橢圓的性質
Author: 藍邦偉  老師

a²=b²+c²

a

橢圓的性質

b

c

$ \frac{x²}{a²} +\frac{y²}{b²}=1$

a²=b²+c²

k = 0 (slider from -20 to 20, step 1)
h = 0 (slider from -20 to 20, step 1)
b = 3 (slider from 0 to 36, step 0.5)
a = 5 (slider from 0 to 36, step 1)
Γ: (x - h)^(2) / a^(2) + (y - k)^(2) / b^(2) = 1
d: x = h
e: y = k
M: (h, k)
A_1: (h + a, k)
A', A = Intersect(Γ, e)
B, B' = Intersect(Γ, d)
F: (h + (a^(2) - b^(2))^(1 / 2), k)
F': (h - (a^(2) - b^(2))^(1 / 2), k)
K_1: (x(Corner[1]), 0)
K_2: (x(Corner[2]), 0)
K_3: (0, y(Corner[2]))
K_4: (0, y(Corner[3]))
u = Vector(K_1, K_2)
v = Vector(K_3, K_4)
t = 4.2 (slider from 4 to 5, step 0.1)
c: sqrt(a^(2) - b^(2))
s = Segment(A, F)
A_{a}: (h + t, k)
A_{b}: (h, k + (5 - t) * 3)
l = Segment(A_{b}, A_{a})
多邊形1, a_{a}, a_{b}, m = Polygon(A_{b}, M, A_{a})
n = LineBisector(A_{b}, A_{a})
G = Midpoint(A_{b}, A_{a})
p = Circle(G, A_{b})
H = Intersect(p, n, 1)
G' = Dilate(G, -0.2, H)
q = CircleArc(G', A_{a}, A_{b})
T_1 = Corner(1)
T_4 = Corner(4)
T_2: (4 * Corner[1] + Corner[2]) / 5
T_3: (4 * Corner[4] + Corner[3]) / 5
多邊形2, t_1, t_2, t_3, t_4 = Polygon(T_1, T_2, T_3, T_4)
r = Segment(T_3, T_2)
a_1 = LineBisector(M, A)
b_1 = LineBisector(M, B)
J = Point(b_1)
d_1 = CircleArc(J, B, M)
c_1 = LineBisector(M, F)
I = Point(c_1)
e_1 = CircleArc(I, M, F)