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JEE Main & Advanced Mathematics Conic Sections Question Bank Parabola

  • question_answer
    The point on the parabola \[{{y}^{2}}=8x\] at which the normal is inclined at 60o to the x-axis  has the co-ordinates      [MP PET 1993]

    A)            \[(6,\ -4\sqrt{3})\]                    

    B)            \[(6,\ 4\sqrt{3})\]

    C)            \[(-6,\ -4\sqrt{3})\]                  

    D)            \[(-6,\ 4\sqrt{3})\]

    Correct Answer: A

    Solution :

               Normal at \[(h,k)\]to the parabola \[{{y}^{2}}=8x\] is                    \[y-k=-\frac{k}{4}(x-h)\]                    Gradient\[=\tan 60{}^\circ =\sqrt{3}=-\frac{k}{4}\]Þ \[k=-4\sqrt{3}\] and \[h=6\]                    Hence required point is \[(6,-4\sqrt{3})\].


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