查看源代码：
`new-problem.md`

---
title: 椭圆上从焦点等角的点的回归直线居然过定点？
createTime: 2026/8/9
categories: ['study']
tags: ['maths']
---

:::center
![题图](./new-problem/image.png){width=500px}
:::

:::problem
椭圆 $\Gamma: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}=1$，$F(c,0)$ 为 $\Gamma$ 的右焦点，$P_1, \dots, P_n \in \Gamma$ 满足 $\measuredangle P_iFP_{i+1} = \dfrac{2\pi}{n}$，$l$ 是 $P_1, \dots, P_n$ 的最小二乘回归直线，证明：当 $P_1$ 在 $\Gamma$ 上运动时，$l$ 过定点 $C$，并求出定点的坐标（用含 $n, a,b,c$ 的代数式表示）。
:::

不妨将 $F$ 平移到原点，将焦准距位似为 $\dfrac{1}{2}$。设 $\Gamma$ 的离心率为 $e$。

设 $\theta_k = \dfrac{2k\pi + \varphi}{n}, P_k(x_k,y_k)$，根据焦半径公式，取

$$
\begin{align*}
  r_k &= \frac{e}{2+2e\cos\theta_k} \\
  x_k &= r_k\cos(-\theta_k) \\
  y_k &= r_k\sin(-\theta_k)
\end{align*}
$$

（负号是让式子更简单的小巧思）

记 $\omega_k = \exp\left(\mathrm{i}\theta_k\right)$，设 $z_k=x_k+\mathrm{i}y_k$，则

$$
z_k = \frac{e\bar{\omega}_k}{2+2e\Re \omega_k} = \frac{e\omega_k^{-1}}{2+ e(\omega_k+\omega_k^{-1})} = \dfrac{1}{\omega_k^2 + \dfrac{2}{e}\omega_k + 1}
$$

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

$$
z_k = \dfrac{1}{(\omega_k-\alpha)(\omega_k-\beta)} = \dfrac{A}{\omega_k-\alpha} - \dfrac{A}{\omega_k-\beta}
$$

其中实数 $A = \dfrac{1}{\alpha - \beta}$。

:::lemma 一个韦达定理的简单推论
若多项式 $P(x)$ 所有根为 $x_1, \dots, x_n$，则

$$
\sum_{k=1}^{n} \frac{1}{x_k - a} = -\frac{P'(a)}{P(a)}
$$

两边对 $a$ 求导，得到

$$
\sum_{k=1}^{n} \frac{1}{(x_k - a)^2} = \frac{P'(a)^2 - P''(a)P(a)}{P(a)^2}
$$
:::

因为 $\omega_k$ 是方程 $\omega^n = \exp(\mathrm{i}\varphi) = u$ 的 $n$ 个根，记 $P(x) = x^n - u$，则 $P'(x) = nx^{n-1}, P''(x) = n(n-1)x^{n-2}$，于是记

$$
\begin{align*}
  S_1(a) &= \sum_{k=1}^{n} \dfrac{1}{\omega_k-a} = -\frac{na^{n-1}}{a^n-u} % = \frac{B}{C-u}
  \\
  S_2(a) &= \sum_{k=1}^{n} \dfrac{1}{(\omega_k-a)^2} = \frac{(na^{n-1})^2-n(n-1)a^{n-2}(a^n-u)}{(a^n-u)^2} \\
  % &= \dfrac{n\alpha^{2n-2} + n(n-1)u\alpha^{n-2}}{(\alpha^n-u)^2} = \frac{D+Eu}{(C-u)^2}
\end{align*}
$$

