A) \[\frac{{{x}^{2}}}{36}+\frac{{{y}^{2}}}{11}=1\]
B) \[\frac{{{x}^{2}}}{6}+\frac{{{y}^{2}}}{\sqrt{11}}=1\]
C) \[\frac{{{x}^{2}}}{6}+\frac{{{y}^{2}}}{11}=1\]
D) None of these
Correct Answer: A
Solution :
Foci \[(\pm 5,\,0)\equiv (\pm ae,\,0)\]. Directrix \[\left( x=\frac{36}{5} \right)\equiv x=\frac{a}{e}\] So, \[\frac{a}{e}=\frac{36}{5},\ ae=5\] Þ \[a=6\] and \[e=\frac{5}{6}\] Therefore, \[b=6\sqrt{1-\frac{25}{36}}=6\frac{\sqrt{11}}{6}=\sqrt{11}\] Hence equation is\[\frac{{{x}^{2}}}{36}+\frac{{{y}^{2}}}{11}=1\].
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