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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    The tangent at the point (2,-2) to the curve\[{{x}^{2}}{{y}^{2}}-2x=4(1-y)\]does not pass through the point: [JEE Online 08-04-2017]

    A)  (-2,-7)                 

    B)  (8, 5)

    C) (-4,-9)                  

    D) \[\left( 4,\frac{1}{3} \right)\]

    Correct Answer: A

    Solution :

                    \[{{x}^{2}}{{y}^{2}}-2x=4-4y\]                 \[2x{{y}^{2}}+2y.{{x}^{2}}.\frac{dy}{dx}-2=-4.\frac{dy}{dx}\]                 \[\frac{dy}{dx}(2y.{{x}^{2}}+4)=2-2x.{{y}^{2}}\]                 \[{{\left. \frac{dy}{dx} \right|}_{2,-2}}=\frac{2-2\times 2\times 4}{2(-2)\times 4+4}=\frac{+14}{+12}=\frac{7}{6}\]                 \[(y+2)=\frac{7}{6}(x-2)\Rightarrow 7x-6y=26\]


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