Elliptical Pizza Theorem

So to find the ellipse you find 2 additional points with the formulas above and then the well known method for drawing a conic through 5 points: conic through 5 points.. I think it does a nice job of describing how a pizza will look with perspective if lines perpendicular to the viewing plane are drawn to the right, top, and left of the pizza, calling those P1, P2, and P3 respectively...

**UPDATE**

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:

f(t)=A+(B-A)*sin(t)+(C-A)*cos(t), 

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...

f(pi) gives the opposing point on the long axis that C is on...


Alternative Ellipse: