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I'm considering an arbitrary, non-degenerate ellipse here, i.e., without assuming that it's centred on the origin or either axis, nor oriented at any specific angle.

I know either 5 points on the perimeter $(x_1,y_1)$ through $(x_5,y_5)$, or the general cartesian equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ where $F = 0$ or $1$. The two forms are equivalent as it's easy to solve the simultaneous equations for 5 values of $(x,y)$ to get $A$ through $F$ when the ellipse isn't degenerate. These uniquely define one ellipse - or a conic at any rate ;-)

But what I'm really after are the cartesian foci and its eccentricity (I can get everything else I might need from these: the major/minor axes, rotate the ellipse to align with the axes, or find its equation in polar form if needed).

I can't find a way to do that......

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marked as duplicate by David K, drhab, jameselmore, Micah, Umberto P. Oct 20 '15 at 17:31

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  • $\begingroup$ That looks like a solution - I'll try it out late, and report back. In the meantime it looks good - thanks! $\endgroup$ – Stilez Oct 20 '15 at 16:41