# Ellipse implicit equation coefficients

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Wajahat 2012 年 5 月 1 日
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
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leon 2013 年 2 月 27 日

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### 採用された回答

Richard Brown 2012 年 5 月 1 日
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: http://www.mathworks.com/matlabcentral/fileexchange/15125-fitellipse-m/ 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)]
where
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
or
(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 2012 年 5 月 2 日
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
Wajahat

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