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JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
    The radius of the director circle of the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\], is              [MP PET 1999]

    A)            \[a-b\]                                       

    B)            \[\sqrt{a-b}\]

    C)            \[\sqrt{{{a}^{2}}-{{b}^{2}}}\]    

    D)            \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]

    Correct Answer: C

    Solution :

               Equation of director-circle of the hyperbola                   \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]is \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}-{{b}^{2}}\]                    So, radius \[=\sqrt{{{a}^{2}}-{{b}^{2}}}\].


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