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

    Answer:

    7


You need to login to perform this action.
You will be redirected in 3 sec spinner