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Find the equation of the normal of the curve xy-x^2+xlog(y)=4 at the point (2,1) in the form ax+by+c=0

differentiating: xy'+y-2x+(x/y)y'+log(y)=0rearranging: y'=y(2x-y-log(y))/x(1+y)at (2,1): y'=3/4 so gradient of normal at (2,1) is -4/3so the equation of the normal is y-1=(-4/3)(x-2)which is equivalent to 4x+3y-11=0

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Find the equation of the normal of the curve xy-x^2+xlog(y)=4 at the point (2,1) in the form ax+by+c=0

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