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
  $$(Ax+Cy)^2+Dx+Ey+F=0$$
  ?


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:
$$\left(\frac{{t^*}^2-\frac EA t^*+F}{E-D}, -\frac{{t^* }^2-\frac DA t^*+F}{E-D}\right)$$
