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The foci of the ellipse X^2/16 +y^2/b^2=...

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.

A

1

B

5

C

7

D

9

Text Solution

Verified by Experts

The correct Answer is:
C
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