Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - Chapter Review Exercises - Page 639: 40

If ${f_1}\left( \theta \right) = - {f_2}\left( {\theta + \pi } \right)$, then the two polar equations represent the same points in rectangular coordinates. Therefore, $r = {f_1}\left( \theta \right)$ and $r = {f_2}\left( \theta \right)$ define the same curves in polar coordinates. Using this result, therefore $r = {f_1}\left( \theta \right) = \frac{{de}}{{1 - e\cos \theta }}$ and $r = {f_2}\left( \theta \right) = \frac{{ - de}}{{1 + e\cos \theta }}$ define the same conic section.

Case 1. Consider the polar equation $r = {f_1}\left( \theta \right)$. In rectangular coordinates, we have (1) ${\ \ \ }$ $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right) = {f_1}\left( \theta \right)\left( {\cos \theta ,\sin \theta } \right)$ Case 2. Consider the polar equation $r = {f_2}\left( \theta \right)$. In rectangular coordinates, we have $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right) = {f_2}\left( \theta \right)\left( {\cos \theta ,\sin \theta } \right)$ However, shifting $\theta$ by $\pi$ we get ${f_2}\left( {\theta + \pi } \right)\left( {\cos \left( {\theta + \pi } \right),\sin \left( {\theta + \pi } \right)} \right)$ also defines the same point in polar coordinates. Thus, $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right) = {f_2}\left( \theta \right)\left( {\cos \theta ,\sin \theta } \right)$ $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right) = {f_2}\left( {\theta + \pi } \right)\left( {\cos \left( {\theta + \pi } \right),\sin \left( {\theta + \pi } \right)} \right)$ $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right) = {f_2}\left( {\theta + \pi } \right)\left( { - \cos \theta , - \sin \theta } \right)$ (2) ${\ \ \ }$ $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right) = - {f_2}\left( {\theta + \pi } \right)\left( {\cos \theta ,\sin \theta } \right)$ If ${f_1}\left( \theta \right) = - {f_2}\left( {\theta + \pi } \right)$, then equations (1) and (2) represent the same points in rectangular coordinates. Therefore, $r = {f_1}\left( \theta \right)$ and $r = {f_2}\left( \theta \right)$ define the same curves in polar coordinates. Case 3. Consider the conic section: $r = \frac{{de}}{{1 - e\cos \theta }}$. Write $r = {f_1}\left( \theta \right) = \frac{{de}}{{1 - e\cos \theta }}$. Suppose ${f_1}\left( \theta \right) = - {f_2}\left( {\theta + \pi } \right)$. Then $r = - {f_2}\left( {\theta + \pi } \right) = \frac{{de}}{{1 - e\cos \theta }}$ $r = {f_2}\left( {\theta + \pi } \right) = - \frac{{de}}{{1 - e\cos \theta }}$ Shifting $\theta$ by $ - \pi $, we get $r = {f_2}\left( \theta \right) = - \frac{{de}}{{1 - e\cos \left( {\theta - \pi } \right)}} = - \frac{{de}}{{1 + e\cos \theta }}$ $r = {f_2}\left( \theta \right) = \frac{{ - de}}{{1 + e\cos \theta }}$. Using previous results, therefore $r = {f_1}\left( \theta \right) = \frac{{de}}{{1 - e\cos \theta }}$ and $r = {f_2}\left( \theta \right) = \frac{{ - de}}{{1 + e\cos \theta }}$ define the same conic section.