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

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).