JEE Math Practice Question
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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