I think that i'm over looking something with this problem. Below is the equation of an hyperbola in polar form.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]R=\frac{1}{1 + 2cos{\theta}}[/tex]

when [tex]\theta =\pi[/tex] shouldn't [tex]R = -1[/tex]? And not [tex]R= 1[/tex]

Am I over looking some property of the [tex]\cos[/tex] function?

Even when i evalute this expression at [tex]\theta=\pi[/tex]in my ti-89 i get that R is = to 1. What am i not seeing?

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# Homework Help: Conics in polar equation form

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