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I was playing with the setup for Desargues Theorem in Geogebra today and got a very odd-looking (to me) result. I imagine I could grind through this analytically and get an ugly-looking parametric representation of it but I'm more interested to know whether anyone recognises these curves by sight and can attach a name to them. This is mostly a matter of idle curiosity at this point:

enter image description here

The red one looks like an elliptic curve to me, but the green one doesn't.

The setup is a bit convoluted to describe but here goes in case you want to try this at home:

  1. Take an arbitrary disjoint pair of ellipses. Join their foci in pairs (orange lines). Pick any point on the right ellipse (A) and project it to the point where those two orange lines intersect (P) (red line). One of the points where this intersects the left ellipse is identified.

The foci of each ellipse, together with one of the two points on its edge thus identified, form a triangle (black lines). By construction, the two triangles are in perspective. Each triangle fully determines its corresponding ellipse.

  1. Construct points as in Desargues Theorem: produce all the sides of the triangles (dark blue dotted lines for the right-hand triangle, light blue for the left-hand one). Place a point were corresponding lines intersect. We'll call these the D-points. As the theorem predicts, they are collinear (dotted purple line).

  2. Turn on "trace" for the three points. This will plot the locus of each point when it moves.

  3. Take the point on the right ellipse and move it around the ellipse.

The result is the thick green and red lines. The red curve is the locus of one of D-points, the green is the two-branched locus of another; the third D-point (blue) remains stationary (this is not surprising; it's the point of intersection of the two lines joining the foci, which aren't moving).

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    $\begingroup$ The line $AP$ does not always intersect the second ellipse. So you can't put points on one ellipse into $1:1$ relation with points on the other. Which means I'd consider both points on the other ellipse, and the combined locus from these, which should be a single algebraic curve. I guess one can compute the degree of said curve, e.g. using a couple of examples in generic position, but I doubt you'll get more of a name than “algebraic curve of degree $d$”. $\endgroup$ – MvG May 7 '17 at 8:56

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