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

Step-by-step solution

Given hyperbola is, x ² 144 - y ² 81 = 1 25 array l x ² 144 / 25 - y ² 81 / 25 =1 a ²= 144 25 , ~b ²= 81 25 , e = 1+ b ² a ² = 1+ 81 144 = 15 12 array foci of hyperbola are ( ae , 0) i.e ( 3,0) Now given ellipse is x² 16 + y² b² =1 a ²=16 Assume eccentricity of this ellipse is e then its foci are ( ae ^ , 0 ) i . e ( 4 e ^ , 0 ) Given foci of given hyperbola and ellipse coincide 4 e ^ =3 e ^ = 3 4 For ellipse, using eccentricity relationship, e ^ 2 =1- b ² a ² 9 16 =1- b ² 16 b ²=7