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椭圆上从焦点等角的点的回归直线居然过定点？

约 1720 字大约 6 分钟

maths

2026-8-9

题图

问题

椭圆 Γ:x2a2+y2b2=1\Gamma: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}=1，F(c,0)F(c,0) 为 Γ\Gamma 的右焦点，P1,…,Pn∈ΓP_1, \dots, P_n \in \Gamma 满足 ∡PiFPi+1=2πn\measuredangle P_iFP_{i+1} = \dfrac{2\pi}{n}，ll 是 P1,…,PnP_1, \dots, P_n 的最小二乘回归直线，证明：当 P1P_1 在 Γ\Gamma 上运动时，ll 过定点 CC，并求出定点的坐标（用含 n,a,b,cn, a,b,c 的代数式表示）。

不妨将 FF 平移到原点，将焦准距位似为 12\dfrac{1}{2}。设 Γ\Gamma 的离心率为 ee。

设 θk=2kπ+φn,Pk(xk,yk)\theta_k = \dfrac{2k\pi + \varphi}{n}, P_k(x_k,y_k)，根据焦半径公式，取

rk=e2+2ecos⁡θkxk=rkcos⁡(−θk)yk=rksin⁡(−θk)\begin{align*} r_k &= \frac{e}{2+2e\cos\theta_k} \\ x_k &= r_k\cos(-\theta_k) \\ y_k &= r_k\sin(-\theta_k) \end{align*}

（负号是让式子更简单的小巧思）

记 ωk=exp⁡(iθk)\omega_k = \exp\left(\mathrm{i}\theta_k\right)，设 zk=xk+iykz_k=x_k+\mathrm{i}y_k，则

zk=eωˉk2+2eℜωk=eωk−12+e(ωk+ωk−1)=1ωk2+2eωk+1z_k = \frac{e\bar{\omega}_k}{2+2e\Re \omega_k} = \frac{e\omega_k^{-1}}{2+ e(\omega_k+\omega_k^{-1})} = \dfrac{1}{\omega_k^2 + \dfrac{2}{e}\omega_k + 1}

记 α,β\alpha, \beta 为 x2+2ex+1=0x^2 + \dfrac{2}{e}x + 1 = 0 的两根，则

zk=1(ωk−α)(ωk−β)=Aωk−α−Aωk−βz_k = \dfrac{1}{(\omega_k-\alpha)(\omega_k-\beta)} = \dfrac{A}{\omega_k-\alpha} - \dfrac{A}{\omega_k-\beta}

其中实数 A=1α−βA = \dfrac{1}{\alpha - \beta}。

一个韦达定理的简单推论

若多项式 P(x)P(x) 所有根为 x1,…,xnx_1, \dots, x_n，则

∑k=1n1xk−a=−P′(a)P(a)\sum_{k=1}^{n} \frac{1}{x_k - a} = -\frac{P'(a)}{P(a)}

两边对 aa 求导，得到

∑k=1n1(xk−a)2=P′(a)2−P′′(a)P(a)P(a)2\sum_{k=1}^{n} \frac{1}{(x_k - a)^2} = \frac{P'(a)^2 - P''(a)P(a)}{P(a)^2}

因为 ωk\omega_k 是方程 ωn=exp⁡(iφ)=u\omega^n = \exp(\mathrm{i}\varphi) = u 的 nn 个根，记 P(x)=xn−uP(x) = x^n - u，则 P′(x)=nxn−1,P′′(x)=n(n−1)xn−2P'(x) = nx^{n-1}, P''(x) = n(n-1)x^{n-2}，于是记

S1(a)=∑k=1n1ωk−a=−nan−1an−uS2(a)=∑k=1n1(ωk−a)2=(nan−1)2−n(n−1)an−2(an−u)(an−u)2\begin{align*} S_1(a) &= \sum_{k=1}^{n} \dfrac{1}{\omega_k-a} = -\frac{na^{n-1}}{a^n-u} % = \frac{B}{C-u} \\ S_2(a) &= \sum_{k=1}^{n} \dfrac{1}{(\omega_k-a)^2} = \frac{(na^{n-1})^2-n(n-1)a^{n-2}(a^n-u)}{(a^n-u)^2} \\ % &= \dfrac{n\alpha^{2n-2} + n(n-1)u\alpha^{n-2}}{(\alpha^n-u)^2} = \frac{D+Eu}{(C-u)^2} \end{align*}

注意 ωkzk∈R\omega_k z_k \in \mathbb{R}，所以 ∣zk∣2=ωk2zk2,zˉ=ωk2zk|z_k|^2 = \omega_k^2 z_k^2, \bar{z} = \omega_k^2z_k，令

