Determining foci of an ellipse

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datapioneer

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I am trying to determine if there is a way to determine the foci of an ellipse given three points that lie on the ellipse: one of the endpoints of the major axis, and two other points on the ellipse. Is this even possible? I've attached a graph where I have defined an ellipse with points C, H, and an endpoint of the major axis F. The two foci are shown but are to be determined. They are not given. My hypothesis is that if one can determine point D, the endpoint of the minor axis of the ellipse (as shown in the attached graph), then one can construct a circle with center D and the circle will intersect the major axis of the ellipse at the foci. The problem is how to go about detemining the point D from only points C, H, and F.

ellipse.png
 
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Seems like it might be a reasonable premise IF you can assume the attitude of the ellipse.
If the major axis is not parallel to the x-axis, then what's your plan?
 
General equation of a conic section involves six (6) constants. So to have an unique answer you should have six points to solve for those constants.
 
Are you assuming that the major axis is horizontal, and the center is at the origin? Or could you be given the points without a coordinate system (and therefore without knowing the center?
 
tkhunny said:
Seems like it might be a reasonable premise IF you can assume the attitude of the ellipse.
If the major axis is not parallel to the x-axis, then what's your plan?
Click to expand...

I can assume the major axis is parallel to the x-axis. So, to answer my question more completely, can one theoretically determine the equation for the ellipse if they know only two points that lie on the ellipse and one of the endpoints of the major axis closest to them? If so, how can this be accomplished? As I said earlier in my initial post, if I were able to determine the position of point D on the graph that I presented, then if I determine the distance from the center of the ellipse to one of the endpoints of the major axis, I can then construct a circle from point D as the center of that circle with the radius of that circle being the semi-major axis distance and where the circle crosses the major axis will be the two foci. Can you assist in helping me determine the position of point D from the three points that I mentioned earlier?
 
I can confirm that it is theoretically possible (without making the addition I asked about based on your picture, whether the center was known).

It is known that a conic section is determined by any 5 points. Given that F is a vertex and the major axis is horizontal, you can reflect the other points G and H in the horizontal line through F, giving 5 points, which therefore determine a conic. Of course, that conic could be a parabola or hyperbola, or the horizontal axis might turn out to be the minor axis; but it the ellipse you want exists, it will be fully determined.

Now you just have to find it!

I'll look into that next; but existence is a good start, since it motivates effort! Be aware, though, that it might require solving a fourth-degree equation or so, and therefore may be solvable only numerically (for all practical purposes).
 
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Dr.Peterson said:
I can confirm that it is theoretically possible (without making the addition I asked about based on your picture, whether the center was known).

It is known that a conic section is determined by any 5 points. Given that F is a vertex and the major axis is horizontal, you can reflect the other points G and H in the horizontal line through F, giving 5 points, which therefore determine a conic. Of course, that conic could be a parabola or hyperbola, or the horizontal axis might turn out to be the minor axis; but it the ellipse you want exists, it will be fully determined.

Now you just have to find it!

I'll look into that next; but existence is a good start, since it motivates effort! Be aware, though, that it might require solving a fourth-degree equation or so, and therefore may be solvable only numerically (for all practical purposes).
Click to expand...

Dr. Peterson, so that I fully understand what you're saying here, if I redraw the diagram showing the ellipse with points C, D, and E as the only points that are given for the ellipse and if I then reflect or mirror the points C and D in the Cartesian plane about the X-asis so that points F and G are drawn, you're saying that we now have 5 points available to us and thus using synthetic Geometry we should be able to determine the equation for this ellipase in the 2-dimensional plane. Is that correct? Now, if I can figure out how to solve the equation for y in terms of x for this conic, I have my solution. I had not considered mirroring the two points C and D nor did I realize that 5 points will determine the conic. See the updated graph attached.
 

Attachments

  • conicSection.png
    conicSection.png
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Need to make a slight adjustment to the graph previously submitted. Here is the updated graph with corrected coordinates of the mirrored points F and G for D and C, respectively. I was unable to simply replace the previous graph for some reason.
 

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  • conicSection.png
    conicSection.png
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The mirrored points and the use of five points was primarily for the sake of showing that there is sufficient information. I would not necessarily use that as part of my solution method. (On the other hand, I did construct this in GeoGebra using that method, as it has a tool for constructing a conic from five points; if you are allowed to use that program or one like it, the problem is solved.)

What I have in mind for solving it is to write a general equation for the ellipse with given vertex, using two parameters, a and b, and plug in the two other given points. Then I would solve the resulting system of two equations.

Are you saying you need a synthetic method, not an analytic method? What is the context of your question, and what tools are available?
 
