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 $