JEE Main & Advanced Mathematics Conic Sections Question Bank Conic section - General

  • question_answer
    The centre of the conic represented by the equation \[2{{x}^{2}}-72xy+23{{y}^{2}}-4x-28y-48=0\] is

    A)            \[\left( \frac{11}{15},\ \frac{2}{25} \right)\]               

    B)            \[\left( \frac{2}{25},\ \frac{11}{25} \right)\]

    C)            \[\left( \frac{11}{15},\ -\frac{2}{25} \right)\]              

    D)            \[\left( -\frac{11}{25},\ -\frac{2}{25} \right)\]

    Correct Answer: A

    Solution :

                 Centre of conic is \[\left( \frac{hf-bg}{ab-{{h}^{2}}},\,\frac{gh-af}{ab-{{h}^{2}}} \right)\]                    \[=\frac{(-36)(-14)-(23)(-2)}{(2)(23)-{{(36)}^{2}}},\ \frac{(-2)(-36)-(2)(-14)}{(2)(23)-{{(-36)}^{2}}}\]                    \[=\left( -\frac{11}{25},\,-\frac{2}{25} \right)\].


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