I'm not a math guy, sorry. I read posts on the subject but couldn't find the answer to my problem (or didn't understood the answers). I'd like to get a simple answer.
I know a generic parabola formula
$$(Pa.x+Pb.y)^2+Pc.x+Pd.y+Pe=0$$
it focus, $(Fx,Fy)$
and the cartesian equation of it's directrix
$$Da.x+Db.y+Dc=0$$
What would be the easiest way to draw a portion of it using a quadratic bezier curve (say maybe from it summit up to $2p$ or like) ?
Choose at will start point $P_0$ and end point $P_2$ on your parabola.
Find the equations of tangent lines at $P_0$ and $P_2$.
Find the intersection point $P_1$ of the two tangents.
The quadratic Bézier $B(t)=(1-t)^2P_0+2t(1-t)P_1+t^2P_2$ is what you want.