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Let the foci of the ellipse x^2 16+ y^2 7=1 and the hyperbola x^2 144- y^2 α = 1 25 coincide. Then the length of the latus rectum of the hyperbola is:

JEE Main 2022 — Mathematics Coordinate Geometry

2022mcqeasy

Let the foci of the ellipse x216+y27=1\frac{{x}^{2}}{16}+\frac{{y}^{2}}{7}=116x2​+7y2​=1 and the hyperbola x2144−y2α=125\frac{{x}^{2}}{144}-\frac{{y}^{2}}{\alpha }=\frac{1}{25}144x2​−αy2​=251​ coincide. Then the length of the latus rectum of the hyperbola is:

Official previous-year question

Held on 25 Jul 2022 · Verified 6 Jul 2026.

Options

  1. A

    329\frac{32}{9}932​

  2. B

    185\frac{18}{5}518​

  3. C

    274\frac{27}{4}427​

  4. D

    2710\frac{27}{10}1027​

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