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?

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

(B) The equation of the line joining $P$ and $Q$ is $2 x+y=3(1+\sqrt{3})$

(D) If $N_{1}=\left(x_{1}, 0\right)$, then $9\left|N_{1} P\right|=4\left|N_{1} R\right|$

Ellipse — JEE Advanced · 2025 · 18 May 2025 · Paper 2 Question with Solution

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

R

, and the line

x = x_{2}

intersects

C

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

Answer: (A), (C)

Solution

From the circle

x^{2} + y^{2} = 9

with

∠ R O M = \frac{π}{6}

and

∠ S O M = \frac{π}{3}

R \equiv (3 cos 3 0^{\circ}, 3 sin 3 0^{\circ}) = (\frac{3 3}{2}, \frac{3}{2})

S \equiv (3 cos 6 0^{\circ}, 3 sin 6 0^{\circ}) = (\frac{3}{2}, \frac{3 3}{2})

For the ellipse

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

with same

x

-coordinates:

P \equiv (\frac{3 3}{2}, 2 sin 3 0^{\circ}) = (\frac{3 3}{2}, 1)

Q \equiv (\frac{3}{2}, 2 sin 6 0^{\circ}) = (\frac{3}{2}, 3)

Slope of

P Q

m = \frac{3 - 1}{\frac{3}{2} - \frac{3 3}{2}} = - \frac{2}{3}

Equation:

y - 3 = - \frac{2}{3} (x - \frac{3}{2})

, simplifying to

2 x + 3 y = 3 (3 + 1)

. (A) ✓

For (C):

N_{2} = (\frac{3}{2}, 0)

∣ N_{2} Q ∣ = 3

∣ N_{2} S ∣ = \frac{3 3}{2}

. So

33 = 2 \cdot \frac{3 3}{2}

✓

For (D):

N_{1} = (\frac{3 3}{2}, 0)

∣ N_{1} P ∣ = 1

∣ N_{1} R ∣ = \frac{3}{2}

. So

9 \cdot 1 \neq = 4 \cdot \frac{3}{2} = 6

✗

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · One or more correct options

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Ellipse

Solution

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