Mathematics · Ellipse

JEE Advanced 2025 — Paper 2 — Question 23

Let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ be two distinct points on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ such that $y_{1}>0$ , and $y_{2}>0$ . Let C denote the circle $x^{2}+y^{2}=9$ , and $M$ be the point $(3,0)$ . Suppose the line $x=x_{1}$ intersects $C$ at $R$ , and the line $x=x_{2}$ intersects $C$ at $S$ , such that the $y$ -coordinates of $R$ and $S$ are positive. Let $\angle R O M=\frac{\pi}{6}$ and $\angle S O M=\frac{\pi}{3}$ , where $O$ denotes the origin $(0,0)$ . Let $|X Y|$ denote the length of the line segment $X Y$ . Then which of the following statements is (are) TRUE?

Option A:
The equation of the line joining $P$ and $Q$ is $2 x+3 y=3(1+\sqrt{3})$
Correct
Option B:
The equation of the line joining $P$ and $Q$ is $2 x+y=3(1+\sqrt{3})$
Option C:
If $N_{2}=\left(x_{2}, 0\right)$ , then $3\left|N_{2} Q\right|=2\left|N_{2} S\right|$
Correct
Option D:
If $N_{1}=\left(x_{1}, 0\right)$ , then $9\left|N_{1} P\right|=4\left|N_{1} R\right|$

Answer: A, C

Step-by-step solution

$\mathrm{P} \equiv\left(3 \cos 30^{\circ}, 2 \sin 30^{\circ}\right) \equiv\left(\frac{3 \sqrt{3}}{2}, 1\right)$

$\mathrm{Q} \equiv\left(3 \cos 60^{\circ}, 2 \sin 60^{\circ}\right) \equiv\left(\frac{3}{2}, \sqrt{3}\right)$

$\mathrm{R}\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right), \mathrm{S}\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)$

Slope of $\mathrm{PQ}=\mathrm{m}_{\mathrm{PQ}}=\frac{\sqrt{3}-1}{\frac{3}{2}-\frac{3 \sqrt{3}}{2}}=-\frac{2}{3}$

Equation of line PQ $\mathrm{y}-\sqrt{3}=-\frac{2}{3}\left(\mathrm{x}-\frac{3}{2}\right)$

$\Rightarrow 2 x+3 y=3(\sqrt{3}+1) \quad$ option (A) is correct

Now if $\mathrm{N}_{2}=\left(\mathrm{x}_{2}, 0\right)=\left(\frac{3}{2}, 0\right)$

$\left|\mathrm{N}_{2} \mathrm{Q}\right|=\sqrt{3}$ and $\left|\mathrm{N}_{2} \mathrm{~S}\right|=\frac{3 \sqrt{3}}{2}$

$\Rightarrow 3\left|\mathrm{~N}_{2} \mathrm{Q}\right|=2\left|\mathrm{~N}_{2} \mathrm{~S}\right| \quad$ option (C) is correct

Now, if $\mathrm{N}_{1}=\left(\mathrm{x}_{1}, 0\right) \Rightarrow \mathrm{N}_{1}=\left(\frac{3 \sqrt{3}}{2}, 0\right)$

$\Rightarrow\left|\mathrm{N}_{1} \mathrm{P}\right|=1,\left|\mathrm{~N}_{1} \mathrm{R}\right|=\frac{3}{2} \quad$ option (D) is incorrect

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Ellipse
Topic: Tangents & Normals to ellipse, chord of conatct