4.1 阿波羅尼奧斯與圓錐曲線（約西元前 200 年）#

import sympy as sp

x, y, h = sp.symbols('x y h', real=True)
alpha, beta = sp.symbols('alpha beta', real=True, positive=True)

# 圓錐: x^2 + y^2 = z^2 * tan(a)^2 ; 平面: z = x/tan(b) + h
z_expr = x / sp.tan(beta) + h
eq = sp.Eq(x**2 + y**2, z_expr**2 * sp.tan(alpha)**2)

poly = sp.expand(x**2 + y**2 - z_expr**2 * sp.tan(alpha)**2)
# 依 x 次方整理 (y^2 視為常數項): 二次項係數決定曲線類型
coeff_x2 = sp.expand(poly).coeff(x, 2)
print("x^2 係數:", coeff_x2)
print("化簡:", sp.simplify(coeff_x2))
print("y^2 係數:", sp.expand(poly).coeff(y, 2))

x^2 係數: -tan(alpha)**2/tan(beta)**2 + 1
化簡: -tan(alpha)**2/tan(beta)**2 + 1
y^2 係數: 1

import sympy as sp

x = sp.symbols('x', positive=True)
l = sp.Symbol('ell', positive=True)      # latus rectum 半長: 2l = 正焦弦長
a = sp.Symbol('a', positive=True)        # 頂點到中心的距離 (橢圓半長軸)

# 拋物線: y^2 恰好等於 應用矩形面積
y2_parabola = 2 * l * x
print("parabola   y^2 =", y2_parabola, "  -> 應用矩形 2l*x, 差 =",
      sp.expand(y2_parabola - 2 * l * x))

# 橢圓: 不足
y2_ellipse = 2 * l * x - (2 * l / (2 * a)) * x**2
rect = 2 * l * x
shortfall = rect - y2_ellipse
print("ellipse    不足量 =", sp.simplify(shortfall))   # = l*x**2/a
print("與應用矩形的比(不足比例):", sp.simplify(shortfall / rect),
      " (= x/2a, 隨 x 增大)")

# 雙曲線: 超出
y2_hyperbola = 2 * l * x + (2 * l / (2 * a)) * x**2
excess = y2_hyperbola - rect
print("hyperbola  超出量 =", sp.simplify(excess))       # = l*x**2/a

parabola   y^2 = 2*ell*x   -> 應用矩形 2l*x, 差 = 0
ellipse    不足量 = ell*x**2/a
與應用矩形的比(不足比例): x/(2*a)  (= x/2a, 隨 x 增大)
hyperbola  超出量 = ell*x**2/a

import sympy as sp

x = sp.symbols('x', positive=True)
l = sp.Symbol('ell', positive=True)
x0 = sp.Symbol('x0', positive=True)
y0 = sp.sqrt(2 * l * x0)      # y0^2 = 2 l x0

y = sp.sqrt(2 * l * x)
dy_dx = sp.diff(y, x)
slope_at_P = sp.simplify(dy_dx.subs(x, x0))
print("斜率 m =", slope_at_P)          # = ell/y0  (因 y0 = sqrt(2 l x0))

# 切線方程 yy0 = l(x + x0) 驗證: 斜率 = l/y0, 且過 (x0, y0)
m_claim = l / y0
print("m == ell/y0 ?", sp.simplify(slope_at_P - m_claim) == 0)
# 切線與 x 軸的交點:
X = sp.symbols('X')
T = sp.solve(sp.Eq(y0, m_claim * (X - x0) + 0), X)  # y=0 代入切線
# y*y0 = l*(x+x0) => 當 y=0: 0 = l*(x+x0) => x = -x0
print("切線在 x 軸交點 x =", -x0, " 次切距 = x0-(-x0) =", 2 * x0)

斜率 m = ell/sqrt(2*ell*x0)
m == ell/y0 ? True
切線在 x 軸交點 x = -x0  次切距 = x0-(-x0) = 2*x0

import sympy as sp

x, t = sp.symbols('x t', positive=True)
l = sp.Symbol('ell', positive=True)
x0 = sp.symbols('x0', positive=True)

P = (x0, sp.sqrt(2 * l * x0))
Q = (x0 + t, sp.sqrt(2 * l * (x0 + t)))

secant_slope = sp.simplify((Q[1] - P[1]) / t)
print("割線斜率 =", sp.radsimp(secant_slope))
limit_slope = sp.limit(secant_slope, t, 0)
print("t->0 極限 =", sp.simplify(limit_slope))   # = ell / sqrt(2*l*x0)
print("等於 ell/y0 ?", sp.simplify(limit_slope - l/sp.sqrt(2*l*x0)) == 0)

割線斜率 = (sqrt(ell)*(-sqrt(x0) + sqrt(t + x0)))/(sqrt(2)*t)
t->0 極限 = sqrt(2)*ell/(2*sqrt(ell*x0))
等於 ell/y0 ? True

import sympy as sp

x, y = sp.symbols('x y', real=True)
l, y0 = sp.symbols('ell y0', positive=True)
x0 = y0**2 / (2 * l)

