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Question about semi-major axis of an ellipse

  1. Jul 3, 2010 #1
    I hope this the right place to post my question...

    2.jpg

    1-2.jpg

    should it be, "we can define a as half the sum of distances......"?
    please correct and explain if I'm mistaken
    thanks
     
    Last edited: Jul 3, 2010
  2. jcsd
  3. Jul 3, 2010 #2

    HallsofIvy

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    Science Advisor

    You are mistaken.

    When [itex]\theta= \theta'[/itex], [itex]1/r= C(1- \epsilon cos(\theta- \theta')=[/itex][itex] C(1- \epsilon cos(0))=[/itex][itex] C(1- \epsilon)[/itex] so that [itex]r= 1/(C(1-\epsilon)[/itex].

    When [itex]\theta= \theta'+ \pi[/itex], [itex]1/r= C(1+\epsilon cos(\theta- \theta- \pi)=[/itex][itex] C(1- \epsilon cos(\pi))=[/itex][itex] C(1+ \epsilon)[/itex] so that [itex]r= 1/(C(1+\epsilon)[/itex].

    The total distance between those points is [itex]1/C(1+\epsilon)+[/itex][itex] 1/C(1- \epsilon)[/itex][itex]= (1/C)(1/(1+\epsilon)+ 1/(1- \epsilon))[/itex]. Getting the common denominator, [itex](1+\epsilon)(1-\epsilon)= 1-\epsilon^2[/itex], we have
    [tex](1/C)\frac{1- \epsilon+ 1+ \epsilon}{1- \epsilon^2}= (1/C)\frac{2}{1- \epsilon^2}[/tex]

    Half of that is
    [tex]a= \frac{1}{C(1- \epsilon^2)}[/tex]
     
    Last edited by a moderator: Jul 3, 2010
  4. Aug 11, 2010 #3
    and that's what I've said :confused:
    thank you
     
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