Answer
(a) $y=2$
(b) $x=\sqrt 3$
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:
$LHS=(\sqrt 3)^2-\sqrt 3(\sqrt 3)(2)+2(2)^2=3-6+8=5=RHS$
(a) Given the equation $x^2-\sqrt 3xy+2y^2=5$, differentiate both sides with respect to $x$: $2x-\sqrt 3y-\sqrt 3xy'+4yy'=0$, which gives $y'=\frac{2x-\sqrt 3y}{\sqrt 3x-4y}$
Thus the slope of the tangent line at the given point is $m=y'=\frac{2\sqrt 3-2\sqrt 3}{3-4(2)}=0$ and the equation for the tangent line is $y=2$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-\infty$, and the line equation is $x=\sqrt 3$
