TITLE: How to parametrize the curve $y^2 = x^2(x+3)$
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: 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 $t=\pm \sqrt 3$.