JEE Main 2023 — Mathematics Coordinate Geometry

Given,

$P(\frac{2\sqrt{3}}{\sqrt{7}},\frac{6}{\sqrt{7}}),Q,R$P(7​23​​,7​6​),Q,R and $S$S be four points on the ellipse $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$4x2​+9y2​=1

Now, $OP={r}_{1}=\sqrt{{(\frac{2\sqrt{3}}{\sqrt{7}})}^{2}+{(\frac{6}{\sqrt{7}})}^{2}}=\sqrt{\frac{48}{7}}$OP=r1​=(7​23​​)2+(7​6​)2​=748​​ where $O$O is origin,

Let $P$P be $({r}_{1}\mathrm{cos}\theta ,{r}_{1}\mathrm{sin}\theta )$(r1​cosθ,r1​sinθ)

$P$P lies on ellipse, so we get,

$$\frac{{{r}_{1}}^{2}{\mathrm{cos}}^{2}\theta }{4}+\frac{{{r}_{1}}^{2}{\mathrm{sin}}^{2}\theta }{9}=1$$4r1​2cos2θ​+9r1​2sin2θ​=1

$$\Rightarrow \frac{{\mathrm{cos}}^{2}\theta }{4}+\frac{{\mathrm{sin}}^{2}\theta }{9}=\frac{7}{48}...(i)$$⇒4cos2θ​+9sin2θ​=487​...(i)

Let $R$R be $(-{r}_{2}\mathrm{sin}\theta ,{r}_{2}\mathrm{cos}\theta )$(−r2​sinθ,r2​cosθ) as PQ&RS are perpendicular and pass through origin,

So, $\frac{{{r}_{2}}^{2}{\mathrm{sin}}^{2}\theta }{4}+\frac{{{r}_{2}}^{2}{\mathrm{cos}}^{2}\theta }{9}=1$4r2​2sin2θ​+9r2​2cos2θ​=1

$$\Rightarrow \frac{{\mathrm{sin}}^{2}\theta }{4}+\frac{{\mathrm{cos}}^{2}\theta }{9}=\frac{1}{{{r}_{2}}^{2}}...(\mathrm{ii})$$⇒4sin2θ​+9cos2θ​=r2​21​...(ii)

Now adding equation (i)&(\mathrm{ii}) we get,

$$\frac{1}{{{r}_{2}}^{2}}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}$$r2​21​=41​+91​−487​=14431​

Now solving,

$$\frac{1}{P{Q}^{2}}+\frac{1}{R{S}^{2}}=\frac{1}{4}(\frac{1}{O{P}^{2}}+\frac{1}{O{R}^{2}})$$PQ21​+RS21​=41​(OP21​+OR21​)

$$\Rightarrow \frac{1}{P{Q}^{2}}+\frac{1}{R{S}^{2}}=\frac{1}{4}(\frac{1}{{{r}_{1}}^{2}}+\frac{1}{{{r}_{2}}^{2}})$$⇒PQ21​+RS21​=41​(r1​21​+r2​21​)

$$\Rightarrow \frac{1}{P{Q}^{2}}+\frac{1}{R{S}^{2}}=\frac{1}{4}(\frac{7}{48}+\frac{31}{144})=\frac{13}{144}=\frac{p}{m}$$⇒PQ21​+RS21​=41​(487​+14431​)=14413​=mp​

$$\therefore p+m=157$$∴p+m=157