# Elllipse given 3 points

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• Jan 7th 2011, 10:23 AM
kepler
Elllipse given 3 points
Hi,

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

Kind regards,

Kepler
• Jan 7th 2011, 10:38 AM
snowtea
3 points is not enough information to completely determine an ellipse.
For example, you can always find a circle through 3 noncollinear points.
• Jan 7th 2011, 10:49 AM
FernandoRevilla
Quote:

Originally Posted by kepler
Is there a way of calculating the equation of an ellipse given 3 knowned points?

Fernando Revilla
• Jan 7th 2011, 11:34 AM
Opalg
• Jan 7th 2011, 12:06 PM
kepler
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
• Jan 7th 2011, 12:47 PM
mr fantastic
Quote:

Originally Posted by kepler
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.
• Jan 7th 2011, 02:40 PM
kepler
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
• Jan 7th 2011, 05:46 PM
kepler
Hi again,

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

Kind regards,

Kepler
• Jan 7th 2011, 06:01 PM
snowtea
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).
• Jan 7th 2011, 07:44 PM
kepler
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
• Jan 8th 2011, 12:33 AM
mr fantastic
Quote:

Originally Posted by kepler
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.
• Jan 8th 2011, 05:07 AM
kepler
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
• Jan 8th 2011, 06:46 AM
snowtea
If its an astronomy problem with angles, I suggest using the equation of a conic in polar form.
• Jan 8th 2011, 08:52 AM
kepler
What's the equation of a conic in polar form? I only have it in special cases.

Regards,

Kepler
• Jan 8th 2011, 09:51 AM
snowtea
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