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JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

  • question_answer
    If the area of auxiliary circle of the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=l\,;\,(a>b)\] is twice the area of the ellipse, then the eccentricity of the ellipse is

    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 \[\Rightarrow a=2b\]           \[\therefore \,\,{{e}^{2}}=1\,-\frac{{{b}^{2}}}{{{a}^{2}}}\Rightarrow \,{{e}^{2}}=1\,\,-\frac{1}{4}\,=\frac{3}{4}\,\Rightarrow \,\,e=\frac{\sqrt{3}}{2}\]


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