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     AIEEE  Solved  Paper-2003

A) 1                             

B) 5                             

C) 7                             

D) 9

Correct Answer: C

Solution :

The foci of an ellipse  are \[(\pm \,ae,\,0)\] and foci of a hyperbola  are . Given equation of the hyperbola is                     Eccentricity is given by                                                 Then, foci of a hyperbola are     Also, given equation of the ellipse is                     Foci of an ellipse are . But given focus of ellipse and hyperbola coincide, then                     Also,