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Conics in polar equation form

  1. Apr 30, 2005 #1
    I think that i'm over looking something with this problem. Below is the equation of an hyperbola in polar form.

    [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?
     
  2. jcsd
  3. Apr 30, 2005 #2
    You have made a mistake,im afraid.
    Even if you find that R=-1,it is just a mathmatic form.
    R is always positive,you can just say R=|1/2cos(theta)|
    If still don't understand,contact me at wangkehandsome@hotmail.com,I will be glad to anwser it for you and even be more glad if you point out my fallacy.
     
  4. Apr 30, 2005 #3
    Well, then he hasn't made a mistake: the given expression is just wrong!
     
  5. Apr 30, 2005 #4
    I see the error. Both of you are correct. the expression is not explicit enough. when i plot these points in the xy plane they are (-1,0) BUT the same coordinates in the
    [tex](r,\theta)[/tex] coordinate system are [tex](1,\pi)[/tex] because the radius is always positive
     
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