# Folium of Descartes

Today, actually just an hour or so ago, another math teacher asked me if I knew a way to parametrize the following:

$x^3+y^3=3xy$

It is also known as the Folium of Descartes and looks like:

Her purposes was just trying to find a quick and easy way to graph it on the calculator. It was just a small, unimportant question. She didn’t need to know the answer, if it wasn’t easily doable. But to me, I needed to know! How do we find the parametric equations which define this? I just don’t know. The answer I found online is:

$x(t)=\frac{3t}{1+t^3}$ and $y(t)=\frac{3t^2}{1+t^3}$

And it’s pretty easily verifiable when you work backwards with the parametrization.

Is there something I’m missing? Is there a method to working from implicitly defined 2D functions to parametrizations for them?

1. CalcDave says:

I don’t know about “in general,” but if you make the substitution y = tx here, you can get that one.

2. sumidiot says:

Coincidentally, I was just reading about this yesterday. As @CalcDave indicates, you think about y=tx. Which is to say, the curve is parametrized by slopes of lines through the origin. Besides the origin, any line through the origin (and not an axis) hits the curve at exactly one point. One you set y=tx, you can plug that in the equation and solve for x and obtain the formulas you mention above.

3. Jaime says:

Hmmm… The “parametrize by slopes of lines through the origin” is a standard trick to get rational or integral points on an elliptic curve. You have to substitute it by “parametrize by rational slopes of lines through a rational point on the curve” and the origin is a rational (and integral) point of the Folium of the Descartes. A trick that comes in very handy with some Project Euler problems.

Take for instance the unit circle, x²+y²=1, and consider lines of slope t going through point (0,1), after some math you can parametrize it as:

x = -2t/(t²+1) , y =(1-t²)/(t²+1)

If you now limit yourself to rational slopes, t=-m/n, the above can be worked out to:

x = 2mn/(m²+n²), y = (m²-n²)/(m²+n²)

We are now just naming z = m²+n² away from obtaining that integral solutions of x² + y² = z² can be parametrized as

x = 2mn , y = m²-n² , z = m²+n²

which is Euclid’s parametrization of Pythagorean Triples.

4. Jaime says:

Applying the above to the Folium of Descartes, given that the origin is an integral point, by naming z = 1+t³, and chosing only rational slopes, t=m/n, you can readily obtain integral solutions to the equation x³+y³=3xyz with the parametrization:

x = 3mn² , y = 3m²n , z = m³+n³

5. Hi everyone! Thanks for the solution and discussion. It totally works out, substituting $y=tx$. I haven’t had time to really think about how by doing this we are parametrizing the curve by slopes of lines through the origin. But I’m going to think a bit more about it.

@Jamie: Interesting connection! I remember reading that derivation to find all rational points on a circle.

Thanks again. I’ll report back once I really sort out in my mind what that substitution is actually doing, and can explain it to — say — a high school student.

6. Joshua Zucker says:

Something totally different occurred to me here. I wonder if it works?

x^3 + y^3 = (x+y)(x^2 – xy + y^2) = (x+y) ((x+y)^2 – 3xy).

I thought of this expression because it matches the 3xy on the other side.

Now let’s let s = x+y and p = xy.
We have s * (s-3p) = 3p
s^2 – 3ps = 3p
s^2 – 3ps – 3p = 0.

Hm.

Well, at least I’ve reduced a cubic to a quadratic. That has to be worth something?

Wait, it’s quadratic in s, but just linear in p!

3p(1+s) = s^2
3p = s^2/(1+s)

There, that has to help a lot.

I’m fearing that there’s still some quadratic solving to do down the road here, though.