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find equation of the normal line …

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find equation of the normal line to the curve y(x-1)(x-3) -x +7 =0 at the point where it meets x axis
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Ritss ?? 5 years, 8 months ago

If the point meets at x axis ; point will be (x,0) Now the putting the value of y =0 and the value of x comes to be 7 Point be (7,0) Equation of curve = y (x-1) (x-3) - x+7=0 y(x^2-4x+3)=7-x Now differentiate wrt x Dy/dx(x^2-4x+3) + y( 2x-4) = -1 Now,, dy/dx= 4y-1-2xy /x^2-4x+3 Dy/dx of pt (7,0) = the slope of tangent =-1/24 Therfore the slope of normal be 24 Now the equation of normal be (y-y')=slope (x-x') Y =24(x-7) 24x-y= 168
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