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If the foci of the ellipse x^2 16+ y^2 b^2=1 coincide with the foci of the hyperbola x^2 144- y^2 81= 1 25, then b^2 is equal to

JEE Main 2012 — Mathematics Coordinate Geometry

2012mcqeasy

If the foci of the ellipse x216+y2b2=1\frac{x^2}{16}+\frac{y^2}{b^2}=116x2​+b2y2​=1 coincide with the foci of the hyperbola x2144−y281=125\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}144x2​−81y2​=251​, then b2b^2b2 is equal to

Official previous-year question

Held on 19 May 2012 · Verified 6 Jul 2026.

Options

  1. A

    8

  2. B

    10

  3. C

    7

  4. D

    9

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