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← JEE Advanced 2025 · Paper 2
MathematicsMultiple correct+4 / -2Question 7 of 48

JEE Advanced 2025 · Paper 2, Mathematics question 7

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\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 such that y1>0y_1 > 0, and y2>0y_2 > 0. Let CC denote the circle x2+y2=9x^2 + y^2 = 9, and MM be the point (3,0)(3, 0).

Suppose the line x=x1x = x_1 intersects CC at RR, and the line x=x2x = x_2 intersects CC at SS, such that the yy-coordinates of RR and SS are positive. Let ∠ROM=π6\angle ROM = \dfrac{\pi}{6} and ∠SOM=π3\angle SOM = \dfrac{\pi}{3}, where OO denotes the origin (0,0)(0, 0). Let ∣XY∣|XY| denote the length of the line segment XYXY.

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

Choose one or more options
Answer and worked solution

Official answer: A, C

RR and SS lie on the auxiliary circle, so PP and QQ have eccentric angles π6\frac\pi6 and π3\frac\pi3:
P=(332,1),Q=(32,3).P=\left(\tfrac{3\sqrt3}2,1\right),\qquad Q=\left(\tfrac32,\sqrt3\right).

Line PQ. The slope is 3−132(1−3)=−23\frac{\sqrt3-1}{\frac32(1-\sqrt3)}=-\frac23, giving
2x+3y=3(1+3).2x+3y=3(1+\sqrt3).
So (A) is true and (B) is false.

Ordinate ratios.

  • (C) ∣N2Q∣=3|N_2Q|=\sqrt3 and ∣N2S∣=332|N_2S|=\frac{3\sqrt3}2, so 3∣N2Q∣=2∣N2S∣3|N_2Q|=2|N_2S|. ✓
  • (D) ∣N1P∣=1|N_1P|=1 and ∣N1R∣=32|N_1R|=\frac32, so 9≠69\ne6. ✗