# Difference of Hyperbolic foci and Spivak's Solution

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 $$r if $$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?
• You can try that with Desmos: desmos.com/calculator/2enzltgitn Jul 7, 2020 at 11:12
• @Aretino After trying both the equations to Desmos it has certainly brought some insight since it showed that both equations represent the same hyperbola with a different choice for the focus that is on the origin. So thanks for the tip. However I am still puzzled as to why choosing lets say the left focus to be on the origin, means that R1 > R2. Why do the distances from the foci, depend on which focus is on the origin, it shouldn't be true. Jul 7, 2020 at 14:32
• How is $\Lambda$ defined? Jul 7, 2020 at 16:25
• $Λ=(1-\epsilon ^2)a$. I believe that $r=\frac{ Λ}{1-\epsilon cos\theta}$ is also a conic. The only difference between $r=\frac{ Λ}{1-\epsilon cos\theta}$ and $r=\frac{ Λ}{1+\epsilon cos\theta}$ is how the directrix is chosen. However, I am not 100% sure. Jul 7, 2020 at 16:34
• @torontohrb I have checked and indeed $r=Λ/(1-ϵ*cosθ)$ is a conic section and infact a hyperbola. From what I know hyperbolas have two directrixes, and in the two hyperbolas that form - one for $r=Λ/(1-ϵ*cosθ)$ and one for $r=Λ/(1+ϵ*cosθ)$ - both directrixes are different. What happens is, the focus that is not on the origin gets mirrored parallel to the y'y (vertical) axis. Jul 7, 2020 at 20:27

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.