A) 1
B) 5
C) 7
D) 9
Correct Answer: C
Solution :
Hyperbola is \[\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\] \[a=\sqrt{\frac{144}{25}},\,\,b=\sqrt{\frac{81}{25}},\,\,{{e}_{1}}=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4}\] Therefore, foci \[=(a{{e}_{1}},0)=\left( \frac{12}{5}.\frac{5}{4},0 \right)=(3,\,0)\] Therefore, focus of ellipse \[=(4e,0)\] i.e. \[(3,\,0)\] Þ \[{{x}^{2}}+\] Hence\[{{b}^{2}}=16\left( 1-\frac{9}{16} \right)=7\].You need to login to perform this action.
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