A) 1
B) 5
C) 7
D) 9
E) 36
Correct Answer: C
Solution :
Foci of ellipse\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[(\pm \,4e,\,0)\] and foci of hyperbola \[\frac{{{x}^{2}}}{\frac{144}{25}}-\frac{{{y}^{2}}}{\frac{81}{85}}=1\]is\[\left( \pm \frac{12}{5}e,0 \right)\] According to the given condition\[4e=12\text{ }e\] \[4\sqrt{1-\frac{{{b}^{2}}}{16}}=\frac{12}{5}\sqrt{1+\frac{81}{144}}\] \[\Rightarrow \] \[\sqrt{1-\frac{{{b}^{2}}}{16}}=\frac{3}{5}\times \frac{15}{12}=\frac{3}{4}\] \[\Rightarrow \] \[\frac{{{b}^{2}}}{16}=1-\frac{9}{16}=\frac{7}{16}\] \[\Rightarrow \] \[{{b}^{2}}=7\]You need to login to perform this action.
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