U=∑k=1nzk=∑k=1n1(ωk−α)(ωk−β)=∑k=1n(Aωk−α−Aωk−β),A=1α−β=AS1(α)−AS1(β)V=∑k=1nzk2=∑k=1n(Aωk−α−Aωk−β)2=∑k=1n(A2(ωk−α)2+A2(ωk−β)2−2A2(ωk−α)(ωk−β))=A2S2(α)+A2S2(β)−2A3S1(α)+2A3S1(β)W=∑k=1n∣zk∣2=∑k=1n(ωk(ωk−α)(ωk−β))2=∑k=1n(Aαωk−α−Aβωk−β)2=A2α2S2(α)+A2β2S2(β)−2A3αβS1(α)+2A3αβS1(β)\begin{align*} U = \sum_{k=1}^{n} z_k &= \sum_{k=1}^{n} \frac{1}{(\omega_k-\alpha)(\omega_k-\beta)} \\ &= \sum_{k=1}^{n} \left(\frac{A}{\omega_k-\alpha} - \frac{A}{\omega_k-\beta}\right), \qquad A=\frac{1}{\alpha-\beta} \\ &= AS_1(\alpha) - AS_1(\beta) \\ V = \sum_{k=1}^{n} z_k^2 &= \sum_{k=1}^{n} \left(\frac{A}{\omega_k-\alpha} - \frac{A}{\omega_k-\beta}\right)^2 \\ &= \sum_{k=1}^{n} \left( \frac{A^2}{(\omega_k-\alpha)^2} + \frac{A^2}{(\omega_k-\beta)^2} - \frac{2A^2}{(\omega_k-\alpha)(\omega_k-\beta)} \right) \\ &= A^2S_2(\alpha) + A^2S_2(\beta) - 2A^3S_1(\alpha) + 2A^3S_1(\beta)\\ % A^2\frac{D+Eu}{(C-u)^2} + A^2\frac{H+Iu}{(G-u)^2} - \frac{2A^3 B}{C-u} + \frac{2A^3 F}{G-u} \\ W = \sum_{k=1}^{n} |z_k|^2 &= \sum_{k=1}^{n} \left( \frac{\omega_k}{(\omega_k-\alpha)(\omega_k-\beta)} \right)^2 \\ &= \sum_{k=1}^{n} \left(\frac{A\alpha}{\omega_k-\alpha} - \frac{A\beta}{\omega_k-\beta}\right)^2 \\ &= A^2\alpha^2S_2(\alpha) + A^2\beta^2S_2(\beta) - 2A^3\alpha\beta S_1(\alpha) + 2A^3\alpha\beta S_1(\beta) % A^2\alpha^2\frac{D+Eu}{(C-u)^2} + A^2\beta^2\frac{H+Iu}{(G-u)^2} - \frac{2A^3\alpha\beta B}{C-u} + \frac{2A^3\alpha\beta F}{G-u} \end{align*}

设回归直线 y=a+bxy=a+bx 与 xx 轴的交点为 XX，则

X=xˉ−yˉb=1n(∑k=1nxk)−1n(∑k=1nyk)n(∑k=1nxk2)−(∑k=1nxk)2n(∑k=1nxkyk)−(∑k=1nxk)(∑k=1nyk)=(∑k=1nxk)(∑k=1nxkyk)−(∑k=1nyk)(∑k=1nxk2)n(∑k=1nxkyk)−(∑k=1nxk)(∑k=1nyk)=ℜUℑV/2−ℑU(ℜV+W)/2nℑV/2−ℜUℑU=ℑ(UˉV−UW)ℑ(nV−U2)=UˉV−UW−(UVˉ−UˉW)nV−U2−(nVˉ−Uˉ2)\begin{align*} X &= \bar{x} - \frac{\bar{y}}{b} \\ &= \frac{1}{n} \left(\sum_{k=1}^{n}x_{k}\right) - \frac{1}{n} \left(\sum_{k=1}^{n}y_{k}\right)\frac{\displaystyle n\left(\sum_{k=1}^{n}x_{k}^{2}\right)-\left(\sum_{k=1}^{n}x_{k}\right)^{2}}{\displaystyle n\left(\sum_{k=1}^{n}x_{k}y_{k}\right)-\left(\sum_{k=1}^{n}x_{k}\right)\left(\sum_{k=1}^{n}y_{k}\right)} \\ &= \frac{\displaystyle \left(\sum_{k=1}^{n}x_{k}\right) \left(\sum_{k=1}^{n}x_{k}y_{k}\right) - \left(\sum_{k=1}^{n}y_{k}\right) \left(\sum_{k=1}^{n}x_{k}^{2}\right)}{\displaystyle n \left(\sum_{k=1}^{n}x_{k}y_{k}\right) - \left(\sum_{k=1}^{n}x_{k}\right) \left(\sum_{k=1}^{n}y_{k}\right)} \\ &= \frac{\Re U \Im V/2 - \Im U(\Re V+W)/2}{n\Im V/2 - \Re U \Im U} \\ &= \frac{\Im(\bar{U}V-UW)}{\Im(nV-U^2)} \\ &= \frac{\bar{U}V-UW - (U\bar{V}-\bar{U}W)}{nV-U^2 - (n\bar{V} - \bar{U}^2)} \end{align*}

