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Math Help - Elllipse given 3 points

  1. #1
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    Elllipse given 3 points

    Hi,

    Is there a way of calculating the equation of an ellipse given 3 knowned points?

    Kind regards,

    Kepler
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  2. #2
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    3 points is not enough information to completely determine an ellipse.
    For example, you can always find a circle through 3 noncollinear points.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by kepler View Post
    Is there a way of calculating the equation of an ellipse given 3 knowned points?

    We need more information. See snowtea's post.


    Fernando Revilla
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  4. #4
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    See also this thread.
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  5. #5
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    Hi,

    I see your point - the only condition I have is that the line of the major axis must pass through the center of the xy axis (0,0)(not the center of the ellipse). What about with four points? Is it too complicated? What I really want to know are the coordinates of the center of the ellipse.

    Kind regards,

    Kepler
    Last edited by kepler; January 7th 2011 at 12:37 PM.
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  6. #6
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    Quote Originally Posted by kepler View Post
    Hi,

    I see your point - the only condition I have is that the line of the major axis must pass through the center of the xy axis (0,0)(not the center of the ellipse). What about with four points? Is it too complicated? What I really want to know are the coordinates of the center of the ellipse.

    Kind regards,

    Kepler
    Please post the whole question you are working on, not just the bits you think are important.
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  7. #7
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    That's it. And the problem is general, to be inserted in a C/C++ program. I have two choices:
    1) The one I've explained
    2) Or I could use an aprox. eccentrity - but I rather not use this choice because it would place the problem in 3D and the eccentrity is aprox. only.

    Kind regards,

    Kepler
    Last edited by kepler; January 7th 2011 at 02:56 PM.
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  8. #8
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    Hi again,

    What "if" I consider the origin axis (0,0) as one of the focus? Could I use 3 points?

    Kind regards,

    Kepler
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  9. #9
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    If you have one focus is at (0,0), and you know the other focus is on the x-axis, then you only need 2 points (assuming they are not reflections of each other through the x-axis).
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  10. #10
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    Hi,

    No; asssuming just that one focus is at (0,0). The other one can be elsewhere. With 3 points, I think it's possible. But how? - my distance equations with square roots are not looking good...

    Kind regards,

    Kepler
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  11. #11
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    Quote Originally Posted by kepler View Post
    Hi,

    No; asssuming just that one focus is at (0,0). The other one can be elsewhere. With 3 points, I think it's possible. But how? - my distance equations with square roots are not looking good...

    Kind regards,

    Kepler
    Again I am asking you to post the whole question. Each time you have asked something and been given an answer, you change what you previously asked. What exactly are you trying to do?

    If you cannot give a straight answer I will be closing this thread because all I have seen so far are people wasting time giving answers to questions that keep getting changed.
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  12. #12
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    Hi again,

    Ok. My answers are an aproximation to a problem. This is an astronomy problem.
    I can generate the geocentric positions of the moon regarding the ecliptic (xy plane). What I need to know is the angle between the equinox (y=0) and the perigee. The ideal would be to construct the equation of an ellipse in 3D given the n points needed and then calculate the osculating elements (which I don't know how to do). The problem is that the osculating elements change from point to point, and the more points I use, the less precise I'll be.

    What I was thinking is, since I want to calculate the perigee, was to project the 3 points into the xy axis ( it would be an ellipse too but I would have to disregard the eccentricity, which is aprox. too) and assume the barycenter (focus) as the Earth (0,0) - wich isn't accurate. But maybe I could get a certain precision. With the 4 points, I would need only the coordinates of the center or one of the focus to obtain the longitude of the perigee - but I loose precision by adding one extra point.

    It's a little confusing...

    Kind regards,

    Kepler
    Last edited by kepler; January 8th 2011 at 05:48 AM. Reason: Complete information
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  13. #13
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    If its an astronomy problem with angles, I suggest using the equation of a conic in polar form.
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  14. #14
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    What's the equation of a conic in polar form? I only have it in special cases.

    Regards,

    Kepler
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