# Find eqn for "normal" line to the tangent- folium

Gold Member

## Homework Statement

Back for more questions. This section has been pretty tricky.

The graph of a folium with equation $2x^3+2y^3-9xy=0$ is given.

Find the equation for the tangent line at the pont $(2,1)$

Find the equation of the normal line to the tangent line in the last question at the point (2,1).

## The Attempt at a Solution

From part a, the tangent line at the point $(2,1)$ is $y=\frac{5}{4}x-\frac{3}{2}$

Now by "normal" line, I assume that the question is talking about the folium itself.
But since we're talking about the line at that point, wouldn't the equation for the normal line at that exact point just be the tangent line? Otherwise, how could we determine how far out to choose our x values?

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fresh_42
Mentor
A normal is something perpendicular. So you are looking for the line through $(2,1)$ which is perpendicular to the tangent. Do you know what the slope of such a perpendicular line is, given that the tangent has slope $\frac{5}{4}\,?$

Gold Member
From geometry that would just be the reciprocal right?

fresh_42
Mentor
From geometry that would just be the reciprocal right?
In a way, yes. But it is negative, since $\frac{5}{4}>0$. It is as if $x$ and $y$ axis were swapped. The formula for an angle between two vectors $\vec{v}=(v_1,v_2)\; , \;\vec{w}=(w_1,w_2)$ is
$$\cos \sphericalangle (\vec{v},\vec{w}) = \dfrac{\vec{v}\cdot \vec{w}}{|\vec{v}|\cdot |\vec{w}|}$$
where the product is $\vec{v}\cdot \vec{w} = v_1w_1+v_2w_2$ and $|\vec{v}|=\sqrt{v_1^2+v_2^2}$.

Here we have $\vec{v}=(1,\frac{5}{4})$ and the cosine of $90°$ is zero.

Gold Member
So then we would have the equation for the normal line is $y=\frac{-4}{5}x+\frac{13}{5}$?

fresh_42
Mentor
So then we would have the equation for the normal line is $y=\frac{-4}{5}x+\frac{13}{5}$?
Yes - as long as part a) is correct, which I haven't checked.

Gold Member
Ok cool stuff! Thanks again!