Answer
The point $(\sqrt3,2)$ is on the curve.
a) $y=2$.
b) $x=\sqrt3$.
Work Step by Step
$$x^2-\sqrt3xy+2y^2=5$$
Verify that the point $(\sqrt3,2)$ is on the curve
We substitute the coordinates of the point $(\sqrt3,2)$ into the formula of the function to see if it equals $5$ or not:
$$(\sqrt3)^2-\sqrt3\times\sqrt3\times2+2\times2^2=3-6+8=5$$
Since it equals $5$, the point is on the curve.
a) Find the tangent at $(\sqrt3,2)$:
- Find the derivative of the function with implicit differentiation:
$$2x-\sqrt3(y+xy')+4yy'=0$$ $$2x-\sqrt3y-\sqrt3xy'+4yy'=0$$ $$y'(4y-\sqrt3x)=\sqrt3y-2x$$ $$y'=\frac{\sqrt3y-2x}{4y-\sqrt3x}$$
- The slope of the tangent line at $(\sqrt3,2)$ is $$y'=\frac{\sqrt3\times2-2\sqrt3}{4\times2-\sqrt3\times\sqrt3}=\frac{0}{8-3}=0$$
- The tangent line to the given curve at $(\sqrt3,2)$ is
$$y-2=0(x-\sqrt3)$$ $$y-2=0$$ $$y=2$$
b) Find the normal line at $(\sqrt 3,2)$:
Since the tangent line at $(\sqrt3,2)$ is the horizontal line $y=2$, the normal line at $(\sqrt3,2)$, being perpendicular with the tangent line, would be a vertical line with the form $x=x_0$
And as the normal line crosses the point $(\sqrt3,2)$, $x_0=\sqrt3$, meaning the normal line in need to find is $$x=\sqrt3$$
