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! 

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).
