# Proving that for each two parabolas, there exists a transformation taking one to the other

Lets $p_1, p_2$ be two prabolas on the plane. Prove there is a transformation $T:\mathbb R^2 \rightarrow \mathbb R^2$, $T(\vec x)=A \vec x+\vec b$, $A$ being a multiplication of scalar and orthogonal matrices and $\vec b, \vec x \in \mathbb R^2$, taking $p_1$ to $p_2$.

I have no idea how to prove this really… Thanks in advance for any help!

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As @rschwieb mentions, we know that a parabola is defined by its focus and directrix.

Associate with parabola $p_i$ the (unique) equilateral triangle $T_i$, with one vertex at the focus of $p_i$, and the other vertices on the directrix of $p_i$. Clearly, there exists a transformation of the desired form that moves $T_1$ onto $T_2$.

(If you prefer, replace “equilateral triangle $T_i$” with, say, “isosceles right triangle $T_i$” with right angle at the focus, so as to establish a bijection between parabolas $p$ and associated triangles $T$. Breaking the triangles’ rotational symmetry may make it just a bit easier to see that you can get $T_1$ and $T_2$ to align focus-to-focus and directrix-to-directrix.)

Keep in mind at all times that a parabola is defined by its focus and directrix.

Conceptually, it is just a matter of translating $p_1$ so that its vertex lies on the origin, rotating it so that it’s aligned with $p_2$, then performing a dilation which changes the space between its focus and directrix to match that space of $p_2$, and then translating everything out onto $p_2$.

So the transformation goes:

$x\mapsto x+t_1\mapsto R(x+t_1)\mapsto DR(x+t_1)\mapsto DR(x+t_1)+t_2=DR(x)+DR(t_1)+t_2$

The right two terms are your $\vec{b}$ and the $DR$ is your $A$. Now you just need to compute each component ðŸ™‚

It is claimed that a parametrization of the general 2-D parabola is given by:
$$x(t) = \frac{1}{2} a_x.t^2 + v_x.t + s_x \\ y(t) = \frac{1}{2} a_y.t^2 + v_y.t + s_y$$
Let this be the parabola $p_1$ . And let the proposed linear transformation be given by:
$$\left[\begin{array}{c} x’ \\ y’ \end{array}\right] = \left[\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] + \left[\begin{array}{c} b_x \\ b_y \end{array}\right]$$
Giving:
$$x’ = a_{11}\left[ \frac{1}{2} a_x t^2 + v_x t + s_x \right] + a_{12}\left[ \frac{1}{2} a_y t^2 + v_y t + s_y \right] + b_x\\ y’ = a_{21}\left[ \frac{1}{2} a_x t^2 + v_x t + s_x \right] + a_{22}\left[ \frac{1}{2} a_y t^2 + v_y t + s_y \right] + b_y$$
Or:
$$x’ = \frac{1}{2} \left[ a_{11} a_x + a_{12} a_y \right] t^2 + \left[ a_{11} v_x + a_{12} v_y \right] t + \left[ a_{11} s_x + a_{12} s_y + b_x \right] \\ y’ = \frac{1}{2} \left[ a_{21} a_x + a_{22} a_y \right] t^2 + \left[ a_{21} v_x + a_{22} v_y \right] t + \left[ a_{21} s_x + a_{22} s_y + b_y \right]$$
Or:
$$x’ = \frac{1}{2} a’_x t^2 + v’_x t + s’_x \\ y’ = \frac{1}{2} a’_y t^2 + v’_y t + s’_y$$
Let this be the parabola $p_2$ . Then we have:
$$a_x a_{11} + a_y a_{12} = a’_x \\ v_x a_{11} + v_y a_{12} = v’_x \\ a_x a_{21} + a_y a_{22} = a’_y \\ v_x a_{21} + v_y a_{22} = v’_y \\$$
From these the matrix coefficients can be solved:
$$a_{11} = \frac{v_y a’_x – a_y v’_x}{a_x v_y – v_x a_y} \\ a_{12} = \frac{a_x v’_x – v_x a’_x}{a_x v_y – v_x a_y} \\ a_{21} = \frac{v_y a’_y – a_y v’_y}{a_x v_y – v_x a_y} \\ a_{22} = \frac{a_x v’_y – v_x a’_y}{a_x v_y – v_x a_y}$$
Provided that the denominators $(a_x v_y – v_x a_y)$ are nonzero,
meaning that the parabolas are not degenerated i.e. they are no straight lines. Then, at last, the displacement vector is found by:
$$b_x = s’_x – \left[a_{11} s_x + a_{12} s_y\right] \\ b_y = s’_y – \left[a_{21} s_x + a_{22} s_y\right]$$
In the article Parabolic Curves
it is shown how to eliminate the parameter $t$ to get more familiar representation of the parabolas.

