使用这个问题：https ://stackoverflow.com/questions/57312462/how-to-find-and-draw-the-intersection-points-of-contour-shapes在圆锥表示为双曲线的情况下（p,q,v,k,b=5,7,2,3,4 #hyperbole） ，程序输出通过点 V 的直线与二次曲线的交点。夸张的程序的结果：
但是，在将二次曲线表示为椭圆 ( p,q,v,k,b=5,0.7,2,3,4 # ellipse) 的情况下，除了位于二次曲线边界上的点之外，还会显示位于椭圆内部的点。椭圆程序的结果：
我的代码：
import numpy as np
import matplotlib.pyplot as plt
def find_roots(x, y):
s = np.abs(np.diff(np.sign(y))).astype(bool)
return x[:-1][s] + np.diff(x)[s] / (np.abs(y[1:][s] / y[:-1][s]) + 1)
# p,q,v,k,b=5,7,2,3,4 #hyperbole
p,q,v,k,b=5,0.7,2,3,4 # ellipse
if(((2 * b * k * p) - (2 * b * k * q) + (k * k) - (b * b * q * q)) > 0):
print("conic is an ellipse")
elif (((2 * b * k * p) - (2 * b * k * q) + (k * k) - (b * b * q * q)) == 0):
print("conic is a parabola")
else:
print("conic is a hyperbole")
X = np.arange(-50, 50, 0.05)
plt.plot(-v, 0)
plt.scatter(-v, 0, color='red', marker='o')
plt.text(-v, 0.8, "V", horizontalalignment="center")
xmin, xmax, ymin, ymax = -10, 10, -10, 10
ax = plt.gca()
ax.get_xlim()
ax.set_xlim([xmin, xmax])
ax.set_ylim([ymin, ymax])
ax.spines['left'].set_position('center')
ax.spines['bottom'].set_position('center')
ax.spines['top'].set_visible(False)
ax.spines['right'].set_visible(False)
x, y = np.meshgrid(X, X)
l = b * x + b * v - v * y
vb = plt.contour(x, y, l, [0], colors='red')
conic = x * x * b * 2 * p * k - x * x * b * 2 * q * k + x * x * k * k + y * y - b * 2 * y + 2 * b * q * x * y
cnc = plt.contour(x, y, (conic), [0], colors='blue')
c = cnc.collections[0].get_paths()[-1]
v0 = c.vertices
x1 = np.sort(v0[:, 0])
y1 = v0[np.argsort(v0[:, 0]), 1]
vb1 = vb.collections[0].get_paths()[0]
v1 = vb1.vertices
x2 = np.sort(v1[:, 0])
y2 = v1[np.argsort(v1[:, 0]), 1]
x = np.linspace(max(x1.min(), x2.min()), min(x1.max(), x2.max()), 1000)
y1i = np.interp(x, x1, y1)
y2i = np.interp(x, x2, y2)
x_intersect = find_roots(x, y1i - y2i)
y_intersect = np.interp(x_intersect, x, y1i)
plt.plot(x_intersect, y_intersect, marker="X", ms=5, color="limegreen")
plt.show()
如何只显示一条线的交点。通过点 V 和 圆锥（椭圆）的边界？
如果您使用二维对象，任务可以大大简化：