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This is all related to Spivak's Calculus book 3rd Edition, Chapter 4, Appendix III Polar Coordinates, Exercise 5.

Here is the exercise:

enter image description here

Here is his solution:

enter image description here

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?
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    $\begingroup$ You can try that with Desmos: desmos.com/calculator/2enzltgitn $\endgroup$ Jul 7, 2020 at 11:12
  • $\begingroup$ @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. $\endgroup$
    – Stamatis
    Jul 7, 2020 at 14:32
  • $\begingroup$ How is $\Lambda$ defined? $\endgroup$ Jul 7, 2020 at 16:25
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    $\begingroup$ $ 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. $\endgroup$ Jul 7, 2020 at 16:34
  • $\begingroup$ @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. $\endgroup$
    – Stamatis
    Jul 7, 2020 at 20:27

1 Answer 1

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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.

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  • $\begingroup$ Great, you answer cleared things up for me. Thank you $\endgroup$
    – Stamatis
    Jul 10, 2020 at 9:13

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