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JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
     Centre of hyperbola \[9{{x}^{2}}-16{{y}^{2}}+18x+32y-151=0\] is

    A)            (1, ?1)                                       

    B)            (?1, 1)

    C)            (?1, ?1)                                     

    D)            (1, 1)

    Correct Answer: B

    Solution :

                       Centre is given by                    \[\left( \frac{hf-bg}{ab-{{h}^{2}}},\,\frac{gh-af}{ab-{{h}^{2}}} \right)=\left( \frac{+16.9}{-9.16},\,\frac{-9(16)}{-9(16)} \right)=(-1,\,1)\].


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