Extra. Let’s start again with the above parametrization, written as:
$$x – s_x = v_x \cdot t + a_x \cdot \frac{1}{2} t^2 \\ y – s_y = v_y \cdot t + a_y \cdot \frac{1}{2} t^2$$
Now put:
$$\xi = t \quad ; \quad \eta = \frac{1}{2} t^2 \quad \Longrightarrow \quad \eta = \frac{1}{2} \xi^2$$
And:
$$x – s_x = v_x \, \xi + a_x \, \eta \\ y – s_y = v_y \, \xi + a_y \, \eta$$
Two equations with two unknowns:
$$\xi = \frac{+ a_y (x-s_x) – a_x (y-s_y)}{v_x a_y – v_y a_x} \qquad ; \qquad \eta = \frac{- v_y (x-s_x) + v_x (y-s_y)}{v_x a_y – v_y a_x}$$
Substitute into the “normed parabola” $\;\eta = \frac{1}{2} \xi^2\;$ or $\;\frac{1}{2} \xi^2 – \eta = 0$ :
$$\frac{1}{2} \left[ \frac{a_y (x-s_x) – a_x (y-s_y)}{a_y v_x – a_x v_y} \right]^2 + \left[ \frac{v_y (x-s_x) – v_x (y-s_y)}{a_y v_x – a_x v_y} \right] = 0$$
Which is the equation of an arbitrary parabola in $x$ and $y$ alone.

Update.
But, if I read the question well, it is required that the transformation
taking the parabola $p_1$ into $p_2$ is represented by a scalar times an
orthogonal matrix. This is most easily achieved by a parameter shift in:
$$x(t) = \frac{1}{2} a_x.t^2 + v_x.t + s_x \\ y(t) = \frac{1}{2} a_y.t^2 + v_y.t + s_y$$
As follows:
$$\overline{x}(t) = \frac{1}{2} a_x(t-\tau)^2 + v_x(t-\tau) + s_x \\ \overline{y}(t) = \frac{1}{2} a_y(t-\tau)^2 + v_y(t-\tau) + s_y$$
$$\overline{x}(t) = \frac{1}{2} a_x.t^2+\left(v_x-a_x.\tau\right)\,t +\left(\frac{1}{2}a_x.\tau^2-v_x.\tau+s_x\right) \\ \overline{y}(t) = \frac{1}{2} a_y.t^2+\left(v_y-a_y.\tau\right)\,t +\left(\frac{1}{2}a_y.\tau^2-v_y.\tau+s_y\right)$$
$$\overline{x}(t) = \frac{1}{2} a_x.t^2 + \overline{v}_x.t + \overline{s}_x \\ \overline{y}(t) = \frac{1}{2} a_y.t^2 + \overline{v}_y.t + \overline{s}_y$$
Now it’s always possible to choose $\tau$ in such a way that $(\overline{v}_x,\overline{v}_y)$ is perpendicular to $(a_x,a_y)$ :
$$a_x.\left(v_x-a_x.\tau\right) + a_y \left(v_y-a_y.\tau\right) = 0 \quad \Longrightarrow \quad \tau = \frac{v_x a_x + v_y a_y}{a^2_x + a^2_y}$$
This means that the local coordinate systems $(\vec{v},\vec{a})$ of our parabolas have become orthogonal.
And a transformation that takes an orthogonal coordinate system into another orthogonal coordinate system is itself orthogonal (apart from some scalar eventually).