The foci of hyperbola coincide with foci of ellipse
$ \frac{x^2}{25} + \frac{y^2}{9} = 1 $.
So the equation of hyperbola if $e = 2$ is
(a) $ \frac{x^2}{4} - \frac{y^2}{12} = 1 $
(b) $ \frac{x^2}{16} - \frac{y^2}{12} = 1 $
(c) $ \frac{x^2}{12} - \frac{y^2}{4} = 1 $
(d) $ \frac{x^2}{2} - \frac{y^2}{9} = 1 $