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JEE Advanced 2025 · Conic Sections

Let P(x1,y1)P(x_{1}, y_{1}) and Q(x2,y2)Q(x_{2}, y_{2}) be two distinct points on the ellipse x29+y24=1\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 such that y1>0y_{1} > 0 and y2>0y_{2} > 0. Let C denote the circle x2+y2=9x^{2} + y^{2} = 9, and M be the point (3,0)(3,0). Suppose the line x=x1x = x_{1} intersects C at R, and the line x=x2x = x_{2} intersects C at S, such that the y-coordinates of R and S are positive. Let ∠ROM=π6\angle ROM = \frac{\pi}{6} and ∠SOM=π3\angle SOM = \frac{\pi}{3}, where O denotes the origin (0,0)(0, 0). Let ∣XY∣|XY| denote the length of the line segment XY. Then which of the following statements is (are) TRUE?

Answers: A. The equation of the line joining P and Q is 2x+3y=3(1+3)2x + 3y = 3(1 + \sqrt{3}) · C. If N2=(x2,0)N_{2} = (x_{2}, 0), then 3∣N2Q∣=2∣N2S∣3|N_{2}Q| = 2|N_{2}S|

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Written solution

Let P(x₁, y₁) and Q(x₂, y₂) be two distinct points on the ellipse x²/9 + y²/4 = 1 such that y₁ > 0 and y₂ > 0. Let C denote the circle x² + y² = 9, and M be the point (3,0). Suppose the line x = x₁ intersects C at R, and the line x = x₂ intersects C at S, such that the y-coordinates of R and S are positive. Let ∠ ROM = π/6 and ∠ SOM = π/3, where O denotes the origin (0, 0). Let |XY| denote the length of the line segment XY. Then which of the followin

Ellipse: x29+y24=1,P(x1,y1), Q(x2,y2), y1,y2>0\text{Ellipse: } \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1,\quad P(x_1,y_1),\ Q(x_2,y_2),\ y_1,y_2>0

C: x2+y2=9,M(3,0),O(0,0)C:\ x^2 + y^2 = 9,\quad M(3,0),\quad O(0,0)

∠ ROM = π/6, ∠ SOM = π/3

To find: which statements (1)–(4) are TRUE\text{To find: which statements } (1)\text{–}(4) \text{ are TRUE}

R=(3cos⁡π6, 3sin⁡π6)=(332, 32)R = \left(3\cos\dfrac{\pi}{6},\ 3\sin\dfrac{\pi}{6}\right) = \left(\dfrac{3\sqrt3}{2},\ \dfrac{3}{2}\right)

S=(3cos⁡π3, 3sin⁡π3)=(32, 332)S = \left(3\cos\dfrac{\pi}{3},\ 3\sin\dfrac{\pi}{3}\right) = \left(\dfrac{3}{2},\ \dfrac{3\sqrt3}{2}\right)

x1=332,x2=32x_1 = \dfrac{3\sqrt3}{2},\qquad x_2 = \dfrac{3}{2}

x129=34⇒y12=4(1−34)=1⇒y1=1\dfrac{x_1^2}{9} = \dfrac34 \Rightarrow y_1^2 = 4\left(1 - \dfrac34\right) = 1 \Rightarrow y_1 = 1

P=(332, 1)P = \left(\dfrac{3\sqrt3}{2},\ 1\right)

x229=14⇒y22=4(1−14)=3⇒y2=3\dfrac{x_2^2}{9} = \dfrac14 \Rightarrow y_2^2 = 4\left(1 - \dfrac14\right) = 3 \Rightarrow y_2 = \sqrt3

Q=(32, 3)Q = \left(\dfrac{3}{2},\ \sqrt3\right)

m=y2−y1x2−x1=3−132−332=−23m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{\sqrt3 - 1}{\tfrac{3}{2} - \tfrac{3\sqrt3}{2}} = -\dfrac{2}{3}

y−1=−23(x−332)y - 1 = -\dfrac23\left(x - \dfrac{3\sqrt3}{2}\right)

⇒ 2x+3y=3+33=3(1+3)\Rightarrow\ 2x + 3y = 3 + 3\sqrt3 = 3(1 + \sqrt3)

(1) 2x+3y=3(1+3) ✓(1)\ 2x + 3y = 3(1+\sqrt3)\ \checkmark

(2) 2x+y=3(1+3) ×(2)\ 2x + y = 3(1+\sqrt3)\ \times

N2(32,0): ∣N2Q∣=3,∣N2S∣=332N_2\left(\dfrac32, 0\right):\ |N_2Q| = \sqrt3,\quad |N_2S| = \dfrac{3\sqrt3}{2}

3∣N2Q∣=33=2∣N2S∣ ✓3|N_2Q| = 3\sqrt3 = 2|N_2S| \ \checkmark

N1(332,0): ∣N1P∣=1,∣N1R∣=32N_1\left(\dfrac{3\sqrt3}{2}, 0\right):\ |N_1P| = 1,\quad |N_1R| = \dfrac32

9∣N1P∣=9 ≠ 6=4∣N1R∣ ×9|N_1P| = 9 \ \ne\ 6 = 4|N_1R| \ \times

PQ: 2x+3y=3(1+3)\boxed{PQ:\ 2x + 3y = 3(1 + \sqrt3)}

Correct statements: (1) and (3)\boxed{\text{Correct statements: } (1)\ \text{and}\ (3)}

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