The foci of a hyperbola coincide with the foci of the ellipse x^2 25 + y^2 9 =1 . The equation of the hyperbola with eccentricity 2 is

Step-by-step solution

Hyperbola Equation from Ellipse Foci and Given Eccentricity The ellipse equation x^2 25 + y^2 9 =1 has a_e^2=25 and b_e^2=9 , so its focal distance is c_e= a_e^2 - b_e^2 = 25-9 =4 . The foci are located at ( 4, 0) . Since the hyperbola shares these foci and is centered at the origin, its focal distance is c_h=4 . Given the eccentricity e_h=2 , and since e_h= c_h a_h , we solve 2= 4 a_h to find a_h=2 and a_h^2=4 . The hyperbola parameter b_h^2 satisfies c_h^2=a_h^2 + b_h^2 . Substituting c_h^2=16 and a_h^2=4 gives b