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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 8

  • question_answer
    If the eccentricity of \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is\[\frac{5}{4},\] then the eccentricity of \[\frac{{{x}^{2}}}{{{b}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}}=1\] is

    A) \[\frac{5}{4}\]              

    B) \[\frac{5}{3}\]

    C) \[\frac{4}{3}\]              

    D) \[\frac{9}{4}\]

    Correct Answer: B

    Solution :

    [b]: Clearly, for the first curve \[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)={{a}^{2}}\left( \frac{25}{16}-1 \right)=\frac{9{{a}^{2}}}{16}\] If\[{{e}_{1}}\] is the eccentricity of the second curve, then \[{{a}^{2}}={{b}^{2}}(e_{1}^{2}-1)\] or \[\frac{16}{9}=e_{1}^{2}-1\] \[\Rightarrow \]\[e_{1}^{2}=\frac{16}{9}+1=\frac{25}{9}\]\[\Rightarrow \]\[{{e}_{1}}=\frac{5}{3}\].


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