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The radius of an ellipse from the origin.

  1. Mar 21, 2008 #1
    Hello,
    given (x^2)/(a^2) + (y^2)/(b^2) = 1.
    and using polar coordinates x=rcos(phi) , y=rsin(phi),
    equating gives r^2 = 1/[(cos^2(phi)/a^2) + (sin^2(phi)/b^2)].
    or if we leave b in the nominator :
    r= b/[(sin^2(phi)+(b^2/a^2)cos^2(phi)]^1/2.

    -could someone give a hint as to how the demoninator of the last expression can be turned into [1 - (e^2)cos^2(phi)]^1/2 where e is the eccentricity of the ellipse?
    and what is the value of e?


    thank you.
     
    Last edited: Mar 21, 2008
  2. jcsd
  3. Mar 21, 2008 #2
    [tex]e^2=1-\frac{b^2}{a^2}[/tex]

    And by the way this is introductory physics section.
     
  4. Mar 21, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi Erez! Welcome to PF! :smile:

    Hint: use cos^2(phi) + sin^(phi) = 1.

    Then … ? :smile:
     
  5. Mar 21, 2008 #4
    Hello,
    I am surprised I recived a reply so quick,
    thank you.
    p.s. in what category/section would this post belong ?
     
  6. Mar 21, 2008 #5
    Pre calculus mathematics, perhaps:smile:
     
  7. Mar 21, 2008 #6

    tiny-tim

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    Hi Erez :smile:

    Well, it was a short question, clearly stated, without loads of irrelevant gumph to read through … and some of us give questions like that priority! :wink:

    Well, this is just geometry, so it should really have gone into "Precalculus Mathematics", which is defined as "All math courses prior to calculus" :smile:
     
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