Daily Archives: May 15, 2020

An Ellipse in Its Polar Form

In this appendix to my previous post “From Dancing Planet to Kepler’s Laws“, we derive the polar form for an ellipse that has a rectangular coordinate system’s origin as one of its foci.

Fig. 1

We start with the ellipse shown in Fig. 1. Namely,

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.

Clearly,

f^2=a^2-b^2\implies f< a.\quad\quad\quad(1)

After shifting the origin O to the right by f, the ellipse has the new origin O' as one of its foci (Fig. 2).

Fig. 2

Since x = x' + f, y=y', the ellipse in x'O'y' is

\frac{(x'+f)^2}{a^2}+\frac{y'^2}{b^2}=1.\quad\quad\quad(2)

Substituting x', y' in (2) by

x'=r\cos(\theta), y'= r\sin(\theta)

yields equation

a^2r^2\sin^2(\theta)+b^2r^2\cos^2(\theta)+2b^2fr\cos(\theta)+b^2f^2-a^2b^2=0.

Replacing \sin^2(\theta), f^2 by 1-\cos^2(\theta), a^2-b^2 respectively, the equation becomes

b^2r^2\cos^2(\theta)+a^2r^2(1-\cos^2(\theta)+2b^2fr\cos(\theta)-a^2b^2+b^2(a^2-b^2)=0.\quad\quad\quad(3)

Fig. 3

Solving (3) for r (see Fig. 3) gives

r=\frac{b^2}{f cos(\theta)+a} or r=\frac{b^2}{f \cos(\theta)-a}.

The first solution

r=\frac{b^2}{f\cos(\theta)+a} \implies r= \frac{\frac{b^2}{a}}{\frac{f}{a}\cos(\theta)+1}.

Let

p=\frac{b^2}{a}, e=\frac{f}{a},

we have

r = \frac{p}{1+e\cdot\cos(\theta)}.

The second solution is not valid since it suggests that r < 0:

\cos(\theta)<1 \implies f\cdot\cos(\theta) < f \overset{(1)}{\implies} f\cos(\theta)<a \implies f\cos(\theta)-a<0.