Let P( 2 3 7 , 6 7 ), Q, R and S be four points on the ellipse 9 x...

Given,

$P\left(\dfrac{2\sqrt{3}}{\sqrt{7}}, \dfrac{6}{\sqrt{7}}\right), Q, R and$S$be four points on the ellipse$\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{9} = 1$</p><p>Now,$OP = {r}_{1} = \sqrt{{\left(\dfrac{2\sqrt{3}}{\sqrt{7}}\right)}^{2} + {\left(\dfrac{6}{\sqrt{7}}\right)}^{2}} = \sqrt{\dfrac{48}{7}}$where$O$is origin,</p><p>Let$P$be \left({r}_{1}cos\theta, {r}_{1}sin\theta\right)$

$P$ lies on ellipse, so we get,

$\dfrac{{{r}_{1}}^{2}\cos^{2}\theta}{4} + \dfrac{{{r}_{1}}^{2}\sin^{2}\theta}{9} = 1$

$⇒ \dfrac{\cos^{2}\theta}{4} + \dfrac{\sin^{2}\theta}{9} = \dfrac{7}{48} . . . \left(i\right)$

Let $R$ be \left(- {r}_{2}sin\theta, {r}_{2}cos\theta\right)$as$PQ \text{ and } RS$are perpendicular and pass through origin,</p><p>So,$\dfrac{{{r}_{2}}^{2}\sin^{2}\theta}{4} + \dfrac{{{r}_{2}}^{2}\cos^{2}\theta}{9} = 1$</p><p>$⇒ \dfrac{\sin^{2}\theta}{4} + \dfrac{\cos^{2}\theta}{9} = \dfrac{1}{{{r}_{2}}^{2}} . . . \left(ii\right)$</p><p>Now adding equation \left(i\right) \text{ and } \left(ii\right)$ we get,

$\dfrac{1}{{{r}_{2}}^{2}} = \dfrac{1}{4} + \dfrac{1}{9} - \dfrac{7}{48} = \dfrac{31}{144}$

Now solving,

$\dfrac{1}{P{Q}^{2}} + \dfrac{1}{R{S}^{2}} = \dfrac{1}{4}\left(\dfrac{1}{O{P}^{2}} + \dfrac{1}{O{R}^{2}}\right)$

$⇒ \dfrac{1}{P{Q}^{2}} + \dfrac{1}{R{S}^{2}} = \dfrac{1}{4}\left(\dfrac{1}{{{r}_{1}}^{2}} + \dfrac{1}{{{r}_{2}}^{2}}\right)$

$⇒ \dfrac{1}{P{Q}^{2}} + \dfrac{1}{R{S}^{2}} = \dfrac{1}{4}\left(\dfrac{7}{48} + \dfrac{31}{144}\right) = \dfrac{13}{144} = \dfrac{p}{m}$

$∴ p + m = 157$