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JEE Advanced 2024 Paper 1 · Mathematics · Conic Sections · medium

List-I P. Eccentricity of the conic 9x^2+16y^2-18x+64y-71=0 Q. Eccentricity of the hyperbola conjugate to (x^2/(9)-(y^2/(16)=1 R. Distance between…

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Match the quantities in List-I with the values in List-II.

List-I

P. Eccentricity of the conic 9x2+16y2−18x+64y−71=09x^2+16y^2-18x+64y-71=0

Q. Eccentricity of the hyperbola conjugate to x29−y216=1\dfrac{x^2}{9}-\dfrac{y^2}{16}=1

R. Distance between the directrices of the hyperbola x2−y2=8x^2-y^2=8

S. Length of the latus rectum of the parabola with focus (2,1)(2,1) and directrix x+y=1x+y=1

List-II

(1) 44

(2) 222\sqrt2

(3) 74\dfrac{\sqrt7}{4}

(4) 54\dfrac54

(5) 2\sqrt2
  1. A. P→3, Q→4, R→2, S→1
  2. B. P→4, Q→3, R→1, S→2
  3. C. P→3, Q→4, R→1, S→2
  4. D. P→3, Q→5, R→1, S→2
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Answer: (C) P→3, Q→4, R→1, S→2

P: completing squares, 9(x−1)2+16(y+2)2=1449(x-1)^2+16(y+2)^2=144, so (x−1)216+(y+2)29=1\dfrac{(x-1)^2}{16}+\dfrac{(y+2)^2}{9}=1 and e=1−916=74e=\sqrt{1-\tfrac9{16}}=\dfrac{\sqrt7}{4}, which is (3). Q: the conjugate hyperbola is y216−x29=1\dfrac{y^2}{16}-\dfrac{x^2}{9}=1, with e2=1+916=2516e^2=1+\tfrac{9}{16}=\tfrac{25}{16}, so e=54e=\tfrac54, which is (4). R: x2−y2=8x^2-y^2=8 has a=22a=2\sqrt2 and e=2e=\sqrt2; the directrices are x=±a/e=±2x=\pm a/e=\pm2, so their distance is 44, which is (1). S: the distance of the focus from the directrix is ∣2+1−1∣2=2\dfrac{|2+1-1|}{\sqrt2}=\sqrt2, and the latus rectum equals twice this, 222\sqrt2, which is (2). So P→3, Q→4, R→1, S→2.
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Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake