首先，您应该能够在一个方向上重新缩放椭圆。为此，需要将椭圆的系数计算为二次曲线，重新缩放，然后恢复椭圆的新几何参数：中心、轴、角度。

然后你的问题减少到找到从椭圆到原点的距离。要解决最后一个问题，你需要一些迭代。这里有一个可能的独立实现。。。from math import *

def bisect(f,t_0,t_1,err=0.0001,max_iter=100):

iter = 0

ft_0 = f(t_0)

ft_1 = f(t_1)

assert ft_0*ft_1 <= 0.0

while True:

t = 0.5*(t_0+t_1)

ft = f(t)

if iter>=max_iter or ft

return t

if ft * ft_0 <= 0.0:

t_1 = t

ft_1 = ft

else:

t_0 = t

ft_0 = ft

iter += 1

class Ellipse(object):

def __init__(self,center,radius,angle=0.0):

assert len(center) == 2

assert len(radius) == 2

self.center = center

self.radius = radius

self.angle = angle

def distance_from_origin(self):

"""

Ellipse equation:

(x-center_x)^2/radius_x^2 + (y-center_y)^2/radius_y^2 = 1

x = center_x + radius_x * cos(t)

y = center_y + radius_y * sin(t)

"""

center = self.center

radius = self.radius

# rotate ellipse of -angle to become axis aligned

c,s = cos(self.angle),sin(self.angle)

center = (c * center[0] + s * center[1],

-s* center[0] + c * center[1])

f = lambda t: (radius[1]*(center[1] + radius[1]*sin(t))*cos(t) -

radius[0]*(center[0] + radius[0]*cos(t))*sin(t))

if center[0] > 0.0:

if center[1] > 0.0:

t_0, t_1 = -pi, -pi/2

else:

t_0, t_1 = pi/2, pi

else:

if center[1] > 0.0:

t_0, t_1 = -pi/2, 0

else:

t_0, t_1 = 0, pi/2

t = bisect(f,t_0,t_1)

x = center[0] + radius[0]*cos(t)

y = center[1] + radius[1]*sin(t)

return sqrt(x**2 + y**2)

print Ellipse((1.0,-1.0),(2.0,0.5)).distance_from_origin()