right), Q , R and S be four points on ellipse 9 x^

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10-2. Parabola, Ellipse, Hyperbola

hard

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 $.........$.

$143$

$137$

$157$

$147$

(JEE MAIN-2023)

Solution

Let $R (2 \cos \theta, 3 \sin \theta)$

as $OP \perp OR$

$\text { so } \frac{3 \sin \theta}{2 \cos \theta} \times \frac{\frac{6}{\sqrt{7}}}{\frac{2 \sqrt{3}}{\sqrt{7}}}=-1$

$\Rightarrow \tan \theta=\frac{-2}{3 \sqrt{3}}$

$\Rightarrow R\left(\frac{-6 \sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}}\right) \text { or } R \left(\frac{6 \sqrt{3}}{\sqrt{31}}, \frac{-6}{\sqrt{31}}\right)$

$\text { Now }=\frac{1}{( PQ )^2}+\frac{1}{( RS )^2}=\frac{1}{4}\left(\frac{1}{( OP )^2}+\frac{1}{( OR )^2}\right)$

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

$=\frac{13}{144}$

$\Rightarrow p+q=157$

Standard 11

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