感谢我的老师。

基本思想是把 $y=kx+b$ 变换成 $\frac{y-kx}{b}=1$ 来代换。

¶ 定义

¶ 定义一

$$|F_{1} P|+\left|F_{2} P\right|=2 a\\ \left|F_{1} F_{2}\right|=2 c\\$$

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad(a>b>0)$$

$$\begin{array}{l}{\left|P F_{1}\right|=a+e x} \\ {\left|P F_{2}\right|=a-e x}\end{array}\\$$

¶ 定义二

¶ 定义三

$$d=|AA_1|\\ \frac{|F_2P|}{d}=e<1$$

$$通径 |CD|=2\frac{b^2}{a}$$

$$|F_2M|=p=\frac{a^2}{c}-c=\frac{b^2}{c}$$

$$\left|A F_{2}\right|=\frac{e p}{1+e \cos \theta}\\$$

$$\left|B F_{2}\right|=\frac{e p}{1-e \cos \theta}\\$$

$$\left|AB\right|=\frac{2 e p}{1-e^2 \cos^2 \theta}\\$$

$$\frac{|AF_2|}{|BF_2|}=\frac{1-e\cos \theta}{1+e\cos \theta}$$

注意，通过控制 $\theta$ 可以得出 $|AB|_{min}=2\frac{b^2}{a}$

¶ 乱七八糟

¶ 三角换元

$$\begin{aligned} \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} &= 1 \\ \left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2} &= 1 \end{aligned}\\$$

$${\left\{\begin{aligned} {x=a \cdot \cos \alpha} \\ {y=b \cdot \sin \alpha} \end{aligned}\right.}$$

¶ 点差法

$$\left\{\begin{aligned}{\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1} \\ {\frac{x_{2}^{2}}{a^{2}}+\frac{y_{2}^{2}}{b^{2}}=1}\end{aligned}\right.\\$$

$$\begin{aligned} \frac{x_{1}^{2}-x_{2}^{2}}{a^{2}}&=-\frac{y_{1}^{2}-y_{0}^{2}}{b^{2}} \\ \frac{y_{1}-y_{2}}{x_{1}-x_{2}} \cdot \frac{y_{1}+y_{2}}{x_{1}+x_{2}}&=-\frac{b^{2}}{a^{2}}\\ \frac{y_{1}-y_{2}}{x_{1}-x_{2}} \cdot \frac{y_0}{x_0}&=-\frac{b^{2}}{a^{2}}\\ k_{A B} \cdot k_{OP}&=-\frac{b^{2}}{a^{2}} \end{aligned}\\$$

¶ 切线

$$\begin{gathered} A x^{2}+B y^{2}+C x+D y+E=0\\ A x_{0} x+B y_{0} y+C \frac{x_{0}+x}{2}+D \frac{y_{0}+y}{2}+E=0 \end{gathered}$$

¶ 整体思想

$$\begin{gathered} x_1 + kx_2 = 0\\ \frac{x_1}{x_2} = -k\\ \frac{x_2}{x_1} = -\frac{1}{k}\\ \frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{x_1^2 + x_2^2}{x_1x_2} = \frac{(x_1+x_2)^2}{x_1x_2}-2\\ \frac{(x_1+x_2)^2}{x_1x_2}=-k-\frac{1}{k}+2 \end{gathered}$$

¶ 焦点三角形

¶ 仿射变换