# 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.