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Suppose I have a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. I know that when choosing the right-hand side focus as the pole and the polar axis has the same direction as x-axis, the equation of the hyperbola in polar coordinates is $r=\dfrac{p}{1-\varepsilon\cos\varphi}$, where $p=\frac{b^2}{a}$, $\varepsilon$ is the eccentricity of the hyperbola and $r,\varphi$ are the polar coordinates of a point on hyperbola.

I understand that when using this equation, I only get the right-hand side branch of the hyperbola. However when I graph this for example in Geogebra, it gives me the whole hyperbola. What am I missing here?

Example

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    $\begingroup$ "I understand that when using this equation, I only get the right-hand side branch of the hyperbola." Don't understand that; it isn't so! :) In your example, consider $\theta=0$, which gives $$a = \frac{9/4}{1-\frac54\cdot 1}=-9$$ Negative values are plotted in the opposite direction, so $\theta=0$ and $a=-9$ gives the vertex on the left branch of the hyperbola. Likewise, all negative $a$ values give points on the left branch. The transition from left to right (that is, negative to positive) happens when $1-\epsilon \cos\theta = 0$, which corresponds to the angle of the asymptote. $\endgroup$
    – Blue
    Nov 16, 2018 at 12:09
  • $\begingroup$ Oh, I see it now, thank you so much! I have never used polar coordinates so it was super confusing to me. Can you please also explain why is there a formula with both plus and minus sign given in Wikipedia en.wikipedia.org/wiki/Hyperbola#Polar_coordinates? Does it have to do with the choice of direction of the polar axis? $\endgroup$
    – user570271
    Nov 16, 2018 at 12:41
  • $\begingroup$ The $\pm$ determines which focus appears at the pole. (In the parabola case (where $\epsilon=1$), you can think of it as "which direction the parabola opens".) I personally favor having the left-hand focus there; that way, $\theta=0$ corresponds to the closer branch. In any case, to help wrap your mind around what's happening, check some test points; in particular, $\theta=0$ and $\theta=180^\circ$. (Again, in the parabola case, you'll know which way the curve opens, because one of those values gives a denominator of zero, so that the corresponding point is "at infinity".) $\endgroup$
    – Blue
    Nov 16, 2018 at 12:50
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    $\begingroup$ @Blue: Would you be willing to write up your comments as an Answer, so that we can get this question off the Unanswered queue? $\endgroup$
    – aleph_two
    Dec 25, 2018 at 4:53
  • $\begingroup$ @aleph_two: Done! $\endgroup$
    – Blue
    Dec 25, 2018 at 5:10

1 Answer 1

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(Converting comments to an answer, by request.)

"I understand that when using this equation, I only get the right-hand side branch of the hyperbola." Don't understand that; it isn't so! :)

In your example, consider $\theta=0$, which gives $$a = \frac{9/4}{1-\frac54\cdot 1}=-9$$

Negative values are plotted in the opposite direction, so $\theta=0$ and $a=−9$ give the vertex on the left branch of the hyperbola. Likewise, all negative $a$ values give points on the left branch. The transition from left to right (that is, negative to positive) happens when $1−\varepsilon \cos\theta=0$, which corresponds to the angle of the asymptote.

Regarding signs in the formula in Wikipedia, $$r = \frac{p}{1\mp e\cos\phi}$$

The $\mp$ determines which focus appears at the pole. (In the parabola case (where $\varepsilon=1$, you can think of it as determining "in which direction the parabola opens".) I personally favor having the hyperbola's left-hand focus there (taking $\pm$ to be $+$), so that $\theta=0$ corresponds to vertex on the closer branch. (Taking $\pm$ to be $+$ for all eccentricities also unifies the behavior of all conics near $\theta=0$. The curves "start" at a vertex with $\theta=0$, then wrap around the pole/focus in the same way.)

In any case, to help wrap your mind around what's happening, check some test points; in particular, $\theta=0$ and $\theta=180^\circ$. (Again, in the parabola case, you'll know which way the curve opens, because one of those values gives a denominator of zero, so that the corresponding point is "at infinity".)

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