# Finding the normal to a curve of an implicit function using differentiation

The question asks to find the tangent and normal to the curve of the following equation at the point ($$\sqrt3, 2$$):

$$x^2-\sqrt{3}xy+2y^2=5$$

I began by finding the first derivative of this, which I found to be:

$$\frac{dy}{dx}=\frac{\sqrt3y-2x}{-\sqrt3x+4y}$$

The slope was obviously zero at the given point. Hence,

$$y=mx+c$$ $$y=c$$ $$\therefore c = 2$$ $$\therefore y = 2$$

However, when I cannot seem to find the correct answer for the equation fo the normal. The answer in my book states the equation for the normal is $$x=\sqrt3$$, my attempt at finding the slope of the normal to find its equation was:

$$m_1m_2=-1$$ $$m_1 = 0$$ $$\therefore m_2 = undefined$$

This clearly does not correspond with the book's answer and I do not understand why my method to to find the slope of the normal does not work. Perhaps the formula I used is not feasible for implicit equations? Could someone enlighten me of my mistake and show me how to obtain the correct answer?

Since the slope of the tangent line at $$\left(\sqrt3,2\right)$$ is $$0$$, the tangent line is horizontal and therefore the normal line is vertical. And the vertical line passing through $$\left(\sqrt3,2\right)$$ is $$x=\sqrt3$$.