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BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    If the eccentricity of the hyperbola \[{{x}^{2}}-{{y}^{2}}{{\sec }^{2}}\alpha =5\]is \[\sqrt{3}\]times the eccentricity of the ellipse \[{{x}^{2}}{{\sec }^{2}}\alpha +{{y}^{2}}=25,\]then the value of \[\alpha \] is

    A)  \[\pi /6\]                              

    B)         \[\pi /4\]                              

    C)         \[\pi /3\]                              

    D)         \[\pi /2\]

    Correct Answer: B

    Solution :

    The given condition gives, \[\frac{5+5{{\cos }^{2}}\alpha }{5}=3.\frac{25-25{{\cos }^{2}}\alpha }{25}\] \[\Rightarrow \]  \[{{\cos }^{2}}\alpha =\frac{1}{2}\Rightarrow \alpha =\frac{\pi }{4},\frac{3\pi }{4}\]


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