<!-- $$
\begin{align*}
  \\
  \sum_{k=1}^{n} \dfrac{1}{\omega_k-\beta} &= -\frac{n\beta^{n-1}}{\beta^n-u} = \frac{F}{G-u}
  \\
  \sum_{k=1}^{n} \dfrac{1}{(\omega_k-\beta)^2} &= \frac{(n\beta^{n-1})^2-n(n-1)\beta^{n-2}(\beta^n-u)}{(\beta^n-u)^2} \\
  &= \dfrac{n\beta^{2n-2} + n(n-1)u\beta^{n-2}}{(\beta^n-u)^2} = \frac{H+Iu}{(G-u)^2}
  \\
  \sum_{k=1}^{n} \frac{1}{(\omega_k-\alpha)(\omega_k-\beta)} &= \frac{AB}{C-u} - \frac{AF}{G-u}
  \\
  \sum_{k=1}^{n} \frac{1}{(\omega_k-\alpha)^2(\omega_k-\beta)^2} &= \sum_{k=1}^{n} \left( \frac{A^2}{(\omega_k-\alpha)^2} + \frac{A^2}{(\omega_k-\beta)^2} - \frac{2A^3}{\omega_k-\alpha} + \frac{2A^3}{\omega_k-\beta} \right) \\
  &= A^2\frac{D+Eu}{(C-u)^2} + A^2\frac{H+Iu}{(G-u)^2} - \frac{2A^3 B}{C-u} + \frac{2A^3 F}{G-u}
  \\
  \sum_{k=1}^{n} \frac{\omega_k^2}{(\omega_k-\alpha)^2(\omega_k-\beta)^2} &= \sum_{k=1}^{n} \left( \frac{A^2\alpha^2}{(\omega_k-\alpha)^2} + \frac{A^2\beta^2}{(\omega_k-\beta)^2} - \frac{2A^3\alpha\beta}{\omega_k-\alpha} + \frac{2A^3\alpha\beta}{\omega_k-\beta} \right) \\
  &= A^2\alpha^2\frac{D+Eu}{(C-u)^2} + A^2\beta^2\frac{H+Iu}{(G-u)^2} - \frac{2A^3\alpha\beta B}{C-u} + \frac{2A^3\alpha\beta F}{G-u}
\end{align*}
$$ -->

注意 $\omega_k z_k \in \mathbb{R}$，所以 $|z_k|^2 = \omega_k^2 z_k^2, \bar{z} = \omega_k^2z_k$，令

$$
\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 $y=a+bx$ 与 $x$ 轴的交点为 $X$，则

$$
\begin{align*}
X &= \bar{x} - \frac{\bar{y}}{b} \\
&= \frac{1}{n} \left(\sum_{k=1}^{n}x_{k}\right) - \frac{1}{n} \left(\sum_{k=1}^{n}y_{k}\right)\frac{\displaystyle n\left(\sum_{k=1}^{n}x_{k}^{2}\right)-\left(\sum_{k=1}^{n}x_{k}\right)^{2}}{\displaystyle n\left(\sum_{k=1}^{n}x_{k}y_{k}\right)-\left(\sum_{k=1}^{n}x_{k}\right)\left(\sum_{k=1}^{n}y_{k}\right)} \\
&= \frac{\displaystyle \left(\sum_{k=1}^{n}x_{k}\right) \left(\sum_{k=1}^{n}x_{k}y_{k}\right) - \left(\sum_{k=1}^{n}y_{k}\right) \left(\sum_{k=1}^{n}x_{k}^{2}\right)}{\displaystyle n \left(\sum_{k=1}^{n}x_{k}y_{k}\right) - \left(\sum_{k=1}^{n}x_{k}\right) \left(\sum_{k=1}^{n}y_{k}\right)} \\
&= \frac{\Re U \Im V/2 - \Im U(\Re V+W)/2}{n\Im V/2 - \Re U \Im U} \\
&= \frac{\Im(\bar{U}V-UW)}{\Im(nV-U^2)} \\
&= \frac{\bar{U}V-UW - (U\bar{V}-\bar{U}W)}{nV-U^2 - (n\bar{V} - \bar{U}^2)}
\end{align*}
$$

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

:::details Code
```python
# [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

$$
\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 $u$ 的项全部抵消，

$$
\begin{align*}
X &= - \frac{\alpha^{4} n - \alpha^{2} n + 2 \alpha^{2} - \alpha^{2 n} n + \alpha^{2 n + 2} n - 2 \alpha^{2 n + 2}}{\left(n - 1\right) (\alpha-1)^2(\alpha+1)^2(\alpha^n-1)(\alpha^n+1)} \\
&= - \frac{(n-2)(\alpha^n-\beta^n) - n(\alpha^{n-2}-\beta^{n-2})}{(n-1)(\alpha-\beta)^2(\alpha^n-\beta^n)}
\end{align*}
$$

注意这个式子关于 $\alpha, \beta$ 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：

:::solution 结论
$$
C\Biggl( \frac{cn}{2(n-1)}\left(1+\frac{\alpha^{n-2}-\beta^{n-2}}{\alpha^{n}-\beta^{n}}\right), 0 \Biggr)
$$

其中 $\alpha, \beta = \dfrac{-a\pm b}{c}$。
:::
查看源代码：new-problem.md

查看源代码：
`new-problem.md`
