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JEE Advanced 2025 Paper 1 · Mathematics · Conic Sections · medium

Match each conic in List-I with its eccentricity in List-II. List-I (P) An ellipse whose latus rectum is half of its minor axis (Q) A hyperbola whose…

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Match each conic in List-I with its eccentricity in List-II.

List-I
(P) An ellipse whose latus rectum is half of its minor axis
(Q) A hyperbola whose asymptotes are perpendicular to each other
(R) An ellipse whose foci and the two ends of its minor axis are the vertices of a square
(S) The conjugate hyperbola of a hyperbola of eccentricity 53\dfrac53

List-II
(1) 12\dfrac{1}{\sqrt2} (2) 54\dfrac54 (3) 2\sqrt2 (4) 32\dfrac{\sqrt3}{2} (5) 23\dfrac23
  1. A. P→4, Q→3, R→2, S→1
  2. B. P→3, Q→4, R→1, S→2
  3. C. P→4, Q→3, R→1, S→2
  4. D. P→4, Q→1, R→3, S→2
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Answer: (C) P→4, Q→3, R→1, S→2

(P) For an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 the latus rectum is 2b2a\frac{2b^2}{a} and the minor axis is 2b2b. Setting 2b2a=b\frac{2b^2}{a}=b gives b=a2b=\frac a2, so e=1−14=32e=\sqrt{1-\frac14}=\frac{\sqrt3}{2}: P→4. (Q) Perpendicular asymptotes y=±baxy=\pm\frac bax require b=ab=a, so e=1+1=2e=\sqrt{1+1}=\sqrt2: Q→3. (R) The foci (±ae,0)(\pm ae,0) and the minor-axis ends (0,±b)(0,\pm b) form a square exactly when ae=bae=b; then a2e2=a2(1−e2)a^2e^2=a^2(1-e^2), so e2=12e^2=\frac12 and e=12e=\frac1{\sqrt2}: R→1. (S) If a hyperbola and its conjugate have eccentricities e1,e2e_1,e_2, then 1e12+1e22=1\frac1{e_1^2}+\frac1{e_2^2}=1. With e1=53e_1=\frac53: 1e22=1−925=1625\frac1{e_2^2}=1-\frac9{25}=\frac{16}{25}, so e2=54e_2=\frac54: S→2. The remaining value 23\frac23 in item (5) is a distractor, obtained for example by mistakenly using the ellipse with a=3, b2=5a=3,\ b^2=5.
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