How to find the equation for an ellipse

Question

I am looking to find the equation for an ellipse given five or six points using the general equation for a conic: A x2 + B xy + C y2 + D x + E y + F = 0.

At first I tried using six points. Here is my python code:

    import numpy as np
    def conic_section(p1, p2, p3, p4, p5, p6):
        def row(point):
            return [point[0]*point[0], point[0]*point[1], point[1]*point[1],                         
            point[0], point[1], 1]
        matrix=np.matrix([row(p1),row(p2),row(p3),row(p4),row(p5), row(p6)])
        b=[0,0,0,0,0,0]
        return np.linalg.solve(matrix,b)

    print conic_section(np.array([6,5]), np.array([2,9]), np.array([0,0]),         
    np.array([11, 5.5]), np.array([6, 7]), np.array([-1,-1]))

The problem is that this will return the solution [0,0,0,0,0,0] because the right hand side of my equation is the zero vector.

I then attempted to change the conic by subtracting the F and dividing it through:

A x2 + B xy + C y2 + D x + E y + F = 0

A x2 + B xy + C y2 + D x + E y = -F

A' x2 + B xy + C' y2 + D' x + E' y = -1.

The reason that this doesn't work though is that if one of my point is (0,0), then I would end up with a Matrix that has a row of zeros, yet the right hand side of the equation would have a -1 for the entries in the vector. In other words, if one of my points is (0,0) - then "F" should be 0, and so I can't divide it out.

Any help would be appreciated. Thank You.

Answer 1

Ellipse equation (without translation and rotation):

The goal is to resolve this linear equation in variable A through F:

Use:

from math import sin, cos, pi, sqrt

import matplotlib.pyplot as plt
import numpy as np
from numpy.linalg import eig, inv

# basis parameters of the ellipse
a = 7
b = 4

def ellipse(t, a, b):
    return a*cos(t), b*sin(t)

points = [ellipse(t, a, b) for t in np.linspace(0, 2*pi, 100)]
x, y = [np.array(v) for v in list(zip(*points))]

fig = plt.figure()
plt.scatter(x, y)
plt.show()

def fit_ellipse(x, y):
    x = x[:, np.newaxis]
    y = y[:, np.newaxis]
    D =  np.column_stack((x**2, x*y, y**2, x, y, np.ones_like(x)))
    S = np.dot(D.T, D)
    C = np.zeros([6,6])
    C[0, 2] = C[2, 0] = 2
    C[1, 1] = -1
    E, V = eig(np.dot(inv(S), C))
    n = np.argmax(np.abs(E))
    return V[:, n]

A, B, C, D, E, F = fit_ellipse(x, y)
K = D**2/(4*A) + E**2/(4*C) - F

# a, b
print('a:', sqrt(K/A), 'b:', sqrt(K/C))

Output:

a: 6.999999999999998 b: 4.0

See:

http://mathworld.wolfram.com/ConicSection.html https://fr.wikipedia.org/wiki/Ellipse_(math%C3%A9matiques)#Forme_matricielle http://nicky.vanforeest.com/misc/fitEllipse/fitEllipse.html

Answer 2

It seems that you have exact points on ellipse, don't need approximation, and use Braikenridge-Maclaurin construction for conic sections by some strange way.

Five points(x[i],y[i]) determines ellipse with this explicit equation (mathworld page, eq. 8)

So to find ellipse equation, you can build cofactor expansion of the determinant by minors for the first row. For example, coefficient Ais determinant value for submatrix from x1y1 to the right bottom corner, coefficient B is negated value of determinant for submatrix without xiyi column and so on.