Answer
Tangent $y=2x-1$, normal $y=-\frac{1}{2}x+\frac{3}{2}$
Work Step by Step
Step 1. Given the equation $y^2(2-x)=x^3$, differentiate both sides with respect to $x$: $2yy'(2-x)+y^2(-1)=3x^2$ which gives $y'=\frac{3x^2+y^2}{2y(2-x)}$,
Step 2. The slope of the tangent line at the given point $(1,1)$ is $m=y'=\frac{3(1)^2+1^2}{2(2-1)}=2$ and the equation for the tangent line is $y-1=2(x-1)$ or $y=2x-1$
Step 3. The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-\frac{1}{2}$, and the line equation is $y-1=-\frac{1}{2}(x-1)$ or $y=-\frac{1}{2}x+\frac{3}{2}$
