JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

  • question_answer
    If tangent to the curve \[{{y}^{2}}={{x}^{3}}\] at its point \[\left( {{m}^{2}},{{m}^{3}} \right)\] is also normal to the curve at \[\left( {{M}^{2}},{{M}^{3}} \right),\] then what is the value of mM ?

    A) \[-\frac{4}{9}\]

    B) \[-\frac{2}{9}\]

    C) \[-\frac{1}{3}\]

    D) 1

    Correct Answer: A

    Solution :

    [a] : We have \[{{y}^{2}}={{x}^{3}}\Rightarrow 2y.\frac{dy}{dx}=3{{x}^{2}}\] \[\Rightarrow \frac{dy}{dx}=\frac{3{{x}^{2}}}{2y}\] Given that, slope of tangent at \[({{m}^{2}},{{m}^{3}})\] = slope of normal at \[({{M}^{2}},{{M}^{3}})\] \[\Rightarrow \]\[{{\left( \frac{dy}{dx} \right)}_{({{m}^{2}},{{m}^{3}})}}=\frac{-1}{{{\left( \frac{dy}{dx} \right)}_{({{M}^{2}},{{M}^{3}})}}}\] \[\Rightarrow \]\[\frac{3{{m}^{4}}}{2{{m}^{3}}}=-\frac{2{{M}^{3}}}{3{{M}^{4}}}\Rightarrow mM=-\frac{4}{9}\]

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