The foci of a hyperbola are the same as those of the ellipse with equation 9x^2 + 16y^2 = 144 . If the length of the transverse axis of this hyperbola is 2 , then its equation is:

Step-by-step solution

The equation of the given ellipse is 9x^2 + 16y^2 = 144 , which can be written as x^2 16 + y^2 9 = 1 . Here, a^2 = 16 and b^2 = 9 . The eccentricity e of the ellipse is given by e = 1 - b^2 a^2 = 1 - 9 16 = 7 4 . The foci of the ellipse are ( ae, 0) = ( 4 7 4 , 0 ) = ( 7 , 0) . Since the hyperbola has the same foci as the ellipse, the foci of the hyperbola are also ( 7 , 0) . Let the equation of the hyperbola be x^2 A^2 - y^2 B^2 = 1 . The foci are ( Ae', 0) , where e' is the eccentricity of the hyperbola. Thus, Ae