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JEE Main & Advanced Mathematics Conic Sections Question Bank Parabola

  • question_answer
    The directrix of the parabola \[{{x}^{2}}-4x-8y+12=0\] is  [Karnataka CET 2003]

    A)            \[x=1\]                                      

    B)            \[y=0\]

    C)            \[x=-1\]                                     

    D)            \[y=-1\]

    Correct Answer: D

    Solution :

               Equation of parabola is \[{{x}^{2}}-4x-8y+12=0\]                    Þ\[{{x}^{2}}-4x+4=8y-8\]                    Þ \[{{(x-2)}^{2}}=8(y-1)\]Þ\[{{X}^{2}}=8Y\]                    Comparing with \[{{X}^{2}}=4aY\], we get \[a=2\]                    \ Directrix is \[Y=-a\] Þ \[y-1=-2\]Þ \[y=-1\].


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