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

Question

( A ) 9
( B ) 1
( C ) 5
( D ) 7