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Eccentric Angle of an Ellipse

  1. May 30, 2007 #1
    I'm revising form my A-levels now and I ran into a bit of problem with a question. It looks easy, but I can't get the answer at the back of the book. Could be a typo, but could be me that's wrong.

    Question: The eccentric angle corresponding to the point (2, 1) on the ellipse with equation [tex] x^2 + 9y^2 = 13[/tex] is [tex]\theta[/tex]. Find [tex]\tan \theta[/tex]

    The book isn't very clear on what the eccentric angle is, so could someone maybe explain that to me, please? I understand it as the angle from the middle of the ellipse - in this case the origin - to the point (2, 1). So, [tex]\tan \theta[/tex] would be opposite/adjacent, 1/2. Apparently, it's not.
     
    Last edited: May 30, 2007
  2. jcsd
  3. May 31, 2007 #2

    Dick

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    I just looked it up at mathworld. The ellipse can be parametrized as x=a*cos(t), y=b*sin(t) where a and b are the semi-axes. The angle 't' is the 'eccentric angle'.
     
  4. Jun 2, 2007 #3
    Thank you. I should've done that myself.
     
  5. Jun 19, 2009 #4

    SAC

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    Thanks a lot, Dick. This was causing me a lot of pain when I saw a question asking how I would graphically determine it, seeing as I had no idea what the eccentric angle was. Makes perfect sense now, just have to use a circle that contains the ellipse to determine the new angle.

    Edit: (oh, and sorry to resurrect this thread, just occurred to me that I shouldn't have done so. I'm just so glad now that I couldn't help it :D)
     
  6. Jun 20, 2009 #5

    Dick

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    Well, cheers. Just because it takes two years doesn't mean it's not worth announcing you got it. I guess.
     
  7. Jun 20, 2009 #6

    SAC

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    Oh, I just happened to come across this on Google, so it wasn't long at all for me. Everything is relative.
     
  8. Jun 20, 2009 #7

    Dick

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    Right, sorry, I confused you with to OP.
     
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