e² - 1 的应用

一个圆锥曲线第三定义的例子：

椭圆	双曲线
$k_{AP}\cdot k_{BP}=-\frac{b^2}{a^2}$	$k_{AP}\cdot k_{BP}=\frac{b^2}{a^2}$

在椭圆中，$-\frac{b^2}{a^2}=e^2-1$，而在双曲线中 $\frac{b^2}{a^2}=e^2-1$。使用 $e^2-1$ 可以无视正负性的影响，将椭圆和双曲线的通用理论更好地串联起来。

$$k_1\cdot k_2=e^2-1$$

证明：

$$\begin{aligned} k_1\cdot k_2&=\frac{y_0}{x_0+a}\cdot\frac{y_0}{x_0-a}=\frac{y_0^2}{x_0^2-a^2} \\ &=-\frac{b^2}{a^2}=e^2-1 \end{aligned}$$

$$k_1\cdot k_2=e^2-1$$

证明：

$$\left.\begin{aligned} \disp\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}=1\\ \disp\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1 \end{aligned}\right\} \intro \frac{x_0^2-x_1^2}{a^2}+\frac{y_0^2-y_1^2}{b^2}=0$$

$$\begin{aligned} k_1\cdot k_2&=\frac{y_0-y_1}{x_0-x_1}\cdot\frac{y_0+y_1}{x_0+x_1}=\frac{y_0^2-y_1^2}{x_0^2-x_1^2}\\ &=-\frac{b^2}{a^2}=e^2-1 \end{aligned}$$

$$k_1\cdot k_2=e^2-1$$

证明：

$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

$$\left.\begin{aligned} \disp\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1\\ \disp\frac{x_2^2}{a^2}+\frac{y_2^2}{b^2}=1 \end{aligned}\right\} \intro \frac{x_1^2-x_2^2}{a^2}+\frac{y_1^2-y_2^2}{b^2}=0$$

$$\begin{aligned} k_1\cdot k_2&=\frac{y_1-y_2}{x_1-x_2}\cdot\frac{(y_1+y_2)/2}{(x_1+x_2)/2}=\frac{y_1^2-y_2^2}{x_1^2-x_2^2} \\ &=-\frac{b^2}{a^2}=e^2-1 \end{aligned}$$

$$k_1\cdot k_2=e^2-1$$

证明：

切线：$\frac{x_0x}{a^2}+\frac{y_0y}{b^2}=1,k_1=-\frac{b^2x_0}{a^2y_0}$

$$k_1\cdot k_2=-\frac{b^2x_0}{a^2y_0}\cdot\frac{y_0}{x_0}=-\frac{b^2}{a^2}=e^2-1$$

$$k_1\cdot k_2=e^2-1\intro M \ \text{轨迹为相似椭圆}$$

证明：

$$\begin{aligned} &k_1\cdot k_2=e^2-1\\ \intro&\frac{y_1}{x_1}\cdot\frac{y_2}{x_2}=-\frac{b^2}{a^2}\\ \intro&\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}=0 \end{aligned}$$

联立

$$\begin{cases} \disp\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1, \disp\frac{x_2^2}{a^2}+\frac{y_2^2}{b^2}=1\\ \disp\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}=0 \end{cases}$$

得

$$\frac{x_1^2+x_2^2+2x_1x_2}{a^2}+\frac{y_1^2+y_2^2+2y_1y_2}{b^2}=2$$

$$\frac{(x_1+x_2)^2}{a^2}+\frac{(y_1+y_2)^2}{b^2}=2$$

$$\frac{x_M^2}{\disp\left(\frac{a}{\sqrt{2}}\right)^2}+\frac{y_M^2}{\disp\left(\frac{b}{\sqrt{2}}\right)^2}=1$$