18
$\begingroup$

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.

Improve this question
$\endgroup$
4

3 Answers 3

Reset to default
30
$\begingroup$

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

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$
    – user153012
    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. $\endgroup$
    – Mark McClure
    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. $\endgroup$
    – Jens
    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. $\endgroup$
    – user153012
    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. $\endgroup$
    – Mark McClure
    Sep 27, 2014 at 19:08
11
$\begingroup$

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 $a=0$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Sigis K
    Oct 1, 2014 at 12:02
  • $\begingroup$ it is the same except a is replaced by f. $\endgroup$
    – Basheer Algohi
    Oct 1, 2014 at 13:47
  • 1
    $\begingroup$ 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 $\endgroup$
    – Sigis K
    Oct 1, 2014 at 22:08
2
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.