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

  • question_answer
    The slope of the tangent to the curve \[x=3{{t}^{2}}+1,y={{t}^{3}}-1\] at \[x=1\] is                                           [Karnataka CET 2003]

    A)            0

    B)            \[\frac{1}{2}\]

    C)            \[\infty \]

    D)            \[-2\]

    Correct Answer: A

    Solution :

               \[x=3{{t}^{2}}+1,\,y={{t}^{3}}-1\]            \ \[\frac{dx}{dt}=6t,\] \[\frac{dy}{dt}=3{{t}^{2}}\]            Now \[\frac{dy}{dx}=\left( \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \right)\] = \[\frac{3{{t}^{2}}}{6t}=\frac{t}{2}\]            For \[x=1\], \[3{{t}^{2}}+1=1\Rightarrow t=0\] Þ Slope = \[\frac{0}{2}=0\].


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