为什么使用 e² - 1

一个圆锥曲线第三定义的例子：

椭圆	双曲线
$k_{AP}\cdot k_{BP}=\phantom{0}$$\phantom{0}-\phantom{0}$$\phantom{0}{\displaystyle \frac{b^2}{a^2}}$	$k_{AP}\cdot k_{BP}=\phantom{0}$$\phantom{0}{\displaystyle \frac{b^2}{a^2}}$

在椭圆中，$-{\displaystyle \frac{b^2}{a^2}}=\phantom{0}$$\phantom{0}{e}^2-\phantom{0}$$\phantom{0}1$，而在双曲线中 ${\displaystyle \frac{b^2}{a^2}}=\phantom{0}$$\phantom{0}{e}^2-\phantom{0}$$\phantom{0}1$。使用 $e^2-\phantom{0}$$\phantom{0}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。

应用 1

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

证

$$\begin{array}{rl}{k}_1\cdot k_2& ={\displaystyle \frac{y_0}{x_0+a}}\cdot {\displaystyle \frac{y_0}{x_0-a}}={\displaystyle \frac{y_0^2}{x_0^2-a^2}}\\ & =-{\displaystyle \frac{b^2}{a^2}}=e^2-1\end{array}$$

应用 2

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

证

$$\begin{array}{r}{\displaystyle \frac{x_0^2}{a^2}}+{\displaystyle \frac{y_0^2}{b^2}}=1\\ {\displaystyle \frac{x_1^2}{a^2}}+{\displaystyle \frac{y_1^2}{b^2}}=1\end{array}\}\Rightarrow {\displaystyle \frac{x_0^2-x_1^2}{a^2}}+{\displaystyle \frac{y_0^2-y_1^2}{b^2}}=0$$$$\begin{array}{rl}{k}_1\cdot k_2& ={\displaystyle \frac{y_0-y_1}{x_0-x_1}}\cdot {\displaystyle \frac{y_0+y_1}{x_0+x_1}}={\displaystyle \frac{y_0^2-y_1^2}{x_0^2-x_1^2}}\\ & =-{\displaystyle \frac{b^2}{a^2}}=e^2-1\end{array}$$

应用 3

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

证

$$M({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}})$$$$\begin{array}{r}{\displaystyle \frac{x_1^2}{a^2}}+{\displaystyle \frac{y_1^2}{b^2}}=1\\ {\displaystyle \frac{x_2^2}{a^2}}+{\displaystyle \frac{y_2^2}{b^2}}=1\end{array}\}\Rightarrow {\displaystyle \frac{x_1^2-x_2^2}{a^2}}+{\displaystyle \frac{y_1^2-y_2^2}{b^2}}=0$$$$\begin{array}{rl}{k}_1\cdot k_2& ={\displaystyle \frac{y_1-y_2}{x_1-x_2}}\cdot {\displaystyle \frac{(y_1+y_2)/2}{(x_1+x_2)/2}}={\displaystyle \frac{y_1^2-y_2^2}{x_1^2-x_2^2}}\\ & =-{\displaystyle \frac{b^2}{a^2}}=e^2-1\end{array}$$

应用 4

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

证

切线：${\displaystyle \frac{x_{0}x}{a^2}}+\phantom{0}$$\phantom{0}{\displaystyle \frac{y_{0}y}{b^2}}=\phantom{0}$$\phantom{0}1,\phantom{0}$$\phantom{0}{k}_1=\phantom{0}$$\phantom{0}-\phantom{0}$$\phantom{0}{\displaystyle \frac{b^2{x}_0}{a^2{y}_0}}$

$$k_1\cdot k_2=-{\displaystyle \frac{b^2{x}_0}{a^2{y}_0}}\cdot {\displaystyle \frac{y_0}{x_0}}=-{\displaystyle \frac{b^2}{a^2}}=e^2-1$$

应用 5

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

证

$$\begin{array}{rl}& k_1\cdot k_2=e^2-1\\ \Rightarrow & {\displaystyle \frac{y_1}{x_1}}\cdot {\displaystyle \frac{y_2}{x_2}}=-{\displaystyle \frac{b^2}{a^2}}\\ \Rightarrow & {\displaystyle \frac{x_1{x}_2}{a^2}}+{\displaystyle \frac{y_1{y}_2}{b^2}}=0\end{array}$$

联立

$$\{\begin{array}{l}{\displaystyle \frac{x_1^2}{a^2}}+{\displaystyle \frac{y_1^2}{b^2}}=1,{\displaystyle \frac{x_2^2}{a^2}}+{\displaystyle \frac{y_2^2}{b^2}}=1\\ {\displaystyle \frac{x_1{x}_2}{a^2}}+{\displaystyle \frac{y_1{y}_2}{b^2}}=0\end{array}$$

得

$${\displaystyle \frac{x_1^2+x_2^2+2x_1{x}_2}{a^2}}+{\displaystyle \frac{y_1^2+y_2^2+2y_1{y}_2}{b^2}}=2$$$${\displaystyle \frac{(x_1+x_2{)}^2}{a^2}}+{\displaystyle \frac{(y_1+y_2{)}^2}{b^2}}=2$$$${\displaystyle \frac{x_M^2}{{\left({\displaystyle \frac{a}{\sqrt{2}}}\right)}^2}}+{\displaystyle \frac{y_M^2}{{\left({\displaystyle \frac{b}{\sqrt{2}}}\right)}^2}}=1$$