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a-gaokaoized-proof-of-conic-property.md

---
title: 爱定比分点 TV 之极点极线
createTime: 2025/10/18
categories:
    - study
tags:
    - maths
---

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## 过程

设 $P\left(\dfrac{x_A + \lambda x_B}{1 + \lambda}, \dfrac{y_A + \lambda y_B}{1+\lambda}\right)$，联立 $\begin{cases} Ax_A^2 + Cy_A^2 + F = 0 & (1) \\ Ax_B^2 + Cy_B^2 + F = 0 & (2) \end{cases}$，$\dfrac{(1) - \lambda^2(2)}{1-\lambda^2}$ 得

$$
A \left(\dfrac{x_A + \lambda x_B}{1 + \lambda}\right)
    \left(\dfrac{x_A - \lambda x_B}{1 - \lambda}\right)
+ C \left(\dfrac{y_A + \lambda y_B}{1 + \lambda}\right)
    \left(\dfrac{y_A - \lambda y_B}{1 - \lambda}\right)
+ F = 0$$

设 $M \left(\dfrac{x_A - \lambda x_B}{1 - \lambda}, \dfrac{y_A - \lambda y_B}{1 - \lambda}\right)$，则 $A x_P x_M + C y_P y_M + F = 0$。

同理，设 $P \left(\dfrac{x_C + \mu x_D}{1 + \mu}, \dfrac{y_C + \mu y_D}{1+\mu}\right)$，$N \left(\dfrac{x_C - \mu x_D}{1 - \mu}, \dfrac{y_C - \mu y_D}{1-\mu}\right)$，

则有 $A x_P x_N + C y_P y_M + F = 0$，则 $l_{MN}$ 为定直线 $A x_P x + C y_P x + F = 0$。

因为 $x_P = \dfrac{x_A + \lambda x_B}{1 + \lambda} = \dfrac{x_C + \mu x_D}{1 + \mu}$，通分并移项得

$$(1+\mu) x_A - (1+\lambda) \mu x_D = (1+\lambda) x_C - (1+\mu)\lambda x_B$$

设

$$
\begin{align*}
x_1 &= \dfrac{(1+\mu) x_A - (1+\lambda) \mu x_D}{1 - \lambda\mu} = \dfrac{(1+\lambda)x_C - (1+\mu)\lambda x_B}{1 - \lambda\mu} \\
    & = \dfrac{(1+\mu)(1-\lambda)\left(\dfrac{x_A-\lambda x_B}{1-\lambda}\right) + (1+\lambda)(1-\mu)\left(\dfrac{x_C - \mu x_D}{1-\mu}\right)}{2 - 2\lambda\mu} \\
    & = \dfrac{(1-\lambda\mu+\mu-\lambda) x_M + (1-\lambda\mu+\lambda-\mu) x_N}{2-2\lambda\mu}
\end{align*}
$$

同理可得

$$
\begin{align*}
y_1 &= \dfrac{(1+\mu) y_A - (1+\lambda) \mu y_D}{1 - \lambda\mu} = \dfrac{(1+\lambda) y_C - (1+\mu)\lambda y_B}{1 - \lambda\mu} \\
    &= \dfrac{(1-\lambda\mu+\mu-\lambda) y_M + (1-\lambda\mu+\lambda-\mu) y_N}{2-2\lambda\mu}
\end{align*}
$$

因此 $(x_1, y_1)$ 在 $AD, BC, MN$ 上，即 $Q(x_1, y_1)$ 在 $MN$ 上，所以 $A x_P x_Q + C y_P y_Q + F = 0$。

## 去齐次吧，齐不被定义[+undefined]的次

[+undefined]: 指自己定义的记号，高考不能用

$$
A + \lambda B \sim C + \mu D \\
(1+\mu)(A + \lambda B) = (1+\lambda)(C + \mu D) \\
(1+\mu)A-(1+\lambda)\mu D = (1+\lambda)C-(1+\mu)\lambda B \\= \dfrac{1}{2}((1+\mu)(A-\lambda B) + (1+\lambda)(C-\mu D))
$$