JEE Main 2023 — Ellipse Question with Solution

Given, points $P$ and $R$ are on the ellipse defined by $9x^2+4y^2=36$ which simplifies to $\frac{x^2}{4}+\frac{y^2}{9}=1$. This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis $a=3$ along the $y$-axis and semi-minor axis $b=2$ along the $x$-axis.

OP is the distance from origin O to point P, which is given by :

$$OP=r_1=\sqrt{{\left(\frac{2\sqrt{3}}{\sqrt{7}}\right)}^2+{\left(\frac{6}{\sqrt{7}}\right)}^2}=\sqrt{\frac{12}{7}+\frac{36}{7}}=\sqrt{\frac{48}{7}}=2\sqrt{\frac{12}{7}}.$$

1. Let's represent the given point $P\left(\frac{2\sqrt{3}}{\sqrt{7}},\frac{6}{\sqrt{7}}\right)$ in polar coordinates. We can write $P$ as $(r_1\cos \theta ,r_1\sin \theta )$. Since $P$ lies on the ellipse, it must satisfy the equation of the ellipse. Substituting $x=r_1\cos \theta$ and $y=r_1\sin \theta$ into the equation of the ellipse gives us :

$$\frac{r_1^2{\cos }^2\theta }{4}+\frac{r_1^2{\sin }^2\theta }{9}=1$$

Simplifying this, we obtain :

$$\frac{{\cos }^2\theta }{4}+\frac{{\sin }^2\theta }{9}=\frac{7}{48}\text{— (equation 1)}$$

2. Similarly, if we represent the point $R$ as $(-r_2\sin \theta ,r_2\cos \theta )$, (the negative sign is due to the fact that line RS is perpendicular to line PQ. Since they are perpendicular, the angle between them is 90 degrees or $\pi /2$ radians. In terms of sin and cos, $\sin (\theta +\pi /2)=\cos (\theta )$ and $\cos (\theta +\pi /2)=-\sin (\theta )$.) it too should satisfy the equation of the ellipse. We have :

$$\frac{r_2^2{\sin }^2\theta }{4}+\frac{r_2^2{\cos }^2\theta }{9}=1$$

Simplifying this, we obtain :

$$\frac{{\sin }^2\theta }{4}+\frac{{\cos }^2\theta }{9}=\frac{1}{r_2^2}\text{— (equation 2)}$$

3. From equations (1) and (2), we have :

$$\frac{1}{r_2^2}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}$$

4. Now, note that lines $PQ$ and $RS$ are perpendicular and pass through the origin, so $PQ=2OP$ and $RS=2OR$. Thus,

$$\frac{1}{P{Q}^2}+\frac{1}{R{S}^2}=\frac{1}{4}\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}\right)$$

5. Substituting the values of $r_1$ and $r_2$, we obtain :

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

6. Hence, $p=13$ and $q=144$.

7. So, the final answer $p+q=13+144=157$.