JEE Main & Advanced
Mathematics
Conic Sections
Question Bank
Numerical Value Type Questions - Conic Sections
question_answer
The foci of the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] and the hyperbola \[\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\] coincide. Then the value of \[{{b}^{2}}\] is