This is all related to Spivak's Calculus book 3rd Edition, Chapter 4, Appendix III Polar Coordinates, Exercise 5.
Here is the exercise:
Here is his solution:
My problem is the highlighted part of his solution. From what I know, if $R_1$ is the distance form one focus of a hyperbola and $R_2$ is the distance from the other focus of the hyperbola, to a point on the hyperbola, then: $|R_1-R_2|=c$, where $c$ is constant.
When the point is on one of the two parts of the hyperbola $R_1>R_2$ and vice versa.
However, he chooses $r>s$ if $a>0$ or $r<s$ if $a<0$ for no apparent reason. Since $a$ is constant, he is clearly making a choice. It is like he is constraining the point to only this one part. If this choice did not alter his desired result I would be fine with it.
However if I have not made any mistakes,
By his choice, indeed $r = Λ/(1+ε\cos(θ))$;
By choosing the opposite, $r = Λ/(1-ε\cos(θ))$.
After arriving at these results I was even more confused since it felt like for a point moving on each part of the hyperbola there was a different equation (in polar coordinates) describing it. So finally my questions are,
- Did he, and if he did, why did he make this choice?
- If my results are correct, how do these two polar equations connect?
To understand what is going on it is better to consider a specific example: if we take $a=2$ and $\epsilon=2$ then $\Lambda=-6$ and the equation given by Spivak reads: $$ r={-6\over1+2\cos\theta}. $$ But $r\ge0$, hence this is defined only for $1+2\cos\theta<0$, that is for ${120°<\theta<240°}$. This corresponds to that branch of the hyperbola which is farther from the origin and is consistent with the position $r-s=2a$, which implies $r>s$.
For the other values of $\theta$, that is for ${-120°<\theta<120°}$, the equation gives a negative value of $r$ and we would usually discard those values as "impossible". But we can give a meaning to those values if we stipulate that $(r,\theta)$ corresponds, when $r$ is negative, to the point $(-r,\theta+180°)$ (i.e. a negative radius means that the point is in the opposite direction with respect to $\theta$). In that case we can define $r'=-r$ and $\theta'=\theta+180°$, which inserted into the above equation give: $$ r'={6\over1-2\cos\theta'},\quad\text{with}\quad 60°<\theta'<300°. $$ But this last equation is exactly what you would get starting with $s-r=2a$, hence it describes the other branch of the hyperbola.
I don't know if this extensions of polar coordinates to $r<0$ is widely accepted, but it is certainly enforced in graphing softwares, because they transform a polar equation like $r=f(\theta)$ into the curve $$ \cases{ x=f(\theta)\cos\theta\\ y=f(\theta)\sin\theta\\ } $$ and a negative value of $f(\theta)$ amounts at taking the opposite vector, as described above.