Predicting parameters of simple configured trajectories using RNN
If you know that your trajectory has a certain parametric form then you can use methods that explore the parameter space for that form. Examples of such methods are Hough transform and custom-built moments.
Hough transform maps a point in a real space into a manifold in the parameter space, and vice-versa, it maps a point in the parameter space into a line in the real space. If you have several points in the real space, they are transformed into the same amount of manifolds in the parameter space. These manifolds intersect in a point whose coordinates are the parameters of the line that goes through your data.
The parameter space for the ellipse is five-dimensional, so a naive application of the Hough transform is inefficient. Smarter implementations are known, e.g. [1].
The method of custom-built moments described in [2] allows to find the parameters of a curve expressible as
$$G(x,y) = \sum\limits_np_nz_n(x,y)=0$$
In the case of an ellipse, we have
$$G(x,y)=p_1+p_2x+p_3y+p_4x^2+p_5y^2+p_6xy=0$$
that is $z_1=1$, $z_2=x$, $z_3=y$, $z_4=x^2$, $z_5=y^2$, $z_6=xy$. You want to determine $p_1,...,p_6$ based on your data.
Enclose the portion of the curve where you have your data in a rectangle, and make a coordinate transformation so that your rectangle is now $ S = [-1 \le x \le 1, -1 \le y \le 1]$. The rectangle is separated by the curve into two regions. Assign the value $1$ to all points from one of the regions, and $0$ to all points in the other region. For example, in this Figure, all black points are $1$, and white points are $0$.

The data points are on the boundary. They are not explicitly shown in the figure because their exact position is not important anymore. It is also not important, how many points you have (as long as they clearly determine the shape of the curve segment) and in which order you collect them.
So, now you have a function defined on the rectangle:
$$I(x,y)=\begin{cases} 
      1 &amp; (x,y) \in Region~1 \\
      0 &amp; (x,y) \in Region~2 
\end{cases}$$
The parameters $p_1,...,p_6$ are found as the solution of a system of equations
$$\sum\limits_nu_np_n=0$$
where $u_n$ are custom moments
$$u_n=\iint_S I(x,y)U_n(x,y) \,dx\,dy$$
The functions $U_n$ are defined as $U_n(x,y)=\Upsilon(F,s,z_n)$ where
$$\Upsilon(F,s,z)=\left([sz_y-zs_y]F\right)_x-\left([sz_x-zs_x]F\right)_y$$
with the $x$ and $y$ subscripts indicating partial differentiation. The functions $F(x,y)$ and $s(x,y)$ are arbitrary. They can be anything you like as long as $F$ is  continuously differentiable, $s$ is twice continuously differentiable on an open set containing the rectangle $S$, and $F=0$ on the boundaries of $S$. More details in [2]. If you don't understand that paper during the first reading, read [3] first, and if you don't understand [3], read [4], then read [3], after which [2] should become clearer.
For ellipses, the authors of [2] recommend to generate 18 arrays of functions $U_n = (U_1,...,U_6)$ from combinations of three functions $F_1 = m_0n_0$, $F_2=xF_1$, $F_3=yF_1$ with six functions $s_n=z_n,\, n=1,...,6$, where $m_0=1-x^2$, $n_0=1-y^2$. With the additional notations $m_1=1-2x^2$, $n_1=1-2y^2$, $m_2=1-3x^2$, $n_2=1-3y^2$, these 18 arrays are found to be

