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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2013

  • question_answer
        If the eccentricity of the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}+1}+\frac{{{y}^{2}}}{{{a}^{2}}+2}=1\]be\[\frac{1}{\sqrt{6}},\]then the latusrectum of ellipse is

    A)  \[\frac{5}{\sqrt{6}}\]                                    

    B)  \[\frac{10}{\sqrt{6}}\]

    C)  \[\frac{8}{\sqrt{6}}\]                    

    D)  None of these

    Correct Answer: B

    Solution :

                    Here,     \[{{a}^{2}}+2>{{a}^{2}}+1\] So,       \[{{a}^{2}}+1=({{a}^{2}}+2)(1-{{e}^{2}})\] \[\Rightarrow \]               \[{{a}^{2}}+1=({{a}^{2}}+2).\frac{5}{6}\] \[\Rightarrow \]               \[6{{a}^{2}}+6=5{{a}^{2}}+10\] \[\Rightarrow \]               \[{{a}^{2}}=10-6=4\] \[\Rightarrow \]               \[a=\pm 2\] \[\therefore \] \[Latusrectum=\frac{2({{a}^{2}}+1)}{\sqrt{{{a}^{2}}+2}}=\frac{2\times 5}{\sqrt{6}}=\frac{10}{\sqrt{6}}\]


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