BMAD 63 Report post Posted September 11, 2017 How many points would you need to have to uniquely determine an ellipse given that you know a foci is located at (0,0). Quote Share this post Link to post Share on other sites
0 gavinksong 11 Report post Posted September 13, 2017 (edited) Minor grammar correction: the singular form of foci is focus. Without really thinking about it, I'm going to guess: Spoiler Two points. The intuition is that one point would contain less information than the minimal definition of an ellipse (minus a focus), whereas two points contains more. Edited September 13, 2017 by gavinksong wording Quote Share this post Link to post Share on other sites
0 gavinksong 11 Report post Posted September 13, 2017 After some more thought, Spoiler Three points. With each point, we get the equation: R = |p| + |p - f| where R and f are unknown values, and p is the new point in vector form. There are three unknowns (since f is 2-dimensional), which we need at least three equations to solve. Quote Share this post Link to post Share on other sites
0 bonanova 83 Report post Posted September 13, 2017 Spoiler Ellipse is the locus of points the sum of whose distances from the foci is constant. So, we just need (a) the other locus and (b) any point on the ellipse to determine the sum: two more points. Quote Share this post Link to post Share on other sites
0 Quantum.Mechanic 0 Report post Posted September 15, 2017 Spoiler It's odd that given one focus, and two points on the ellipse, is not enough. But three points on the ellipse, with one focus, should be enough to determine the ellipse and the other focus. And four points on the ellipse, and no foci, would also work. But how to do this is beyond me. Quote Share this post Link to post Share on other sites
0 gavinksong 11 Report post Posted September 17, 2017 (edited) On 9/16/2017 at 12:56 AM, Quantum.Mechanic said: Hide contents It's odd that given one focus, and two points on the ellipse, is not enough. But three points on the ellipse, with one focus, should be enough to determine the ellipse and the other focus. And four points on the ellipse, and no foci, would also work. But how to do this is beyond me. My thoughts exactly... except is it possible with four points? Edited September 17, 2017 by gavinksong Quote Share this post Link to post Share on other sites
0 plasmid 40 Report post Posted September 24, 2017 I don't think it's possible with three points on the ellipse even if you know that one focus is at (0, 0). Spoiler Call F the point where the second focus lies, and call R the sum of the distances from (0, 0) and from F to any point on the ellipse. If you’re given one point on the ellipse, A, then you know that point F must lie some distance from A given by [distance from F to A] = [R] – [distance from (0, 0) to A] so if you knew R then you would know that F lies on a circle of known radius around A. But since you don’t know what R is, F could be anywhere. Suppose you’re given two points on the ellipse, A and B. If you knew R, then you would know that F must lie on a circle of known radius around A and a circle of known radius around B so you would know that F must lie on one of two points. But since you don’t know the value of R, point F isn’t narrowed down to one of two points, but it is narrowed down to lie on some curve defined by the intersection of the circles around A and B as R increases. The figure below probably helps; different color circles around A and B correspond to different values of R, and note that the radius of the circles around A and B for a given value of R will be different if points A and B are at different distances from (0, 0). Now suppose you’re given three points: A, B, and C. In general, F would probably at best be narrowed down to two points but not uniquely defined. And if point C were on or near that curve traced out as potential points for F, then there might be even more points where circles around C of different radii intersect that curve for different values of R. I haven’t yet gone into enough depth to determine whether four points on the ellipse would be enough and if there are any special cases that would make that fail. Quote Share this post Link to post Share on other sites
0 gavinksong 11 Report post Posted September 28, 2017 (edited) On September 25, 2017 at 8:24 AM, plasmid said: I don't think it's possible with three points on the ellipse even if you know that one focus is at (0, 0). Reveal hidden contents Call F the point where the second focus lies, and call R the sum of the distances from (0, 0) and from F to any point on the ellipse. If you’re given one point on the ellipse, A, then you know that point F must lie some distance from A given by [distance from F to A] = [R] – [distance from (0, 0) to A] so if you knew R then you would know that F lies on a circle of known radius around A. But since you don’t know what R is, F could be anywhere. Suppose you’re given two points on the ellipse, A and B. If you knew R, then you would know that F must lie on a circle of known radius around A and a circle of known radius around B so you