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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    Given that the slope of the tangent to a curve \[y=y(x)\] at any point (x,y) is \[\frac{2y}{{{x}^{2}}}.\] If the curve passes through the centre of the circle  \[{{x}^{2}}+{{y}^{2}}2x2y=0,\] then its equation is : [JEE Main 8-4-2019 Afternoon]

    A) \[x{{\log }_{e}}|y|=2(x-1)\]

    B) \[x{{\log }_{e}}|y|=x-1\]

    C) \[{{x}^{2}}{{\log }_{e}}|y|=-2(x-1)\]

    D) \[x{{\log }_{e}}|y|=-2(x-1)\]

    Correct Answer: A

    Solution :

    given \[\frac{dy}{dx}=\frac{2y}{{{x}^{2}}}\] \[\Rightarrow \]\[\int_{{}}^{{}}{\frac{dy}{2y}=\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}}}}\]             \[\Rightarrow \]\[\frac{1}{2}\ell ny=-\frac{1}{x}+c\] passes through centre (1,1) \[\Rightarrow \]\[c=1\] \[\Rightarrow \]\[x\ell ny=2(x-1)\]


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