Mario, here is one way:

(* generation of similar data: *)
dd = .1;
data = Table[{2 + x^2 + RandomReal[{-dd, dd}], x + RandomReal[{-dd, dd}]}, {x, -2, 2, .1}];
(* working on x- and y-component seperately: *)
{datax, datay} = Transpose[data];
(* we then have: *)
ListPlot[{datax, datay}]

(* this then can be fitted in a regular way: *)
modelx = x2 t^2 + x1 t + x0;
modely = y1 t + y0;
paramsx = FindFit[datax, modelx, {x0, x1, x2}, t];
paramsy = FindFit[datay, modely, {y0, y1}, t];
func[t_] = ({modelx, modely} /. paramsx) /. paramsy;
ParametricPlot[func[t], {t, 1, Length[data]}, Epilog -> {Red, Point[data]}]

Does that help? Regards -- Henrik

Well thank you for your questions they helped me to think better about my objective.

The motion is a rotation of a robot about a center point. But there is an error that's why this ellipse is formed. i.e. it should be a point rather than an ellipse.

The data source is from sensor position data, then it is transformed using a transformation matrix to the center.

The objective is to predict the error so I would predict the motion without relaying on the sensor, just the difference between angles.

Can you supply more specifics about your objective in "fitting" ? Here are some examples:

• If the objective is to provide a curve that is an approximation of the the data with a formula (parabola, ellipse, or something else), how close must that approximation be?

• If the objective is to predict future points, what is the mechanism for generating the current data and future data?

• If the objective is "I'll know it when I see it", then know that we can't read your mind. I say this because of your statement "But still, these two approaches are not the 'best fit', it just a better guess."

I have two updates:

I was able to fit equations in this link as follows:

Also, I discovered that I can use the drawing tool to draw ellipses on the points and then convert the cell with the plot (without the original) to input, then it will give the equation for the ellipse in terms of Graphics[Circle[{29.009734682576344, 57.18617339634879}, {0.12473528321493177, 0.2126919292376499}]]

But still, these two approaches are not the "best fit", it just a better guess.