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,