Problem
Let $\xi,\eta$ be the horizontal and vertical coordinates of a plane. Consider the following sequence of point \begin{equation}\begin{aligned} \xi_k &= \xi_0 + kT \dot{\xi}_0 + k^2 \frac{T^2}{2} u_\xi\\ \eta_k &= \eta_0 + kT \dot{\eta}_0 + k^2 \frac{T^2}{2} u_\eta\\ \end{aligned} \qquad k=0,1,2\dots,K\tag{1}\end{equation} where:
- $\xi_0$, $\eta_0$ is a given starting position;
- $\dot{\xi}_0$, $\dot{\eta}_0$ is a given starting velocity;
- $u_{\xi}$, $u_{\eta}$ is a given acceleration, which is constant respect the index $k$;
- $T>0$ is a given sampling time;
- $K>0$ is the index of the last point, and is given.
I'm quite sure that, excluding some special case where the trajectory is a line, in general every point $\{\xi_k,\eta_k\}_{k=0}^K$ lies on a suitable parabola, so the problem is the following:
problem if it exists, find the Cartesian equation of the underling parabola.
My attempt
First of all a quick change of coordinates that simplify the notation \begin{equation*}\begin{aligned} \Delta \xi_k &\triangleq \xi_k-\xi_0\\ \Delta \eta_k &\triangleq \eta_k-\eta_0\\ \end{aligned}\end{equation*} so $(1)$ simplifies in \begin{equation}\begin{aligned} \Delta\xi_k &= kT \dot{\xi}_0 + k^2 \frac{T^2}{2} u_\xi\\ \Delta \eta_k &= kT \dot{\eta}_0 + k^2 \frac{T^2}{2} u_\eta\\ \end{aligned} \qquad k=0,1,2\dots,K\tag{1}\end{equation} now the idea is to solve the first equation with respect $T$, then plug the result in the second equation and hoping to find an expression of the form \begin{equation}\Delta \eta_k=a(\Delta \xi_k)^2+b\Delta \xi_k+c \qquad k=0,1,\dots,K\end{equation} or in the form \begin{equation}\Delta \xi_k=a(\Delta \eta_k)^2+b\Delta \eta_k+c \qquad k=0,1,\dots,K\end{equation} for suitables $a,b,c$ depending on the data $\dot{\xi}_0, \dot{\eta}_0, u_{\xi}, u_{\eta}, T$.
step 1: solving respect $T$
If I'm not wrong, $T$ is given by the two solutions \begin{equation*}T=\frac{-k\dot{\xi}_0\pm\sqrt{(k\dot{\xi}_0)^2-4\frac{k^2 u_{\xi}}{2}(-\Delta\xi_k)}}{2\frac{k^2 u_{\xi}}{2}}=\frac{-\dot{\xi}_0 \pm \sqrt{\dot{\xi}_0^2+2u_{\xi}\Delta \xi_k}}{k u_\xi}\end{equation*}
step 2: plugging $T$ into the second equation
Now I have \begin{equation}\begin{aligned} \Delta \eta_k &= k\dot{\eta}_0\left(\frac{-\dot{\xi}_0 \pm \sqrt{\dot{\xi}_0^2+2u_{\xi}\Delta \xi_k}}{k u_\xi}\right) +\frac{k^2 u_{\eta}}{2}\left(\frac{-\dot{\xi}_0 \pm \sqrt{\dot{\xi}_0^2+2u_{\xi}\Delta \xi_k}}{k u_\xi}\right)^2\\ &=\frac{\dot{\eta}_0}{u_\xi}\left(-\dot{\xi}_0 \pm \sqrt{\dot{\xi}_0^2+2u_{\xi}\Delta \xi_k}\right)+\frac{u_\eta}{2 u_{\xi}^2}\left(-\dot{\xi}_0 \pm \sqrt{\dot{\xi}_0^2+2u_{\xi}\Delta \xi_k}\right)^2 \end{aligned} \end{equation} and this is my stopping point since I'dont know how to rearrange the result.
This is an interesting idea. The limitation of this approach is that $\Delta \eta_k=a(\Delta \xi_k)^2+b\Delta \xi_k+c$ describes a parabola whose axis is parallel to the vertical axis and $\Delta \xi_k=a(\Delta \eta_k)^2+b\Delta \eta_k+c$ describes a parabola whose axis is parallel to the horizontal axis. But in general the axis of a parabola can be parallel to an arbitrary line through the origin, not necessarily one of the chosen axes.
Rather than translate the system of coordinates, try rotating the system of coordinates: $$ \begin{pmatrix} \xi' \\ \eta' \end{pmatrix} = \begin{pmatrix} p & -q \\ q & p \end{pmatrix} \begin{pmatrix} \xi \\ \eta \end{pmatrix} $$ where $p^2 + q^2 = 1$, in such a way that you get a formula $$ \xi_k'=a(\eta_k')^2+b\eta_k'+c. $$
The first step in this direction is to find $p$ and $q$ such that $$ \eta_k' = c_2 + b_2 (kT), $$ that is, eliminate the one of the quadratic terms. Then you can solve for $kT$ as a degree-one polynomial in $\eta_k',$ and having done that, can plug that polynomial in $\eta_k'$ for every value of $kT$ in the formula for $\xi_k'$, which will then turn out to be a quadratic polynomial in $\eta_k'$.
Note that if you use an arbitrary rotation matrix $\begin{pmatrix} p & -q \\ q & p \end{pmatrix},$ the formula you get for $\eta_k'$ is a quadratic polynomial in $kT$ whose quadratic term is $$ \left(\frac12 p u_\xi + \frac12 q u_\eta\right) (kT)^2.$$ So it comes down to finding $p$ and $q$ such that $p u_\xi + q u_\eta = 0$ and $p^2 + q^2 = 1,$ followed by a lot of algebra if you want to find the values of $a,$ $b,$ and $c$ in terms of the original given information.
Less algebra is required if you merely want to convince someone like me that $\xi_k'$ is a quadratic polynomial in $\eta_k',$ that is, $\xi_k'=a(\eta_k')^2+b\eta_k'+c$ for constants $a,$ $b,$ and $c$ that could (in principle) be computed in terms of the given information if you really wanted to do so, which says the points $(\xi_k,\eta_k)$ all like on a single parabola parallel to the $\xi'$ axis.
Incidentally, I find it easier to do the continuous version of this problem, in which we let \begin{align} \xi &= \xi_0 + \dot{\xi}_0 t + \frac12 u_\xi t^2, \\ \eta &= \eta_0 + \dot{\eta}_0 t + \frac12 u_\eta t^2 \end{align} so that $t = kT,$ $\xi = \xi_k,$ and $\eta = \eta_k$ satisfy the system of equations for each $k = 0, 1, \ldots, K.$ Once we have shown a solution for all real values of $t,$ we know it is also valid for the $K+1$ particular values $t = kT.$