Answer
(a) $y=-x-1$
(b) $y=x+3$
Work Step by Step
To verify that the given point is on the curve, plug-in the coordinates to the left side of the equation: $LHS=(1)^2-2(-2)-4(1)-1=0=RHS$
(a) Given the equation $y^2-2x-4y-1=0$, differentiate both sides with respect to $x$: $2yy'-2-4y'=0$ which gives $y'=\frac{1}{y-2}$
Thus the slope of the tangent line at the given point is $m=y'=\frac{1}{1-2}=-1$ and the equation for the tangent line is $y-1=-(x+2)$ or $y=-x-1$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=1$, and the line equation is $y-1=x+2$ or $y=x+3$
