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JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    The equation of the normal to the parabola, \[{{x}^{2}}=8y\]at \[x=4\]is     JEE Main  Online Paper (Held On 19  May  2012)

    A) \[x+2y=0\]                         

    B)                        \[x+y=2\]

    C)                        \[x-2y=0\]                          

    D)                        \[x+y=6\]

    Correct Answer: D

    Solution :

                    \[{{x}^{2}}=8y\]                                                            ...(i) When, x = 4, then y = 2 Now\[{{\left. \frac{dy}{dx}=\frac{2x}{8}=\frac{x}{4},\frac{dy}{dx} \right]}_{x=4}}=1\] Slope of normal \[=-\frac{1}{\frac{dy}{dx}}=-1\] Euqation of normal at x = 4 is \[y=-2=-1(x-4)\] \[\Rightarrow \]\[y=-x+4+2=-x+6\] \[\Rightarrow \]\[x+y=6\]


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