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The foci of the ellipse x^2 16+ y^2 b^2=1 and the hyperbola x^2 144- y^2 81= 1 25 coincide. Then the value of b^2 is

JEE Main 2003 — Mathematics Coordinate Geometry

2003mcqeasy

The foci of the ellipse x216+y2b2=1\frac{x^2}{16}+\frac{y^2}{b^2}=116x2​+b2y2​=1 and the hyperbola x2144−y281=125\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}144x2​−81y2​=251​ coincide. Then the value of b2b^2b2 is

Official previous-year question

Held on 30 Apr 2003 · Verified 6 Jul 2026.

Options

  1. A

    9

  2. B

    1

  3. C

    5

  4. D

    7

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