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

  • question_answer
    The line \[x+y=2\]is tangent to the curve \[{{x}^{2}}=3-2y\] at its point [MP PET 1998]

    A)            (1, 1)

    B)            (?1, 1)

    C)            (\[\sqrt{3}\], 0)

    D)            (3, ?3)

    Correct Answer: A

    Solution :

                       Given curve \[{{x}^{2}}=3-2y\]                ...(i)            Differentiate w.r.t. x,   \[2x=0-2\frac{dy}{dx}\]\[\Rightarrow \]\[\frac{dy}{dx}=-x\]            Slope of the tangent of the curve \[=-x\] From the given line, slope \[=-1\],\[\therefore x=1\] and from equation (i), \[y=1\]. \[\therefore \]Co-ordinate of the point is (1, 1).


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