would know that F must lie on one of two points. But since you don’t know the value of R, point F isn’t narrowed down to one of two points, but it is narrowed down to lie on some curve defined by the intersection of the circles around A and B as R increases. The figure below probably helps; different color circles around A and B correspond to different values of R, and note that the radius of the circles around A and B for a given value of R will be different if points A and B are at different distances from (0, 0). Now suppose you’re given three points: A, B, and C. In general, F would probably at best be narrowed down to two points but not uniquely defined. And if point C were on or near that curve traced out as potential points for F, then there might be even more points where circles around C of different radii intersect that curve for different values of R. I haven’t yet gone into enough depth to determine whether four points on the ellipse would be enough and if there are any special cases that would make that fail. I stand by my answer. Spoiler Although I gave an algebraic answer, my original thought process was exactly yours. I'm a spatial thinker. However, I think it's a dubious claim that when "you’re given three points... [the remaining focus] would probably at best be narrowed down to two points". Given any distinct pair of points on the ellipse, you can map out all the possible locations of the second focus on a hyperbolic curve. Given three points, we can select three distinct pairs and produce three curves. For your claim to be true, all three curves to would have to meet at two locations. With the foci of the hyperbolic curves forming a triangle, my intuition tells me this is probably pretty unlikely. Edited September 28, 2017 by gavinksong Quote Share this post Link to post Share on other sites
0 bonanova 83 Report post Posted October 10, 2017 On 9/13/2017 at 2:21 AM, bonanova said: Hide contents Ellipse is the locus of points the sum of whose distances from the foci is constant. So, we just need (a) the other locus and (b) any point on the ellipse to determine the sum: two more points. I, as well, stand by my answer. Spoiler OP does not say "on" an ellipse. Nor a particular ellipse. To finish my proof, fasten a string to the foci that loops a pencil at the ellipse point. Draw an ellipse. Caveat: Spoiler The OP could be more definitive. Take, for example, the phrase "that you know a foci is located at (0,0)." My interpretation would be rock solid if OP said How many points would you need to have to uniquely determine an ellipse "that has a focus located at (0,0)"? It would be dead wrong if OP said There is an ellipse that has a focus at (0,0). How many points on that ellipse do you need to draw that ellipse? Quote Share this post Link to post Share on other sites
0 plasmid 40 Report post Posted October 12, 2017 I think gavinksong is right. I sat down and did the math for the case where you're given that one focus is at (0, 0), and two points on the ellipse are at (x, 1) and (-x, 1) so the second focus must be somewhere on the Y-axis (since that math is reasonably tractable). As you move the second focus around and calculate the point at the top and/or bottom of the resulting ellipse, there are no repeats. Meaning that there is no point on the Y-axis where you could put a third point and have any ambiguity about the ellipse. My original argument was thinking in terms of a circle around the third point that contacts the curve defined by the first two points at multiple sites, but that doesn't happen if the third point is on the curve (at least in this case) -- if you think in terms of the picture from my previous post and gradually changing the value of R so you have two potential locations for the second focus along the curve which move outward along the curve as R increases, then apparently the circle around the third point grows fast enough that it "outruns" the point moving along the curve and doesn't intersect it again. If you were to move the third point farther away from the curve (when I did the math for the case above, I was covering cases where the third point sitting on the curve of potential focus locations), then the further away it gets from the curve the more distance along the curve will be covered as a circle around the third point increases in radius and the less likely you are to see any second intersection. So since it doesn't happen for points on the Y-axis, it seems unlikely to happen anywhere. But I can't definitively prove that three points will be always be enough, especially for cases where the first two given points are different distances from (0, 0) and the curve of potential spots for the second focus is not linear. (Well, there's the trivial case where all of the points including (0, 0) are co-linear so the "ellipse" is a line segment, but I'm not counting that case.) Quote Share this post Link to post Share on other sites
How many points would you need to have to uniquely determine an ellipse given that you know a foci is located at (0,0).
Share this post
Link to post
Share on other sites