If the foci of the ellipse x² 16 + y² b² =1 and the hyperbola x² 144 - y² 81 = 1 25 coincide then the value of b² is

Step-by-step solution

The equation of the ellipse is x² 16 + y² b² =1 and equation of the hyperbola is x² 144 - y² 81 = 1 25 or x² ( 12 5 )² - y ( 9 5 )² =1 Eccentricity of ellipse, e= 1- b² 16 and eccentricity of hyperbola, e^ = 1+ ( 9 5 )² ( 12 5 )² = 1+ 81 144 = 15 12 Since, foci of ellipse and hyperbola coincide. aligned & & 4 e &= 12 5 e^ & e &= 3 5 e^ & 1- b² 16 &= 3 5 15 12 & & 1- b² 16 &= ( 3 4 )² b²=7 aligned