Let P left(frac{2 sqrt{3}}{sqrt{7}}, frac{6}{ | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$. Since $PQ$ and $RS$ are chords passing through the origin,$P, Q$ are symmetric

Vedclass · Accepted Answer

$157$

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(C) The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$. Since $PQ$ and $RS$ are chords passing through the origin,$P, Q$ are symmetric about the origin,so $O$ is the midpoint of $PQ$ and $RS$. Thus,$PQ = 2OP$ and $RS = 2OR$. We have $\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4(OP)^2} + \frac{1}{4(OR)^2} = \frac{1}{4} \left( \frac{1}{(OP)^2} + \frac{1}{(OR)^2} ight)$. Let $P = (2 \cos \alpha, 3 \sin \alpha)$ and $R = (2 \cos 	heta, 3 \sin 	heta)$. Since $OP \perp OR$,the product of their slopes is $-1$: $\left( \frac{3 \sin \alpha}{2 \cos \alpha} ight) \left( \frac{3 \sin 	heta}{2 \cos 	heta} ight) = -1$ $\Rightarrow 	an \alpha 	an 	heta = -\frac{4}{9}$. Given $P = \left( \frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}} ight)$,we have $\cos \alpha = \frac{\sqrt{3}}{\sqrt{7}}$ and $\sin \alpha = \frac{2}{\sqrt{7}}$,so $	an \alpha = \frac{2}{\sqrt{3}}$. Then $	an 	heta = -\frac{4}{9} 	imes \frac{\sqrt{3}}{2} = -\frac{2 \sqrt{3}}{9}$. Using $(OP)^2 = 4 \cos^2 \alpha + 9 \sin^2 \alpha = 4 \left( \frac{3}{7} ight) + 9 \left( \frac{4}{7} ight) = \frac{12+36}{7} = \frac{48}{7}$. Using $(OR)^2 = 4 \cos^2 	heta + 9 \sin^2 	heta = \frac{4}{\sec^2 	heta} + \frac{9 	an^2 	heta}{\sec^2 	heta} = \frac{4 + 9 	an^2 	heta}{1 + 	an^2 	heta} = \frac{4 + 9(12/81)}{1 + 12/81} = \frac{4 + 4/3}{1 + 4/27} = \frac{16/3}{31/27} = \frac{16}{3} 	imes \frac{27}{31} = \frac{144}{31}$. Substituting these: $\frac{1}{4} \left( \frac{7}{48} + \frac{31}{144} ight) = \frac{1}{4} \left( \frac{21+31}{144} ight) = \frac{52}{4 	imes 144} = \frac{13}{144}$. Thus $p=13, q=144$,so $p+q = 157$.

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