Ellipse Arcana

   Equation of an ellipse in Cartesian coordinates

         (x/a)**2+(y/b)**2=1  ,

       where a is the semi-major axis and b is the semi-minor axis if a>b.


   Let r_plus and r_minus be the distance to the ellipse curve from the
       left and right ellipse respectively:


       r_plus+r_minus=\sqrt{(x+c)**2+y**2}+\sqrt{(x-c)**2+y**2}

       =\sqrt{x**2+2cx+a**2-b**2+b**2-(bx/a)**2}

           +\sqrt{x**2-2cx+a**2-b**2+b**2-(bx/a)**2}

       =\sqrt{(xc/a)**2+2cx+a**2}+\sqrt{(xc/a)**2-2cx+a**2}

       =(xc/a+a)+(a-xc/a)

       =2a

       which verifies the geometrical ellipse construction technique.


   If e goes to 1, with a held fixed then the ellipse flattens
       into a line segment that is 2a in length.


   If e goes to 1 with b**2/a held fixed, you get a parabola.
       Let the distance from the left end of the ellipse to the
       focus be

       c'=a-c=a-\sqrt{a**2-b**2} \approx a-a(1-(1/2)(b/a)**2)

         =(1/2)b**2/a 
      
         to first order in (b/a)**2.
       If we hold  b**2/a fixed while letting a go to infinity, then
       e goes to 1, but c' remains finite.

       The equation of the ellipse in this case should be
       transformed using x'=x+a:

       [(x'-a)/a ]**2 +(y/b)**2=1

       (x'/a)**2-2x'/a+1+(y/b)**2=1

       x'**2*(b/a)**2-2x'(b**2/a)+y**2=0

       -2x'(b**2/a)+y**2=0  in the limit where (b/a)**2 goes to 0
                            while b**2/a stays constant

       -4c'x'+y**2=0

       x'=y**2/4c'   .

       Thus we have found an equation of parabola as the limiting
       form of the ellipse in this case.

   One could go on and on.