Answer
tangent $y=\frac{3}{4}x-\frac{5}{2}$
normal $y=-\frac{4}{3}x+10$
Work Step by Step
Step 1. Given $(y-x)^2=2x+4$, we have $2(y-x)(y'-1)=2$ or $y'=\frac{1}{y-x}+1$ which gives the slope of tangent lines to the curve.
Step 2. At point $(6,2)$, the slope of the tangent line is $m=\frac{1}{2-6}+1=\frac{3}{4}$ and the equation is $y-2=\frac{3}{4}(x-6)$ or $y=\frac{3}{4}x-\frac{5}{2}$
Step 3. The slope of the normal line is $n=-1/m=-\frac{4}{3}$ and the equation is $y-2=-\frac{4}{3}(x-6)$ or $y=-\frac{4}{3}x+10$
