The fixed point on the curve , y=x^2-4x+5 such that the tangent at P is perpendicular to the line x+2y=7 , is

Answers (1)

Solution:    y=x^2-4x+1    ...........(1)

                   fracmathrmd ymathrmd x=2(x-1)         ............(2)

             For the given line ,   y=frac-12x+frac72

            Now ,    m_1m_2=-1Rightarrow 2(x-2)(frac-12)=-1

            Rightarrow                     x=3

          From (1) ,    y=2

         	herefore       The point is (3,2).

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