JEE Advanced 2025 Paper 2, Mathematics Section 2 Q3: Involving Two or More Circles

JEE Advanced2025Paper 2Mathematics Section 2

Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ 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 $∠ R O M = \frac{π}{6}$ and $∠ S O M = \frac{π}{3}$ , 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?

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

Solution

Step 1: Locate the four points.

$R$ lies on $C$ with $∠ R O M = \frac{π}{6}$ : $R = (3 cos 3 0^{\circ}, 3 sin 3 0^{\circ}) = (\frac{3 3}{2}, \frac{3}{2})$ .

Similarly $S = (\frac{3}{2}, \frac{3 3}{2})$ . By construction $x_{1} = \frac{3 3}{2}$ , $x_{2} = \frac{3}{2}$ .

On the ellipse: $P = (3 cos 3 0^{\circ}, 2 sin 3 0^{\circ}) = (\frac{3 3}{2}, 1)$ and $Q = (3 cos 6 0^{\circ}, 2 sin 6 0^{\circ}) = (\frac{3}{2}, 3)$ .

Step 2: Equation of $P Q$ .

Slope $= \frac{3 - 1}{\frac{3}{2} - \frac{3 3}{2}} = - \frac{2}{3}$ .

Line: $y - 3 = - \frac{2}{3} (x - \frac{3}{2}) \Rightarrow 2 x + 3 y = 3 (3 + 1)$ .

Option (A) TRUE; (B) FALSE.

Step 3: Check the length ratios.

$∣ N_{2} Q ∣ = 3$ and $∣ N_{2} S ∣ = \frac{3 3}{2}$ , so $3∣ N_{2} Q ∣ = 33 = 2 \cdot \frac{3 3}{2} = 2∣ N_{2} S ∣$ . Option (C) TRUE.

$∣ N_{1} P ∣ = 1$ and $∣ N_{1} R ∣ = \frac{3}{2}$ , giving $9∣ N_{1} P ∣ = 9$ but $4∣ N_{1} R ∣ = 6$ . Option (D) FALSE.

$∴$ The correct options are (A) and (C).

[ADD IMAGE: Circle $x^{2} + y^{2} = 9$ with ellipse $x^{2} /9 + y^{2} /4 = 1$ inside. Points $R$ and $S$ on the circle in the upper half, with rays $O R$ and $O S$ making $3 0^{\circ}$ and $6 0^{\circ}$ with the $x$ -axis through $M = (3, 0)$ . Points $P$ and $Q$ on the ellipse directly below $R$ and $S$ respectively, with feet $N_{1}, N_{2}$ on the $x$ -axis.]

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JEE Advanced 2025 Paper 2, Mathematics Section 2 Q3: Involving Two or More Circles

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