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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Tangent and Normal

  • question_answer
    The equation of tangent at \[(-4,\,-4)\] on the curve \[{{x}^{2}}=-4y\] is [Karnataka CET 2001: Pb. CET 2000]

    A)            \[2x+y+4=0\]

    B)            \[2x-y-12=0\]

    C)            \[2x+y-4=0\]

    D)            \[2x-y+4=0\]

    Correct Answer: D

    Solution :

      \[{{x}^{2}}=-4y\]Þ\[2x=-4\frac{dy}{dx}\]Þ\[\frac{dy}{dx}=\frac{-x}{2}\]Þ\[{{\left( \frac{dy}{dx} \right)}_{(-4,\,-4)}}=2\].                    We know that equation of tangent is,                     \[(y-{{y}_{1}})\,=\,{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}(x-{{x}_{1}})\] Þ \[y+4=2(x+4)\]                                                                                  Þ \[2x-y+4=0\].


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