如何找到椭圆的方程

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

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 A is 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.

Answer 2

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