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?


1 Answer 1


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 $x=\sqrt3$.


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