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

Given equation of ellipse is x² 16 + y² b² =1 Here, a²=16 a=4 e= 1- b² 16 = 16-b² 4 Foci of ellipse are ( a e, 0) i e, ( 16-b² , 0 ) . Also, given equation of hyperbola is x² 144 - y² 81 = 1 25 Here, a²= ( 12 5 )², b²= ( 9 5 )² e= 1+ b² a² = 1+ 81 144 = 5 4 Foci of the hyperbola are ( a e, 0) i e,( 3,0) . According to the given condition, foci of ellipse = foci of hyperbola 16-b² =3 b²=7