Ellipse
PYP_JEE_ADV_2025_P2
Grade None
Question:
Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two distinct points on the ellipse
$$\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$$
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 $\angle ROM = \dfrac{\pi}{6}$ and $\angle SOM = \dfrac{\pi}{3}$, where $O$ denotes the origin $(0, 0)$. Let $|XY|$ denote the length of the line segment $XY$.
Then which of the following statements is (are) TRUE?
The equation of the line joining $P$ and $Q$ is $2x + 3y = 3(1 + \sqrt{3})$
The equation of the line joining $P$ and $Q$ is $2x + y = 3(1 + \sqrt{3})$
If $N_2 = (x_2, 0)$, then $3|N_2 Q| = 2|N_2 S|$
If $N_1 = (x_1, 0)$, then $9|N_1 P| = 4|N_1 R|$
Step-by-Step Solution
Key Concept: Parametric coordinates of circle and ellipse, and the relationship between ordinate values of points on an auxiliary circle and the corresponding points on the ellipse.
The circle $C: x^2 + y^2 = 9$ has radius $R_c = 3$. Point $M = (3, 0)$ is on the positive $x$-axis.
- For $R$: polar angle is $\pi/6 \implies R = (3\cos(\pi/6), 3\sin(\pi/6)) = \left(\dfrac{3\sqrt{3}}{2}, \dfrac{3}{2}\right)$.
- For $S$: polar angle is $\pi/3 \implies S = (3\cos(\pi/3), 3\sin(\pi/3)) = \left(\dfrac{3}{2}, \dfrac{3\sqrt{3}}{2}\right)$.
The ellipse is $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$. Let $y_1, y_2 > 0$:
- Point $P$: has $x_1 = \dfrac{3\sqrt{3}}{2} \implies \dfrac{27/4}{9} + \dfrac{y_1^2}{4} = 1 \implies y_1 = 1$. Thus, $P = \left(\dfrac{3\sqrt{3}}{2}, 1\right)$.
- Point $Q$: has $x_2 = \dfrac{3}{2} \implies \dfrac{9/4}{9} + \dfrac{y_2^2}{4} = 1 \implies y_2 = \sqrt{3}$. Thus, $Q = \left(\dfrac{3}{2}, \sqrt{3}\right)$.
The line joins $P\left(\dfrac{3\sqrt{3}}{2}, 1\right)$ and $Q\left(\dfrac{3}{2}, \sqrt{3}\right)$. The slope is:
$$m = \dfrac{\sqrt{3}-1}{3/2 - 3\sqrt{3}/2} = -\dfrac{2}{3}$$
Equation: $y - 1 = -\dfrac{2}{3}\left(x - \dfrac{3\sqrt{3}}{2}\right) \implies 2x + 3y = 3(1+\sqrt{3})$ (Option A is True).
Verify the ratio statements:
- For statement C: $N_2 = (x_2, 0) = (3/2, 0)$. $|N_2 Q| = y_2 = \sqrt{3}$ and $|N_2 S| = y_S = \dfrac{3\sqrt{3}}{2}$.
$$3|N_2 Q| = 3\sqrt{3}, \quad 2|N_2 S| = 3\sqrt{3} \implies 3|N_2 Q| = 2|N_2 S| \text{ (True)}.$$
- For statement D: $N_1 = (x_1, 0) = (3\sqrt{3}/2, 0)$. $|N_1 P| = y_1 = 1$ and $|N_1 R| = y_R = 3/2$.
$$9|N_1 P| = 9, \quad 4|N_1 R| = 6 \implies 9|N_1 P| \neq 4|N_1 R| \text{ (False)}.$$
Correct Answer: A, C