The foci of a hyperbola coincide with the foci of the ellipse (x²/25)+(y²/9)=1. Then the equation hyperbola with eccentricity 2 is
a. (x²/12)−(y²/4)=1
b. (x²/4)−(y²/12)=1
c. 3x²−y²+12=0
d. 9x²−25y²−225=0