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Five points are required to define a unique ellipse. An ellipse has five degrees of freedom: the $x$ and $y$ coordinates of each focus, and the sum of the distance from each focus to a point on the ellipse, or alternatively, the $x$ and $y$ coordinates of the center, the length of each radius, and the rotation of the axes about the center.

I need a function, that fits an ellipse, for given five $(x,y)$ pairs. Is there a function in Mathematica to do that? If it's possible I need a plot with the ellipse and the given points, and also the equation of the fitted ellipse.

I need an other function, that could check that if a point is on an ellipse. For example on an ellipse, that we just fitted with the previous function.

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    $\begingroup$ Related: math.stackexchange.com/questions/163920/… $\endgroup$ Commented Sep 27, 2014 at 14:20
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    $\begingroup$ This could be considered to be a special case of Fitting points to tilted, off-center ellipse. But it's different in that this question admits a solution that goes through all points. $\endgroup$ Commented Sep 27, 2014 at 19:02
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    $\begingroup$ There is code here that will best-fit an ellipse. This is useful in cases where you want an ellipse but the given points actually determine a hyperbola. $\endgroup$ Commented Oct 3, 2014 at 21:33
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    $\begingroup$ See also the demonstration Five Points Determine a Conic Section. An interesting related problem is that of Small Lattice Ellipses. $\endgroup$ Commented Oct 6, 2014 at 10:37

3 Answers 3

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The following is based on the fact that the determinant of a matrix is equal to zero when two rows are the same. Thus, if you plug any of the points in, you get a true statement.

SeedRandom[3]; pts = RandomReal[{-1, 1}, {5, 2}]; row[{x_, y_}] := {1, x, y, x*y, x^2, y^2}; eq = Det[Prepend[row /@ pts, row[{x, y}]]] == 0 (* Out: 0.0426805-0.0293168x-0.155097x^2-0.019868y-0.087933x*y-0.061593y^2 == 0 *) ContourPlot[Evaluate[eq], {x, -1, 1}, {y, -1, 1}, Epilog -> Point[pts]]

enter image description here

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  • $\begingroup$ Thank you @Mark McClure! Can you tell me something more? Beside the Evaluate[eq] how can I plot an extra {x_6,y_6} point in the same ContourPlot? Or some extra points? $\endgroup$ Commented Sep 27, 2014 at 17:24
  • $\begingroup$ @user153012 If you'd like to plot more points in a list, say morePoints, simply add another Point primitive containing your list to the Epilog. $\endgroup$ Commented Sep 27, 2014 at 18:53
  • $\begingroup$ @user153012 For more than five points, see my comment to the original question: it becomes a fitting problem that has been asked about before. $\endgroup$ Commented Sep 27, 2014 at 19:05
  • $\begingroup$ @Jens Thank you. The referred question is interesting. I know for $5$ points it is a special case of that. But this last question is a little bit different, because I wanted to plot extra points, after we determined the ellipse. $\endgroup$ Commented Sep 27, 2014 at 19:07
  • $\begingroup$ @Jens My assumption is that the OP simply wants to add more points that are know to be on the same ellipse, rather than find a best fit. $\endgroup$ Commented Sep 27, 2014 at 19:08
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The general equation of an ellipse (here) is given by:

ellipse[x_, y_] = a x^2 + b x y + c y^2 + d x + e y + f == 0;

solving using 5 points results in:

SeedRandom[3]; pts = RandomReal[{-1, 1}, {5, 2}]; sol = Solve[ellipse @@@ pts]; ellipse[x, y] /. sol[[1]] // Simplify (*a (-0.275185 + 1. x^2 + x (0.189022 + 0.566953 y) + 0.1281 y + 0.397124 y^2) == 0*)

all $a$ values result in the same equation except when $a=0$.

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  • $\begingroup$ With MMA V9 I do not get the right answer, for example f (-0.275185 + 1. x^2 + x (0.189022 + 0.566953 y) + 0.1281 y + 0.397124 y^2) == 0 is the equation of the ellipse which is returned. The parameter f is not resolved. $\endgroup$ Commented Oct 1, 2014 at 12:02
  • $\begingroup$ it is the same except a is replaced by f. $\endgroup$ Commented Oct 1, 2014 at 13:47
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    $\begingroup$ Your ellipse definition is in fact the general equation of a conics - it is only an ellipse if the discriminant is negative. How does your answer take this restriction into account? Thanks $\endgroup$ Commented Oct 1, 2014 at 22:08
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Through 5 points we can pass a conic, an ellipse, hyperbola etc. After Algohi's solution coefficients are obtained we can determine choice of conic by sign of the second evaluated invariant $ (b^2 - 4 a c) $ along with standard calculated expression for values of rotation/translation of central conic.A sign change test for a test point chosen inside or outside can be done,it should vanish on the arc.

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