Answer
(a) $y=3x+6$
(b) $y=-\frac{1}{3}x+\frac{8}{3}$
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: $LHS=(-1)^2(3)^2=9=RHS$
(a) Given the equation $x^2y^2=9$, differentiate both sides with respect to $x$: $2xy^2+2x^2yy'=0$, which gives $y'=-\frac{y}{x}$
Thus the slope of the tangent line at the given point is $m=y'=-\frac{y}{x}=-\frac{3}{-1}=3$ and the equation for the tangent line is $y-3=3(x+1)$ or $y=3x+6$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-\frac{1}{3}$, and the line equation is $y-3=-\frac{1}{3}(x+1)$ or $y=-\frac{1}{3}x+\frac{8}{3}$
