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JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
    If the eccentricities of the hyperbolas \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] and  \[\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{{{x}^{2}}}{{{a}^{2}}}=1\] be e and \[{{e}_{1}}\], then \[\frac{1}{{{e}^{2}}}+\frac{1}{e_{1}^{2}}=\] [MNR 1984; MP PET 1995; DCE 2000]

    A)            1     

    B)            2

    C)            3     

    D)            None of these

    Correct Answer: A

    Solution :

                       \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\] Þ \[{{e}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}\]                    \[{{e}_{1}}=\sqrt{1+\frac{{{a}^{2}}}{{{b}^{2}}}}\] Þ \[e_{1}^{2}=\frac{{{b}^{2}}+{{a}^{2}}}{{{b}^{2}}}\] Þ \[\frac{1}{e_{1}^{2}}+\frac{1}{{{e}^{2}}}=1\].


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