A) 8
B) 10
C) 7
D) 9
Correct Answer: C
Solution :
Given equation of ellipse is\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] eccentricity \[=e=\sqrt{1-\frac{{{b}^{2}}}{16}}\] \[foci:\pm ae=\pm 4\sqrt{1-\frac{{{b}^{2}}}{16}}\] Equation of hyperbola is \[\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\] \[\Rightarrow \]\[\frac{{{x}^{2}}}{\frac{144}{25}}-\frac{{{y}^{2}}}{\frac{81}{25}}=1\] eccentricity= \[e=\sqrt{\frac{81}{25}\times \frac{25}{144}}=\sqrt{1+\frac{81}{144}}\] \[=\sqrt{\frac{225}{144}}=\frac{15}{12}\] \[foci:\pm ae=\pm \frac{12}{5}\times \frac{15}{12}=\pm 3\] Since, foci of ellipse and hyperbola coincide \[\therefore \]\[\pm 4\sqrt{1-\frac{{{b}^{2}}}{16}}=\pm 3\Rightarrow {{b}^{2}}=7\]
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