A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{\sqrt{3}}{2}\]
C) \[\frac{1}{\sqrt{3}}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
Given ellipse is \[\frac{{{x}^{2}}}{{{a}^{2}}}\,+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\,\,\,(a>b)\] whose area is \[\pi ab\]. The auxiliary circle to given ellipse is \[{{x}^{2}}+{{y}^{2}}={{a}^{2}},\] whose area is \[\pi {{a}^{2}}\]. Now, \[\pi {{a}^{2}}\,=2\pi ab\] (Given)ss \[\Rightarrow a=2b\] \[\therefore \,\,{{e}^{2}}=1\,-\frac{{{b}^{2}}}{{{a}^{2}}}\Rightarrow \,{{e}^{2}}=1\,\,-\frac{1}{4}\,=\frac{3}{4}\,\Rightarrow \,\,e=\frac{\sqrt{3}}{2}\]You need to login to perform this action.
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