Ellipse Arcana

  1. 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.

  2. 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.

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

  4. 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.
  5. One could go on and on.