Let P(2√3√7,6√7), Q, R, and S be four points on the ellipse 9x^2+4y^2=36.

Question

Let $P\left(\dfrac{2\sqrt3}{\sqrt7},\dfrac6{\sqrt7}\right)$, $Q$, $R$, and $S$ be four points on the ellipse $9x^2+4y^2=36$. Let $PQ$ and $RS$ be mutually perpendicular chords through the origin. If $\dfrac1{(PQ)^2}+\dfrac1{(RS)^2}=\dfrac pq$, where $p,q$ are coprime, then $p+q$ is equal to:

Accepted Answer

Ellipse: $\dfrac{x^2}4+\dfrac{y^2}9=1$. $OP^2=x_P^2+y_P^2=\dfrac{12}7+\dfrac{36}7=\dfrac{48}7$.

$R=(2\cos\theta,3\sin\theta)$ with $OP\perp OR$ gives $\tan\theta=-\dfrac2{3\sqrt3}$, leading to $OR^2=\dfrac{144}{31}$.

Since $PQ=2\,OP,\ RS=2\,OR$:
$$\frac1{(PQ)^2}+\frac1{(RS)^2}=\frac14\left(\frac7{48}+\frac{31}{144}\right)=\frac14\cdot\frac{13}{36}=\frac{13}{144}.$$

$$p=13,\ q=144\ (\text{coprime}),\qquad p+q=157.$$

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