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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    The point(s) on the curve \[{{y}^{3}}+3{{x}^{2}}=12y\] where the tangent is vertical (parallel to y-axis), is (are) [IIT Screening 2002]

    A) \[\left( \pm \frac{4}{\sqrt{3}},-2 \right)\]

    B) \[\left( \pm \frac{\sqrt{11}}{3},1 \right)\]

    C) \[(0,\,0)\]

    D) \[\left( \pm \frac{4}{\sqrt{3}},2 \right)\]

    Correct Answer: D

    Solution :

    • \[{{y}^{3}}+3{{x}^{2}}=12y\]Þ \[3{{y}^{2}}\,.\,\frac{dy}{dx}+6x=12\,.\,\frac{dy}{dx}\]                   
    • Þ \[\frac{dy}{dx}(3{{y}^{2}}-12)+6x=0\]Þ \[\frac{dy}{dx}=\frac{6x}{12-3{{y}^{2}}}\]                   
    • Þ  \[\frac{dx}{dy}=\frac{12-3{{y}^{2}}}{6x}\]           
    • Since tangent is parallel to y-axis           
    • \[\therefore \] \[\frac{dx}{dy}=0\] Þ \[12-3{{y}^{2}}=0\]or \[y=\pm 2\].           
    • Then \[x=\pm \frac{4}{\sqrt{3}}\]. At \[\left( \pm \frac{4}{\sqrt{3}},\,-2 \right);\] the equation of curve doesn?t satisfy.


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