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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    If the eccentricities of the ellipse \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\] and the hyperbola\[\frac{{{x}^{2}}}{64}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]are reciprocals of each other, then\[{{b}^{2}}\]is equal to:

    A)  192                       

    B)         64

    C)  16                         

    D)         32

    E)  128

    Correct Answer: A

    Solution :

    Eccentricity of ellipse\[=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{3}{4}}\] \[=\frac{1}{2}\] \[\therefore \]Eccentricity of hyperbola\[=2\] \[\therefore \]  \[2=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1+\frac{{{b}^{2}}}{64}}\] \[\Rightarrow \]               \[4=1+\frac{{{b}^{2}}}{64}\Rightarrow 3\times 64={{b}^{2}}\] \[\Rightarrow \]               \[192={{b}^{2}}\]            


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