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

  • question_answer
    The tangent to the curve \[y=a{{x}^{2}}+bx\] at \[(2,\,-8)\] is parallel to x-axis. Then       [AMU 1999]

    A)            \[a=2,\,b=-2\]

    B)            \[a=2,\,\,b=-4\]

    C)            \[a=2\,\,b=-8\]

    D)            \[a=4,\,b=-4\]

    Correct Answer: C

    Solution :

               \[y=a{{x}^{2}}+bx\]            \[\frac{dy}{dx}=2ax+b\Rightarrow {{\left( \frac{dy}{dx} \right)}_{(2,\,-8)}}=4a+b\]            \[\because \]  Tangent is parallel to x-axis            \ \[\frac{dy}{dx}=0\Rightarrow b=-4a\]                            ?..(i)            Now, point (2, ?8) is on the curve of \[y=a{{x}^{2}}+bx\]            \  \[-8=4a+2b\]                                            ??(ii)            From (i) and (ii), we get \[a=2,\,\,b=-8\].


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