Given, ellipse is
‌+‌=1‌ or ‌‌+‌=1Compare it with standard equation
‌+‌=1,
we get
a=5,b=4Now, focus of ellipse
=(±e,0) where
c=√a2−b2Put 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
‌−‌=1Since, it passes through
(±3,0), we get
‌−‌=1, gives
a=±3 or
a2=9Also, given that product of eccentrcikes is
1.Now, (Eccentricity of ellipse )
(Eccentricity of hyperbola)
=1 ⇒‌‌(√1−‌)(√1+‌)=1(using formula of eccentricity of ellipse and hyperbola)
⇒‌‌(√‌)(√1+‌)=1 (using formula of eccentricity of ellipse and hyperbola)
⇒(√‌)(√1+‌)=1Squaring on both sides,
1+‌=‌⇒b2=16Thus, equation of hyperbola is
‌−‌=1.