Answer
(a) $y=\frac{6}{7}x+\frac{6}{7}$
(b) $y=-\frac{7}{6}x-\frac{7}{6}$
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: $LHS=6(-1)^2+3(-1)(0)+2(0)^2+17(0)(1)-6=0=RHS$
(a) Given the equation $6x^2+3xy+2y^2+17y-6=0$, differentiate both sides with respect to $x$: $12x+3y+3xy'+2yy'+17y'=0$, which gives $(3x+2y+17)y'=-12x-3y$ or $y'=-\frac{12x+3y}{3x+2y+17}$
Thus the slope of the tangent line at the given point is $m=y'=-\frac{12(-1)+3(0)}{3(-1)+2(0)+17}=\frac{6}{7}$ and the equation for the tangent line is $y=\frac{6}{7}(x+1)$ or $y=\frac{6}{7}x+\frac{6}{7}$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-\frac{7}{6}$, and the line equation is $y=-\frac{7}{6}(x+1)$ or $y=-\frac{7}{6}x-\frac{7}{6}$
