JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
    The equation of the hyperbola whose directrix is \[x+2y=1\], focus (2, 1) and eccentricity 2 will be  [MP PET 1988, 89]

    A)            \[{{x}^{2}}-16xy-11{{y}^{2}}-12x+6y+21=0\]

    B)            \[3{{x}^{2}}+16xy+15{{y}^{2}}-4x-14y-1=0\]

    C)            \[{{x}^{2}}+16xy+11{{y}^{2}}-12x-6y+21=0\]

    D)            None of these

    Correct Answer: A

    Solution :

                       \[{{(x-2)}^{2}}+{{(y-1)}^{2}}=4\left[ \frac{{{(x+2y-1)}^{2}}}{5} \right]\] Þ \[5[{{x}^{2}}+{{y}^{2}}-4x-2y+5]\] \[=4[{{x}^{2}}+4{{y}^{2}}+1+4xy-2x-4y]\] Þ \[{{x}^{2}}-11{{y}^{2}}-16xy-12x+6y+21=0\].


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