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 \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\] whose area is \[2\pi ab.\]The auxiliary circle of the ellipse is \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] whose area is \[\pi {{a}^{2}}\]. Given that \[\pi {{a}^{2}}=2\pi ab\] \[a=2b\] Now, eccentricity of ellipse \[=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{{{b}^{2}}}{4{{b}^{2}}}}\] \[=\sqrt{\frac{4-1}{4}}=\frac{\sqrt{3}}{2}\]
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