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

  • question_answer
    If \[x={{t}^{2}}\]and \[y=2t\], then equation of the normal at \[t=1\]is                       [RPET 1996]

    A)            \[x+y-3=0\]

    B)            \[x+y-1=0\]

    C)            \[x+y+1=0\]

    D)            \[x+y+3=0\]

    Correct Answer: A

    Solution :

                       \[x={{t}^{2}}\]and \[y=2t\]            At \[t=1,\ x=1\]and \[y=2\]            Now \[\left( \frac{dy}{dx} \right)=\frac{dy/dt}{dx/dt}=\frac{2}{2t}=\frac{1}{t}\]\[\Rightarrow \]\[{{\left( \frac{dy}{dx} \right)}_{t=1}}=1\]            \[\therefore \]Equation of the normal at (1, 2) is \[y-2=-\frac{1}{\frac{dy}{dx}}(x-1)\]                    Þ \[y-2=-1(x-1)\]Þ \[x+y-3=0\].


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