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Ellipse implicit equation coefficients

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Hi I have a set of 2D points and I want to know the coefficients of the following implicit equation of the ellipse: Ax^2+2Bxy+Cy^2=1
Any idea?
Moreover, I have found the best fitting ellipse with its major, minor axes, center and orientation. Can I can use this information to find the above mentioned three coefficients (A,B and C)?
The implementations of ellipse fitting already available in Matlab Central use the general form of ellipse, I do not need coefficients for those.
Best Regards Wajahat

Accepted Answer

Richard Brown
Richard Brown on 1 May 2012
First, if the centre is nonzero, then that form of the ellipse equation will not work. You need the more general quadratic form
(*) x^' * A * x + c' * x + d = 0
where A is a 2x2 symmetric matrix, c a 2x1 vector and d a scalar. Multiplying out gives the full equation
a_11 x^2 + 2a_12 xy + a_22 y^2 + c_1 x + c_2 y + d = 0
If you used my code: then what you get when you fit your ellipse is the centre z, the orientation of the major axis as a counterclockwise rotation from the x-axis alpha , and a and b the semimajor and semiminor axes respectively. Any of the other fitting codes probably return similar parameters.
These correspond to the following parametric equation for the ellipse (parametrised by theta in [0, 2\pi])
x = z + Q * [a * cos(theta); b * sin(theta)]
Q = [cos(alpha), -sin(alpha); sin(alpha) cos(alpha)]
All you need to do to get back to the quadratic form is to eliminate theta. First some algebra
diag([1/a, 1/b]) * Q'*(x - z) = [cos(theta); sin(theta)] = u
Using the fact that u'*u = 1,
(x - z)' * Q * diag([1/a^2, 1/b^2]) * Q' * (x - z) = 1
(x - z)' * A * (x - z) = 1
where A is symmetric. Expanding this out gives the required form:
x'*A*x - 2*z'*A*x + z'*A*z - 1 = 0
So linking back to my first equation (*), the MATLAB code for generating the matrices and vectors required are
Q = [cos(alpha), -sin(alpha); sin(alpha) cos(alpha)];
A = Q * diag([a^-2, b^-2]) * Q';
c = 2*A*z;
d = z'*A*z - 1;


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Wajahat on 2 May 2012
Hi Richard,
Do you have some clue how to find the coefficients u,v,a,b,c in a(x-u)(x-u)+2b(x-u)(y-v)+c(y-v)(y-v)=1 directly from a point set. Perhaps using matlab's least squares curve fitting etc?
I have tried the method you describe above, but it is not giving me the correct results. I want to verify if the coefficients are same both way.
Richard Brown
Richard Brown on 2 May 2012
In all honesty, I would recommend using the code that I posted a link to fit your ellipse - it fits the ellipse using nonlinear least squares, i.e. actually minimising the sums of squared perpendicular distances to the ellipse.
If you try to use linear least squares directly on the coefficients of the equation, it is far less clear what you are actually minimising, and depending on your data, your results can be not good.
If you're having trouble, feel free to post your data, and I'll take a look
Wajahat on 2 May 2012
Hi Richard
Thanks for the offer.
I have sent my data to your email( the one mentioned here on MatlabCentral) along with some sample images.
Best Regards

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