# How to parametrize the curve $y^2 = x^2(x+3)$

I am emberassed to ask this, but I couldn't find a way.

I want to write the curve $y^2 = x^2(x+3)$ as

$$y=f(t) \quad \quad x=g(t) \quad \quad t \in \mathbb R$$

I guess I have to do something like $x = t-3$ but I am not sure.

Thanks for any help.

• Note that parameterizations are not unique; there is more than one correct answer, though some answers are easier/nicer than others. Commented Apr 27, 2015 at 20:53

Set $x=t^2-3$, we then obtain $y = (t^2-3)t$.

EDIT

The motivation for this choice is as follows:

Taking the square-root, we have $y=x\sqrt{x+3}$. Hence, for a nice parameterization, we want $x+3$ to be a perfect square.

• Thanks, and how could you get the intuition for this? Commented Apr 27, 2015 at 20:25
• @ThePortakal Taking the square-root, we have $y=x\sqrt{x+3}$. Hence, for a nice parameterization, we want $x+3$ to be a perfect square. Commented Apr 27, 2015 at 20:27
• I think your comment is more impressive than the actual answer. Commented Apr 27, 2015 at 20:37

This is standard: the cubic curve $y^2=x^2(x+3)$ has a double point at(0,0), hence any line through the origin will interset the curve in a single third point. So set $y=tx\,$ and replace in the equation. You'll get the equation for the third point of intersection of the line $y=tx$ with the cubic curve: $$(t^2-3)x^2=x^3,\enspace\text{whence}\enspace\begin{cases}x=t^2-3\\y=t(t^2-3)\end{cases}$$ The double point $(0,0)$ is obtained twice, for $t=\pm \sqrt 3$.