Parametric Form for a General Parabola

It is well known that a parametric form of the parabola $y^2=4ax$ is $(at^2, 2at)$.

What are possible parametric forms of the general parabola $$(Ax+Cy)^2+Dx+Ey+F=0$$ ?

Bézier curves are a convenient way to produce parameterizations of parabolas: a quadratic Bézier is a (part of a) parabola. If $P_0$ and $P_2$ are points on the parabola and $P_1$ the intersection of the tangents at those points, the quadratic Bézier curve they define is given by $$\phi:t\mapsto(1-t)^2P_0+2t(1-t)P_1+t^2P_2.\tag{1}$$ (The parameter $t$ is usually taken to range from $0$ to $1$ for a Bézier patch.)

We can reproduce your parametrization by taking the vertex $P_0(0,0)$ and an end of the latus rectum $P_2(p,2p)$ as the points on the parabola. (Here I use the conventional name $p$ for this parameter instead of the $a$ in the question.) The tangent at the end of the latus rectum meets the parabola’s axis at a 45° angle, so our third control point will be $P_1(0,p)$. Plugging these into (1) we get $$(1-t)^2(0,0)+2t(1-t)(0,p)+t^2(p,2p)=(pt^2,2pt),$$ as required. As described here, parametrization of a parabola by a pair of quadratic polynomials has a nice symmetry about the vertex. Choosing the vertex as our first control point makes this symmetry quite simple.

To obtain the corresponding parameterization for a general parabola, you can either rotate and translate these three points to match the position and orientation of the given parabola, or compute them from other information that you have about the parabola. For example, if we have a parabola with vertex $P_0(x_0,y_0)$, focal length $p$ and axis direction $\theta$, we will have $P_1=P_0+(-p\sin\theta,p\cos\theta)$ and $P_2=P_0+(p\cos\theta-2p\sin\theta,2p\cos\theta+p\sin\theta)$, which gives the parameterization $$\begin{align}x&=x_0-2pt\sin\theta+pt^2\cos\theta \\ y&= y_0+2pt\cos\theta+pt^2\sin\theta.\end{align}$$

I’ll leave working out this parameterization for the general-form equation to you. As a hint, remember that for the parabola $y=ax^2+bx+c$, $p={1\over4a}$ and that a parabola’s vertex is halfway between its focus and directrix.

This solution to my other question on the axis of symmetry of a general parabola gives the following:

Axis of symmetry: $$Ax+Cy+t^*=0$$ Tanget at vertex: $$(D-2At^*)x+(E-2Ct^*)y+F-{t^*}^2=0$$ where $t^* \left(=\frac {AD+CE}{2(A^2+C^2)}\right)$ is chosen for both lines to be perpendicular.

Solving for the intersection of the two lines gives the coordinates of the vertex as $$\left(-\frac{C{t^*}^2-Et^*+CF}{CD-AE}, \frac{A{t^*}^2-Dt^*+AF}{CD-AE}\right)$$

Replacing $t^*$ with the general parameter $t$ gives a parametric form for the general parabola $(Ax+Cy)^2+Dx+Ey+F=0$ as

$$\color{red}{\left(-\frac{Ct^2-Et+CF}{CD-AE}, \frac{At^2-Dt+AF}{CD-AE}\right)}$$ which is the same as $$\color{red}{\left(\frac{Ct^2-Et+CF}{AE-CD}, -\frac{At^2-Dt+AF}{AE-CD}\right)}$$

See graphical implementation here.

For the special case where $A=C$, $$t^*=\frac {D+E}{4A}$$ Axis of Symmetry: $$Ax+Ay+\frac {D+E}{4A}=0$$ or $$x+y+\frac {D+E}{4A^2}=0$$ Vertex: $$\left(\frac{{t^*}^2-\frac EA t^*+F}{E-D}, -\frac{{t^* }^2-\frac DA t^*+F}{E-D}\right)$$