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Hyperbola question

2003 · Shift 0 · Q114

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JEE Main
Mathematics
Hyperbola
2003 · Shift 0 · Q114

Hyperbola question

2003 · Shift 0 · Q114

JEE MainMathematicsHyperbolaMCQ+4 / −1

The foci of the ellipse

{{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1

and the hyperbola

{{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}

coincide. Then the value of

{b^2}

is :

A
$9$
B
$1$
C
$5$
D
$7$

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