If the foci of the ellipse x^2 16 + y^2 b^2 =1 and the foci of the hyperbola x^2 144 - y^2 81 = 1 25 coincide, then the value of b^2 is

Step-by-step solution

The equation of the ellipse is x^2 16 + y^2 b^2 = 1 . For this ellipse, a^2 = 16 . The foci are at ( ae, 0) , where e = 1 - b^2 16 . Thus, ae = 16 - b^2 . The foci are at ( 16 - b^2 , 0) . The equation of the hyperbola is x^2 144 - y^2 81 = 1 25 . Dividing by 1 25 , we get x^2 144/25 - y^2 81/25 = 1 . Here, a^2 = 144 25 and b^2 = 81 25 . For the hyperbola, the eccentricity e_h is given by e_h = 1 + b^2 a^2 = 1 + 81/25 144/25 = 1 + 81 144 = 144 + 81 144 = 225 144 = 15 12 = 5 4 . The foci of the hyperbola are at ( a_