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

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    $\begingroup$ There’s no such thing as “the” parametric form. You might better describe the types of expressions that you’re looking for. In general, however, it is possible to express a parabola in general position parametrically as a pair of quadratics. $\endgroup$ – amd Dec 5 '16 at 21:31
  • $\begingroup$ @amd - OK, "a" parametric form then :) $\endgroup$ – hypergeometric Dec 6 '16 at 2:04
  • $\begingroup$ You ought to work one of these out yourself for a change. It builds character and may give you some insights. $\endgroup$ – amd Dec 6 '16 at 4:14
  • $\begingroup$ @amd Haha been busy working out solutions to other problems on MSE. This came across my mind so thought I would share the problem at the same time that I'm thinking about it. $\endgroup$ – hypergeometric Dec 6 '16 at 10:01
  • $\begingroup$ @amd - After some character-building effort I've worked out the solution! Posted below. :) $\endgroup$ – hypergeometric Dec 7 '16 at 0:56
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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.

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  • $\begingroup$ Thanks for your solution. Will take a closer look. Have just posted mine as well. $\endgroup$ – hypergeometric Dec 7 '16 at 0:58
  • $\begingroup$ Interesting points, in particular the use of Bezier curves. (+1) $\endgroup$ – hypergeometric Dec 7 '16 at 7:13
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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: $$\left(\frac{{t^*}^2-\frac EA t^*+F}{E-D}, -\frac{{t^* }^2-\frac DA t^*+F}{E-D}\right)$$

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  • $\begingroup$ The parameter $t$ measures the distance of the line $Ax+Cy+t=0$ from the origin, so it’s essentially the $x$-coordinate in a frame that’s been rotated so as to make the axis of the parabola vertical. The Bézier parameter has a geometric interpretation, too, but it’s not nearly that simple. $\endgroup$ – amd Dec 7 '16 at 4:20

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