Answer
(a) $y=\frac{7}{4}x-\frac{1}{2}$
(b) $y=-\frac{4}{7}x+\frac{29}{7}$
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: $LHS(2)^2+(2)(3)-(3)^2=4+6-9=1=RHS$
(a) Given the equation $x^2+xy-y^2=1$, differentiate both sides with respect to $x$: $2x+y+xy'-2yy'=0$, which gives $(x-2y)y'=-(2x+y)$
Thus the slope of the tangent line at the given point is $m=y'=\frac{2x+y}{2y-x}=\frac{2(2)+(3)}{2(3)-(2)}=\frac{7}{4}$
And the equation for the tangent line is:
$y-3=\frac{7}{4}(x-2)$ or $y=\frac{7}{4}x-\frac{1}{2}$
(b) The slope of the line normal to the tangent line is:
$n=-\frac{1}{m}=-\frac{4}{7}$
And the line equation is:
$y-3=-\frac{4}{7}(x-2)$ or $y=-\frac{4}{7}x+\frac{29}{7}$
