How to find center of ellipse from two points (these are just points on the ellipse, not related to foci), and two radii ($r_x$ and $r_y$, from standard definition of the ellipse $\frac{x^2}{r_x^2} + \frac{y^2}{r_y^2}=1$) of that ellipse? (there will be two centers, I assume)

  • $\begingroup$ You mean the focal points? $\endgroup$ – Raskolnikov Nov 21 '10 at 13:07
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    $\begingroup$ Your question is very confusing. Is there some relationship between the two points and the two radii? For the radii, are you given the line segments or just the lengths? What exactly do you mean by radii -- the distance from one of the two focal points (not centers) to some point on the ellipse? Which point? $\endgroup$ – Peter Shor Nov 21 '10 at 13:09
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    $\begingroup$ In general one needs 5 points to determine a conic section: en.wikipedia.org/wiki/Five_points_determine_a_conic $\endgroup$ – Joseph Malkevitch Nov 21 '10 at 14:00
  • $\begingroup$ The center of the ellipse is the midpoint of the segment joining the two foci. Alternatively, if you have the implicit Cartesian equation of an ellipse, there are methods for determining the center from the coefficients of the quadratic. $\endgroup$ – J. M. isn't a mathematician Nov 21 '10 at 14:07
  • $\begingroup$ Edited the question. $\endgroup$ – Rogach Nov 21 '10 at 17:43

You seem to be asking about an ellipse whose axes are aligned with the $x$ and $y$ directions. Note that an ellipse which is rotated by an arbitrary angle is still an ellipse. But you cannot uniquely determine such an ellipse from the information given.

If two points $(x_1, y_1)$ and $(x_2, y_2)$ pass through an ellipse centred at $(x_0, y_0)$ with semi-axes along $x$ and $y$ of length $r_x$ and $r_y$ respectively, then the points $(x_1/r_x, y_1/r_y)$ and $(x_2/r_x, y_2/r_y)$ pass through a circle of unit radius with centre $(x_0/r_x, y_0/r_y)$. So if you can find the centre of this circle, you can find the centre of the corresponding ellipse.


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