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Jee Main 2025Medium JEE math MCQ

Let \( E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b \) and \( H: \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 \). Let the distance between the foci of \( E \) and the foci of \( H \) be \( 2\sqrt{3} \). If \( a - A = 2 \), and the ratio of the eccentricities of \( E \) and \( H \) is \( \frac{1}{3} \), then the sum of the lengths of their latus rectums is equal to
  1. A. 10
  2. B. 9
  3. C. 8
  4. D. 7

Solution

The correct option is **C**. (C. 8)

MATH

mediumPYQ Reworded
Question
Read carefully, then pick the best option.
Let E:x2a2+y2b2=1,a>b E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b and H:x2A2y2B2=1 H: \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 . Let the distance between the foci of E E and the foci of H H be 23 2\sqrt{3} . If aA=2 a - A = 2 , and the ratio of the eccentricities of E E and H H is 13 \frac{1}{3} , then the sum of the lengths of their latus rectums is equal to
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Jee Main 2025 — Medium JEE Mathematics MCQ | MyGoalPrep