Is there a way to find out the centre coordinates of an ellipse given the corner points of a trapezoid inscribed within it? The height and the lengths of AB and CD are known. Here is a diagram:

In this, given the points A,B,C and D, is there a way to gain any information about the ellipse outside this trapezoid? I realise that its major and minor axes could vary, but can't the centre be found out?

  • $\begingroup$ In case anyone's wondering, this is a part of a computer vision problem I'm trying to find a solution for. $\endgroup$ – Caife Oct 17 '16 at 10:12

The proposed ellipse is not fully determined. You need to specify some additional information, such as a fifth point or the length of the minor axis, to define a unique ellipse.

  • $\begingroup$ In my problem, I can also find out middle points between A-C and B-D which lie on the ellipse, if needed. Basically, I can find out any point that lies on the ellipse given a straight horizontal line which passes through it - but these additional lines need to be kept at a minimum. Will that help? $\endgroup$ – Caife Oct 17 '16 at 10:11
  • $\begingroup$ Give us one additional point on the ellipse to make a total of five points. You can also specify that the ellipse is tangent to a given line. $\endgroup$ – Oscar Lanzi Oct 17 '16 at 10:17
  • $\begingroup$ You may assume that the major axis will be parallel to the horizontal lines. A fifth and sixth point can also be found out by drawing a third line and finding two edge points. I'm curious, though - how will five points make it possible to find a unique ellipse? $\endgroup$ – Caife Oct 17 '16 at 10:19
  • $\begingroup$ Here is why five points specify an ellipse. You probably know that three points specify a circle. But for an ellipse in a given plane you have two additional degrees of freedom -- the eccentricity (or equivalently the ratio of the major axis to the minor axis) and the orientation of the major or minor axis. Adding two more points, for a total of five instead of three, takes up the two additional degrees of freedom. Also -- a necessary condition to actually have an ellipse instead of a hyperbola is that the five points should be the vertices of a convex pentagon. $\endgroup$ – Oscar Lanzi Oct 17 '16 at 10:33
  • $\begingroup$ I also foind that assuming the axes parallel to sides of the trapezoid does not really work at keaat for an isosceles trapezoid, and eduted the first comment accordingly. $\endgroup$ – Oscar Lanzi Oct 17 '16 at 10:45

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