注意 ∣u∣=1|u|=1，因此 uˉ=u−1\bar{u} = u^{-1}，使用计算机代数系统分别计算并化简分子和分母：

Code
# [Jupyter Notebook]

# %%
import sympy as sp

# %%
n = sp.symbols('n', integer=True, positive=True)
e = sp.symbols('e', real=True, positive=True)
u = sp.symbols('u', nonzero=True)
i = I = sp.I

# %%
alpha = sp.symbols('alpha', real=True)
beta = 1 / alpha

# %%
def S1(a):
    """
    sum 1/(omega-a)
    """
    return -n*a**(n-1)/(a**n-u)


def S2(a):
    """
    sum 1/(omega-a)^2
    """
    return (
        n*a**(2*n-2)
        + n*(n-1)*u*a**(n-2)
    )/(a**n-u)**2

# %%
A = 1/(alpha - beta)

# %%
U = (A*S1(alpha) - A*S1(beta))
U

# %%
V = (
    A**2*S2(alpha)   + A**2*S2(beta)
  - 2*A**3*S1(alpha) + 2*A**3*S1(beta)
)

V

# %%
W = (
    A**2*alpha**2*S2(alpha)
  + A**2*beta**2*S2(beta)
  - 2*A**3*alpha*beta*S1(alpha)
  + 2*A**3*alpha*beta*S1(beta)
)

W

# %%
def conj(expr):

    expr = sp.conjugate(expr)

    expr = expr.subs(
        sp.conjugate(u),
        1/u
    )

    expr = sp.simplify(expr)

    return expr

# %%
Aexpr = conj(U)*V - U*W
Bexpr = n*V-U**2

numerator = Aexpr-conj(Aexpr)
denominator = Bexpr-conj(Bexpr)

# %%
numerator = sp.factor(sp.simplify(sp.together(numerator)))  # 这一步不理想的话可以反复做
numerator

# %%
denominator = sp.factor(sp.simplify(sp.together(denominator)))  # 这一步不理想的话可以反复做
denominator

# %%
X = sp.factor(sp.simplify(numerator / denominator))
X

得到

分子=αnn2u(α2+1)(u−1)(u+1)(α4n+α2n+2n−2α2n+2−α2n+2α2−α2nn)(α−1)3(α+1)3(αn−u)2(αnu−1)2分母=−α2n2u(n−1)(u−1)(u+1)(α2+1)αn−2(αn−1)(αn+1)(α−1)(α+1)(αn−u)2(αnu−1)2\begin{align*} \text{分子} &= \frac{ \alpha^{n} n^{2} \red{u} \left(\alpha^{2} + 1\right) \red{\left(u - 1\right) \left(u + 1\right)} \left(\alpha^{4} n + \alpha^{2n+2} n - 2 \alpha^{2n+2} - \alpha^{2} n + 2 \alpha^{2} - \alpha^{2 n} n\right) }{ \left(\alpha - 1\right)^{3} \left(\alpha + 1\right)^{3} \red{\left(\alpha^{n} - u\right)^{2} \left(\alpha^{n} u - 1\right)^{2}} } \\ \text{分母} &= - \frac{ \alpha^{2} n^{2} \red{u} \left(n - 1\right) \red{\left(u - 1\right) \left(u + 1\right)} (\alpha^2+1)\alpha^{n-2}(\alpha^n-1)(\alpha^n+1) }{ \left(\alpha - 1\right) \left(\alpha + 1\right) \red{\left(\alpha^{n} - u\right)^{2} \left(\alpha^{n} u - 1\right)^{2}} } \\ \end{align*}

注意所有含 uu 的项全部抵消，

X=−α4n−α2n+2α2−α2nn+α2n+2n−2α2n+2(n−1)(α−1)2(α+1)2(αn−1)(αn+1)=−(n−2)(αn−βn)−n(αn−2−βn−2)(n−1)(α−β)2(αn−βn)\begin{align*} X &= - \frac{\alpha^{4} n - \alpha^{2} n + 2 \alpha^{2} - \alpha^{2 n} n + \alpha^{2 n + 2} n - 2 \alpha^{2 n + 2}}{\left(n - 1\right) (\alpha-1)^2(\alpha+1)^2(\alpha^n-1)(\alpha^n+1)} \\ &= - \frac{(n-2)(\alpha^n-\beta^n) - n(\alpha^{n-2}-\beta^{n-2})}{(n-1)(\alpha-\beta)^2(\alpha^n-\beta^n)} \end{align*}

注意这个式子关于 α,β\alpha, \beta 对称，这是因为这两个根在代数层面不可区分。

最后进行收尾工作，将结果变换回一般的标准形式：

结论

C(cn2(n−1)(1+αn−2−βn−2αn−βn),0)C\Biggl( \frac{cn}{2(n-1)}\left(1+\frac{\alpha^{n-2}-\beta^{n-2}}{\alpha^{n}-\beta^{n}}\right), 0 \Biggr)

其中 α,β=−a±bc\alpha, \beta = \dfrac{-a\pm b}{c}。

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