=1, we get a=5,b=4 Now, focus of ellipse =(±C,0) where C=√a2−b2 Put the values of a and b, we get C=√52−42=√25−16=√9 ∴ Focus =(±3,0) According to question, hyperbola passes through the focus of ellipse. Let equation of hyperbola be ‌
x2
a2
−‌
y2
b2
=1 Since, it passes through (±3,0), we get ‌
(±3)2
a2
−‌
0
b2
=1, gives a=±3 or a2=9 Also, given that product of eccentricities is 1. Now, (Eccentricity of ellipse ) (Eccentricity of hyperbola) =1 ⇒‌‌(√1−‌
16
25
)(√1+‌
b2
9
)=1 (using formula of eccentricity of ellipse and hyperbola) ⇒‌‌(√‌