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.

1 thought on “An Ellipse in Its Polar Form

  1. Pingback: From Dancing Planet to Kepler’s Laws | Vroom

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