# 切線斜率 ell/y0; 法線斜率 -y0/ell
normal = sp.Eq(y - y0, -y0 / l * (x - x0))
normal_explicit = sp.expand(normal.lhs - normal.rhs)
print("法線方程:", normal)

# 重演「從點 (X0, 0) 作拋物線法線」的古典問題:
X0 = sp.Symbol('X0', real=True)
sol = sp.solve(normal_explicit.subs({y: 0}).subs(x, X0), y0)
print("從軸上點 X0 作法線, y0 的方程:")
print("  y0 滿足 (化簡):", sp.factor(sp.simplify(sol[0] if isinstance(sol, list) else 0)))
# 直接列多項式: 代 y=0, x=X0 => 0 - y0 = -y0/l*(X0 - y0^2/(2l))
eqn = sp.expand(sp.simplify(normal_explicit.subs({y: 0}).subs(x, X0)))
print("  多項式 form:", sp.factor(eqn))

法線方程: Eq(y - y0, -y0*(x - y0**2/(2*ell))/ell)
  多項式 form: -y0*(2*ell**2 + 2*ell*X0 - y0**2)/(2*ell**2)

import sympy as sp

x, y = sp.symbols('x y', real=True)
p = sp.Symbol('p', positive=True)         # 焦點 (p,0), 準線 x = -p

# e = 1: dist(P, F)^2 = dist(P, L)^2
eq = sp.Eq((x - p)**2 + y**2, (x + p)**2)
gens = sp.solve(eq, y**2)
print("e=1 化簡:", sp.expand(gens[0]))     # => y^2 = 4 p x (與 4.1.2 的 2l 對應, l = 2p)

# e = 1/2, 焦距 c, a 關係驗證 (橢圓): b^2 = a^2 - c^2, e = c/a
a_val, c_val = 2, 1
b_val = sp.sqrt(a_val**2 - c_val**2)
E = (x - 3)**2 / a_val**2 + (y + 5)**2 / b_val**2 - 1
F1, F2 = (3 - c_val, -5), (3 + c_val, -5)
# 焦點性質: 到兩焦點距離和 = 2a —— 用隨機點驗證
import random
check_pts = [(3 + a_val * sp.cos(t), -5 + b_val * sp.sin(t)) for t in [0.3, 1.2, 2.8, 4.5]]
for t, pt in zip([0.3, 1.2, 2.8, 4.5], check_pts):
    d1 = sp.sqrt((pt[0] - F1[0])**2 + (pt[1] - F1[1])**2)
    d2 = sp.sqrt((pt[0] - F2[0])**2 + (pt[1] - F2[1])**2)
    total = sp.N(d1 + d2)
    print(f"t={t}: d1+d2 = {total}")

e=1 化簡: 4*p*x
t=0.3: d1+d2 = 4.00000000000000
t=1.2: d1+d2 = 4.00000000000000
t=2.8: d1+d2 = 4.00000000000000
t=4.5: d1+d2 = 4.00000000000000

from sympy.geometry import Point, Line, Circle, Ellipse, Parabola

# 命題性質 1: 橢圓的焦點、離心率、半正焦弦
E = Ellipse(Point(0, 0), 5, 3)         # 半長軸 5, 半短軸 3
print("焦點:", E.foci)                  # [(-4, 0), (4, 0)]
print("離心率:", E.eccentricity)        # 4/5
print("半正焦弦:", E.semilatus_rectum)  # 9/5 = b^2/a
print("周長 (橢圓積分):", E.circumference)

# 命題性質 2: 雙曲線 (無內建類別, 用隱函數方程處理 — 詳見 4.3 節)
import sympy as sp
x, y = sp.symbols('x y')
h_eq = x**2/16 - y**2/9 - 1            # a=4, b=9/1?
# 漸近線: y = ±(b/a)x, 由此直接讀出斜率
print("雙曲線方程前三項 x^2 係數:", sp.expand(h_eq).coeff(x, 2))
asym = sp.solve(sp.Eq(h_eq, 0), y)
print("漸近線 (遠行為):", asym)         # 無法直接解出? 改斜率比
# 漸近線斜率 = sqrt(b^2/a^2) = 3/4
print("漸近線斜率:", sp.sqrt(9/16))

# 命題性質 3: 過拋物線外一點的切線 (Parabola 類別無 tangent_lines, 用判別式)
P = Parabola(Point(0, 0), Line(Point(0, 4), Point(1, 4)))
print("拋物線方程:", P.equation())      # -x^2 - 8y + 16

焦點: (Point2D(-4, 0), Point2D(4, 0))
離心率: 4/5
半正焦弦: 9/5
周長 (橢圓積分): 15*elliptic_e(16/25)
雙曲線方程前三項 x^2 係數: 1/16
漸近線斜率: 0.750000000000000
拋物線方程: -x**2 - 8*y + 16