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

Step-by-step solution

Eccentricity for x² a² + y² b² =1 is b²=a² (1-e² ) and eccentricity for x² 144 25 - y² 81 25 =1 is Again, foci =a₁ e₁= 12 5 15 12 =3 So, focus of hyperbola is (3,0)=(a e, 0) and focus of ellipse is (a e, 0)=(5 a, 0) As, this foci are same, so aligned & 5 a=3 & e &= 3 5 & So, & e² &=1- b² a² =1- b² 25 & & b² 25 &=1-e²=1- 9 25 & & b² 25 &= 16 25 & & b² &=16 aligned