Let $P (\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}),$ Q, R and S be four points on the ellipse $9 x^{2} + 4 y^{2} = 36 .$ Let PQ 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

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143

137

147

157

answer is D.

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Detailed Solution

Given $r_{1} = \sqrt{\frac{12}{49} + \frac{36}{49}} = \frac{4 \sqrt{3}}{7}$
$\frac{{r_{1}}^{2} \cos^{2} θ}{4} + \frac{{r_{1}}^{2} \sin^{2} θ}{9} = 1 \Rightarrow \frac{\cos^{2} θ}{4} + \frac{\sin^{2} θ}{9} = \frac{1}{{r_{1}}^{2}}$

$\frac{{r_{2}}^{2} \sin^{2} θ}{4} + \frac{{r_{2}}^{2} \cos^{2} θ}{9} = 1 \Rightarrow \frac{\sin^{2} θ}{4} + \frac{\cos^{2} θ}{9} = \frac{1}{{r_{2}}^{2}}$
$∴ \frac{1}{{r_{1}}^{2}} + \frac{1}{{r_{2}}^{2}} = \frac{1}{4} + \frac{1}{9} = \frac{13}{36}$
$\frac{1}{{PQ}^{2}} + \frac{1}{{RS}^{2}} = \frac{1}{4 {r_{1}}^{2}} + \frac{1}{4 {r_{2}}^{2}} = \frac{1}{4} [\frac{13}{36}] = \frac{13}{144} = \frac{p}{q} ∴ p + q = 157$