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I am trying to find a least-squares ellipse fit for a set of 100 data points $(x,y)$.

Now I have found the values of $A,B,C,D,E,F$ according to the conical equation of the ellipse $$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0 $$ I would like to know how to find the points that actually lie on this ellipse. From my basic understanding, if I substitute a value of $x$ in the above equation, it should give me the corresponding value of $y$.

When I do the above, I get a straight line and not really a fitted ellipse. How can I find the fitted ellipse?

My task is to plot these points so that I can see the best possible fit. For reference see [link]. This is the source of ellipse fitting that I am currently using.

I appreciate help from anyone who has experience with this. I am sorry if I am lacking some basic mathematical knowledge, but from what I understand, it isn't all that straightforward.

Regards

Arj

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  • $\begingroup$ Try to google "fitting the ellipse" and maybe this will help: cs.cornell.edu/cv/OtherPdf/Ellipse.pdf $\endgroup$
    – user35603
    Aug 13, 2014 at 10:55
  • $\begingroup$ Formulas look there. Or choose other. math.stackexchange.com/questions/809907/… $\endgroup$
    – individ
    Aug 13, 2014 at 11:15
  • $\begingroup$ "I get a straight line and not really a fitted ellipse" - This doesn't sound right. What were the parameters? What values of x did you use? Can you post how you found the y values? They are not linear in x unless A,B,C are small relative to the other params, I think? $\endgroup$
    – Sanjay Manohar
    Aug 13, 2021 at 11:32

1 Answer 1

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A sure way to determine whether a point lies on the ellipse is to substitute the point's x- and y-coordinates into the equation and see whether the equation is exactly satisfied. (Your method is also valid, but requires more work.)

An example of how to use MATLAB to plot a curve and points is at https://www.mathworks.com/help/curvefit/fit.html. You will need to solve your equation for y. Be careful, though, because a point may appear to be on the ellipse but actually not be on it.

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