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

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).