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JEE Advanced 2018 Paper 2 · Mathematics · Conic Sections · medium

(x^2/(25)+(y^2/(16)=1. List-I (P) The product of the perpendicular distances from the two foci of E to a variable tangent of E. (Q) The radius of the…

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Consider the ellipse E: x225+y216=1E:\ \dfrac{x^{2}}{25}+\dfrac{y^{2}}{16}=1. Match the quantities in List-I with the values in List-II.

List-I

(P) The product of the perpendicular distances from the two foci of EE to a variable tangent of EE.

(Q) The radius of the circle that is the locus of the foot of the perpendicular drawn from a focus of EE to a variable tangent.

(R) The square of the radius of the director circle of EE (the locus of points from which two perpendicular tangents can be drawn to EE).

(S) The least area of the triangle formed by a tangent to EE at a point in the first quadrant and the coordinate axes.

List-II

(1) 2020

(2) 1616

(3) 55

(4) 4141

(5) 99
  1. A. P→2, Q→3, R→4, S→1
  2. B. P→2, Q→3, R→1, S→4
  3. C. P→5, Q→3, R→4, S→1
  4. D. P→2, Q→4, R→3, S→1
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Answer: (A) P→2, Q→3, R→4, S→1

The foci are (±3,0)(\pm3,0) since c2=25−16=9c^{2}=25-16=9. (P) For the tangent xcos⁡θ5+ysin⁡θ4=1\frac{x\cos\theta}{5}+\frac{y\sin\theta}{4}=1 the two focal distances multiply to b2=16b^{2}=16, which is (2). (Q) The foot of the perpendicular from a focus to any tangent lies on the auxiliary circle x2+y2=a2x^{2}+y^{2}=a^{2}, of radius 55, which is (3). (R) The director circle is x2+y2=a2+b2=41x^{2}+y^{2}=a^{2}+b^{2}=41, so the square of its radius is 4141, which is (4). (S) The tangent has intercepts 5cos⁡θ\frac5{\cos\theta} and 4sin⁡θ\frac4{\sin\theta}, so the triangle area is 10sin⁡θcos⁡θ=20sin⁡2θ≥20=ab\frac{10}{\sin\theta\cos\theta}=\frac{20}{\sin2\theta}\ge20=ab, which is (1). Hence P→2, Q→3, R→4, S→1. Numerical checks with random tangents confirmed the product 1616, the radius 55, perpendicular slopes at a point on the circle x2+y2=41x^2+y^2=41, and the minimum area 2020. The value (5) is a distractor.
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