JEE Advanced 2025 Paper 2 Q07 Mathematics Coordinate Geometry Ellipse Hard

JEE Advanced 2025 Paper 2 · Q07 · Ellipse

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 $\angle ROM = \frac{\pi}{6}$ and $\angle SOM = \frac{\pi}{3}$, where $O$ denotes the origin $(0, 0)$. Let $|XY|$ denote the length of the segment $XY$.

Then which of the following statements is (are) TRUE?

  1. A. The equation of the line joining $P$ and $Q$ is $2x + 3y = 3(1 + \sqrt{3})$
  2. B. The equation of the line joining $P$ and $Q$ is $2x + y = 3(1 + \sqrt{3})$
  3. C. If $N_2 = (x_2, 0)$, then $3 |N_2 Q| = 2 |N_2 S|$
  4. D. If $N_1 = (x_1, 0)$, then $9 |N_1 P| = 4 |N_1 R|$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C

Solution

$R$ is on circle of radius $3$ with angle $\angle ROM = \pi/6$ from positive $x$-axis: $R = (3 \cos(\pi/6), 3 \sin(\pi/6)) = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$. So $x_1 = \frac{3\sqrt{3}}{2}$. $P$ on ellipse with this $x$: $y_1^2/4 = 1 - 3/4 = 1/4$ $\to$ $y_1 = 1$. $P = \left(\frac{3\sqrt{3}}{2}, 1\right)$. Similarly $S = (3 \cos(\pi/3), 3 \sin(\pi/3)) = \left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$; $x_2 = \frac{3}{2}$ $\to$ $y_2 = \sqrt{3}$. $Q = \left(\frac{3}{2}, \sqrt{3}\right)$. Slope $PQ = (\sqrt{3} - 1) / (3/2 - 3\sqrt{3}/2) = -2/3$. Line: $y - 1 = -\frac{2}{3}(x - \frac{3\sqrt{3}}{2})$ $\to$ $2x + 3y = 3 + 3\sqrt{3} = 3(1 + \sqrt{3})$. (A) TRUE. (B) FALSE. (C) $|N_2 Q| = \sqrt{3}$; $|N_2 S| = \frac{3\sqrt{3}}{2}$. $3 \cdot \sqrt{3} = 3\sqrt{3} = 2 \cdot \frac{3\sqrt{3}}{2}$. EQUAL $\to$ TRUE. (D) $|N_1 P| = 1$; $|N_1 R| = \frac{3}{2}$. $9 \cdot 1 = 9 \ne 4 \cdot \frac{3}{2} = 6$. FALSE. Correct: (A), (C).

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