JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
    The equation of the hyperbola referred to the axis as axes of co-ordinate and whose distance between the foci is 16 and eccentricity is \[\sqrt{2}\], is                                                                                             [UPSEAT 2000]

    A)            \[{{x}^{2}}-{{y}^{2}}=16\]         

    B)            \[{{x}^{2}}-{{y}^{2}}=32\]

    C)            \[{{x}^{2}}-2{{y}^{2}}=16\]       

    D)            \[{{y}^{2}}-{{x}^{2}}=16\]

    Correct Answer: B

    Solution :

                      According to question, Transverse axis = Conjugate axis            Given that, \[e=\sqrt{2},\,2ae=16\]; \\[a=4\sqrt{2}\]            Therefore, equation of hyperbola is \[{{x}^{2}}-{{y}^{2}}=32.\]


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