随便玩玩

T1

如图1, 3(FC + FA + CA) = BD + BF^{\prime} + DF^{\prime} ,那么椭圆 T 与双曲线 S 的离心率之比为?


图1

解:

\begin{aligned}
2a&=CF+CA+AF^{\prime}\\
\frac43a&=FA+FC+CF\\
\\
\frac23a&=AF^{\prime}-AF=2a^{\prime}
\end{aligned}

所以

e_T:e_S=2a^\prime:2a=1:3

Note1

如图2,过椭圆切线 f 分别做过两个焦点的垂线 F_1E, F_2H, 切点为 D(x_1,y_1).

那么有 \angle F_1DF_2 的角平分线 m\bot f.

证明:

F_1D = a + ex\\F_2D = a – ex

mx 轴交于点 M(x_1,0).

\begin{aligned}
x_1&=2c\frac{F_1P}{F_1P+F_2P}-c\\
&=ae+e^2x_0-c\\
&=e^2x_0
\end{aligned}

\begin{aligned}
k_1&=\frac{y_0}{x_0-ae-e^2x_0+c}\\
&=\frac{y_0}{x_0-e^2x_0}\\
&=\frac{y_0}{x_0(1-e^2)}\\
&=\frac{y_0}{x_0\frac{b^2}{a^2}}\\
&=\frac{y_0a^2}{x_0b^2}\\\
k_2&=-\frac{b^2x_0}{a^2y_0}
\end{aligned}

所以

k_1k_2=-\frac{x_0y_0a^2b^2}{x_0y_0a^2b^2}=-1

m\bot l

则有 f\angle F_1DF 的角平分线。

\begin{aligned}
\therefore F_1E=EF\\
\therefore OE=\frac12F_2F
\end{aligned}

F_2F=F_1D+F_2D=2a\\
\therefore OE=a

E,H 都在圆 x^2+y^2=a^2 上.


图2