Actually I was suggesting my above answer to the people at geogebra.org and they told me about this nice parametric equation that only needs the center point and any 2 points on the ellipse, where A is the center:
at 0 it equals C, at 1/2)*Pi it equals B, then at 2*pi back to C!
This one just wows me after as hard as I worked to figure out the one above...
And from that I derived this one:
f(t) = (1/2)*D+(1/2)*B+((1/2)*B-(1/2)*D)*sin(t)+(C-(1/2)*D-(1/2)*B)*cos(t)
Which given 3 points on the ellipse parameterizes it so that between B and D is the shorter axis...