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I read that the equation of a conic in polar coordinates is $$r=\frac{l}{1+e\cos \theta}.$$

But when I try to reduce the hyperbola $$x^2 - y^2 =1$$ to that form by setting $x=r\cos \theta $, $y=r \sin \theta$, I instead get

$$r^2 = \frac{1}{\cos 2\theta}.$$

Where's the problem?

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    $\begingroup$ The $\theta$ in the first formula is not the angle from the origin, but the angle from one of the foci. $\endgroup$ – Daniel Fischer Sep 30 '13 at 15:10
  • $\begingroup$ Oh, wow. Thanks for clearing that up. So how would I go about transforming the equation? Are there simple identities relating the angles? $\endgroup$ – Spine Feast Sep 30 '13 at 15:49
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As Danil Fischer already pointed out in a comment, the difference is the location of the center: in the first formula, the center of the coordinate system is one of the two foci of the hyperbola. In the other formula you derived yourself, the center of the coordinate system is the center of symmetry of the hyperbola.

This section of the Wikipedia article on the Hyperbola has some details about a conversion. It starts with the hyperbola

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

which is your ellipse but scaled by a factor of $a$ in $x$ direction and by a factor of $b$ in $y$ direction. So you have $a=b=1$.

It then derives an equation of

$$r=\frac{a(e^2-1)}{1+e\cos\theta}$$

which is in a coordinate system around the point $(-ea,0)$. From a different part of that page, one can find

$$e=\frac{\sqrt{a^2+b^2}}{a}$$

So with this you should have all the bits and pieces to do the conversion if needed. Your $l$ does already agree with that formula, since

$$ l = \frac{b^2}{a} = a(e^2-1) $$

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