Dr. Peterson, I had posed a fictitious problem on my blog for a gardener who discovered a partially-remaining circular garden wall and wanted to know how he/she could go about reconstructing the wall. This problem is very easy to accomplish since taking two chords of the circular wall from what is showing on that wall and constructing the perpendicular bisectors to those chords, these perpendicular bisecting lines will intersect in the center of the circle. Then, it is very easy for the gardener to attach a string to the center and to a point on the existing wall and trace a 360-degree path about the center to trace out the wall so it can be reconstructed. So, what I had done on the blog was to extend this problem to one of an ellipse wherein the wall that was showing was points C, D, the endpoint of the wall and a small amount of the elliptical wall beyond the endpoint in the lower quadrant of the imaginary Cartesian plane. So, I posed the problem to my blog readers and asked them to find a solution. No one has up to this point. My hypothesis was that it would be possible to reconstruct the elliptical wall given enough points (one being the endpoint of the major axis of the ellipse). You have shown me that 5 non-collinear points will describe the conic (ellipse) given that we are told the wall was originally elliptical.

I have used GeoGebra to create the graphs that I'm including here in this post. I realize that I can select two points and a point on the ellipse in GeoGebra and construct an ellipse. I guess what I'm doing in the problem on my blog is reverse-engineering that process with the 5 points. Hope this clarifies why I'm asking the question. It's not for homework. I'm not in school. I'm a 66 year-old man with a BA degree in Mathematics and a PhD in IT Education. I really appreciate all the help you've given me so far. Very enlightening.
 
So, it sounds like you expect a synthetic solution, but wouldn't reject an analytic solution.

I had time to pursue the method I suggested, and what at first looked like a quartic equation simplified to a linear equation.

I changed the coordinate system from yours, putting F at the origin and the major axis along the x-axis; G is (p,q) and H is (r,s). Then I have

[MATH]a = \frac{p^2 s^2 - q^2 r^2}{2(ps^2 - rq^2)}[/MATH]​
[MATH]b = a\sqrt{\frac{s^2 - q^2}{(r - p)(2a - p - r)}}[/MATH]​

I put this in a spreadsheet and it works consistently with my GeoGebra construction. Here is an example:

FMH115956.png

With (p,q) = (1,2) and (r,s) = (3,3), I get a = 4.5, b = 3.181980515.

I may see if I can come up with a synthetic construction.
 
Dr. Peterson, I'm not at all opposed to an analytical solution. What you've done here is fantastic! However, if you would like to entertain attempting a synthetic solution, that would be much appreciated. I don't want you spending too much of your valuable time working on this as I feel you have given me the analytical solution that I was looking for. Nice work. I love GeoGebra, which I discovered on my daily driver Arch Linux platform, not Windows 10. Now I have the app installed in Windows 10 Pro as well and simply couldn't do without it.
 
Interesting problem. Here's another take on it. Let's consider a standard ellipse centered at the origin [MATH]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/MATH]. We can write this parametrically as [MATH]\vec R(t) =\langle a\cos t,b\sin t\rangle [/MATH]. Lets assume the right vertex is given, so we know [MATH]a[/MATH], and suppose we are given a point [MATH]\langle p, q\rangle[/MATH] on the ellipse in the first quadrant. So we have [MATH]\vec R(t_1) = \langle a\cos t_1, b\sin t_1\rangle = \langle p,q \rangle[/MATH]. This tells us that [MATH]t_1 = \arccos(\frac p a)[/MATH] so [MATH]b= \frac q {\sin(t_1)} = \frac q {\sin(\arccos \frac p a)}=\frac q {\sqrt{a^2-p^2}} [/MATH]. So it seems that if we know the major vertex and one point in the first octant, that's enough to determing the ellipse. To check it I took the example given by Dr. Peterson in post #11, but I moved it to the origin. So the major vertex is [MATH]\langle a,0\rangle =\langle 4.5,0\rangle[/MATH] and the point [MATH]\langle p, q\rangle = \langle 1.5,3 \rangle[/MATH]. This gives the parametric equation [MATH]\vec R(t) = \langle 4.5\cos t, 3.181980515\sin t\rangle[/MATH]. I have attached the graph, which is just the graph in post #11 translated to the origin.
ellipse.jpg
 
LCKurtz said:
Let's consider a standard ellipse centered at the origin ...
Lets assume the right vertex is given, so we know [MATH]a[/MATH], and suppose we are given a point [MATH]\displaystyle \langle p, q\rangle[/MATH] on the ellipse in the first quadrant.
Click to expand...
Note that in the problem as stated (and as solved by me), we do not know the center, so we do not know a.
 
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