The foci of the ellipse (x²)/16 + (y²)/(b²) = 1 and the hyperbola (x²)/144 - (y²)/81 = 1/25 coincide. Then the value of b² is:

Coordinate Geometry: JEE Main 2003 Mathematics question

JEE Main 2003 · Annual, Paper 1 · Q3957

JEE Main 2003Annual, Paper 1MathematicsMediumCoordinate Geometry478× since 2002

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 :

JEE Main 2003 · Annual, Paper 1 · Q3957

JEE Main 2003Annual, Paper 1MathematicsMediumCoordinate Geometry478× since 2002

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 :