$F=F_1,~s=1\Rightarrow % 1
U_1=0,~
U_2=2ym_0,~
U_3=-2xn_0,~
U_4=4xym_0,~
U_5=-4xyn_0,~
U_6=2(y^2-x^2)
$
$F=F_1,~s=x\Rightarrow % 2
U_1=-2ym_0,~
U_2=0,~
U_3=2y^2(3x^2-2)+2m_1,~
U_4=2x^2ym_0,~
U_5=2y(y^2(5x^2-3)+2m_1),~
U_6=2xm_1n_0
$
$F=F_1,~s=y\Rightarrow % 3
U_1=2xn_0,~
U_2=2(y^2(2-3x^2)-m_1),~
U_3=0,~
U_4=2x(y^2(4-5x^2)+3x^2-2),~
U_5=-2xy^2n_0,~
U_6=-2ym_0n_1
$
$F=F_1,~s=x^2\Rightarrow % 4
U_1=-4xym_0,~
U_2=-2x^2ym_0,~
U_3=2x(y^2(5x^2-4)-3x^2+2),~
U_4=0,~
U_5=4xy(y^2(4x^2-3)+2-3x^2),~
U_6=2x^2(y^2(4x^2-3)+2-3x^2)
$
$F=F_1,~s=y^2\Rightarrow % 5
U_1=4xyn_0,~
U_2=2y(y^2(3-5x^2)-2m_1),~
U_3=2xy^2n_0,~
U_4=4xy(y^2(3-4x^2)+3x^2-2),~
U_5=0,~
U_6=2y^2(y^2(3-4x^2)+3x^2-2)
$
$F=F_1,~s=xy\Rightarrow % 6
U_1=2(x^2-y^2),~
U_2=-2xm_1n_0,~
U_3=2ym_0n_1,~
U_4=2x^2(y^2(3-4x^2)+3x^2-2),~
U_5=2y^2(y^2(4x^2-3)+2-3x^2),~
U_6=0
$
$F=F_2,~s=1\Rightarrow % 7
U_1=0,~
U_2=2xym_0,~
U_3=m_2n_0,~
U_4=4x^2ym_0,~
U_5=2ym_2n_0,~
U_6=x(y^2(x^2+1)+m_2)
$
$F=F_2,~s=x\Rightarrow % 8
U_1=-2xym_0,~
U_2=0,~
U_3=x(y^2(7x^2-5)-5x^2+3),~
U_4=2x^3ym_0,~
U_5=2xy(2y^2(3x^2-2)+3-5x^2),~
U_6=x^2(3-5x^2)n_0
$
$F=F_2,~s=y\Rightarrow % 9
U_1=-m_2n_0,~
U_2=x(y^2(5-7x^2)+5x^2-3),~
U_3=0,~
U_4=x^2(y^2(9-11x^2)+7x^2-5),~
U_5=y^2m_2n_0,~
U_6=-2xym_0n_1
$
$F=F_2,~s=x^2\Rightarrow % 10
U_1=-4x^2ym_0,~
U_2=-2x^3ym_0,~
U_3=x^2(y^2(11x^2-9)-7x^2+5),~
U_4=0,~
U_5=2x^2y(y^2(9x^2-7)+5-7x^2),~
U_6=x^3(y^2(9x^2-7)-7x^2+5)
$
$F=F_2,~s=y^2\Rightarrow % 11
U_1=-2ym_2n_0,~
U_2=2xy(2y^2(2-3x^2)+5x^2-3),~
U_3=-y^2m_2n_0,~
U_4=2x^2y(y^2(7-9x^2)+7x^2-5),~
U_5=0,~
U_6=xy^2(y^2(7-9x^2)+7x^2-5)
$
$F=F_2,~s=xy\Rightarrow % 12
U_1=-x(m_2+y^2(x^2+1)),~
U_2=x^2(5x^2-3)n_0,~
U_3=2xym_0n_1,~
U_4=x^3(y^2(7-9x^2)+7x^2-5),~
U_5=xy^2(y^2(9x^2-7)+5-7x^2),~
U_6=0
$
$F=F_3,~s=1\Rightarrow % 13
U_1=0,~
U_2=-m_0n_2,~
U_3=-2xyn_0,~
U_4=-2xm_0n_2,~
U_5=-4xy^2n_0,~
U_6=y(y^2(3-x^2)-x^2-1)
$
$F=F_3,~s=x\Rightarrow % 14
U_1=m_0n_2,~
U_2=0,~
U_3=y(y^2(7x^2-5)+3-5x^2),~
U_4=-x^2m_0n_2,~
U_5=y^2(y^2(11x^2-7)+5-9x^2),~
U_6=2xym_1n_0
$
$F=F_3,~s=y\Rightarrow % 15
U_1=2xyn_0,~
U_2=y((5-7x^2)y^2+5x^2-3),~
U_3=0,~
U_4=2xy(y^2(5-6x^2)+4x^2-3),~
U_5=-2xy^3n_0,~
U_6=y^2m_0(5y^2-3)
$
$F=F_3,~s=x^2\Rightarrow % 16
U_1=2xm_0n_2,~
U_2=x^2m_0n_2,~
U_3=2xy((6x^2-5)y^2+3-4x^2),~
U_4=0,~
U_5=2xy^2(y^2(9x^2-7)+5-7x^2),~
U_6=x^2y(y^2(9x^2-7)+5-7x^2)
$
$F=F_3,~s=y^2\Rightarrow % 17
U_1=4xy^2n_0,~
U_2=y^2(y^2(7-11x^2)+9x^2-5),~
U_3=2xy^3n_0,~
U_4=2xy^2((7-9x^2)y^2+7x^2-5),~
U_5=0,~
U_6=y^3(y^2(7-9x^2)+7x^2-5)
$
$F=F_3,~s=xy\Rightarrow % 18
U_1=y((x^2-3)y^2+x^2+1),~
U_2=-2xym_1n_0,~
U_3=y^2m_0(3-5y^2),~
U_4=x^2y((7-9x^2)y^2+7x^2-5),~
U_5=y^3((9x^2-7)y^2+5-7x^2),~
U_6=0
$

After taking the integrals of these $U_n$ with $I(x,y)$ you get a $18 \times 6$ matrix $U$ of the moments $u_n$. Taking the eigenvector associated with the smallest eigenvalue of $U^TU$ as described here, you will get the parameters $p_1, ... p_6$.
I tested them on the curve segment in the figure above. The segment is from an ellipse
$$\frac{((x-1)\cos(\frac{\pi}{6})+(y+1)\sin(\frac{\pi}{6}))^2}{2^2} + \frac{((x-1)\sin(\frac{\pi}{6})-(y+1)\cos(\frac{\pi}{6}))^2}{1^2} = 1$$
I got the following result
[ 1.         -1.70012217  2.53636068  0.49314339  0.91927365 -0.74693632]

which is not very accurate but I used a 2D-trapezoid rule for integration as described here and took a uniform $600 \times 600$ grid. You will definitely get better results with a more sophisticated grid and integration rule.
References

D.H. Ballard, Generalizing the Hough transform to detect arbitrary shapes, Pattern Recognition, Volume 13, Issue 2, 1981, Pages 111-122, ISSN 0031-3203,
https://doi.org/10.1016/0031-3203(81)90009-1
Popovici, I., &amp; Withers, W. D. (2009). Curve Parametrization by Moments. IEEE Transactions on Pattern Analysis and Machine Intelligence, 31(1), 15–26. https://doi.org/10.1109/TPAMI.2008.54
Popovici, I., &amp; Withers, W. D. (2006). Locating Thin Lines and Roof Edges by Custom-Built Moments. 2006 International Conference on Image Processing. https://doi.org/10.1109/ICIP.2006.312501
Popovici, I., &amp; Withers, W. D. (2006). Custom-built moments for edge location. IEEE Transactions on Pattern Analysis and Machine Intelligence, 28(4), 637–642. https://doi.org/10.1109/TPAMI.2006.75