1707

已知椭圆 $C : \frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$ 的右焦点为 $F$ , 直线 $l$ 交 $C$ 于 $A, B$ 两点, $O$ 为坐标原点, 且 $O A ⊥ O B$ . 过 $O$ 作 $O H ⊥ A B$ , 垂足为 $H$ , 求直线 $F H$ 的斜率的取值范围.

解

这是一个经典的结论:

引理

设椭圆 $C : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , $A, B, H$ 如题述, 则 $O H = \frac{a b}{\sqrt{a^{2} + b^{2}}}$ 为定值.

于是 $O H = \frac{2}{\sqrt{3}}$ 为定值, $F H$ 斜率的最值其实就是 $F H$ 与圆 $x^{2} + y^{2} = \frac{4}{3}$ 相切时取到, 如图, $k_{max} = \frac{O H^{'}}{H^{'} F} = \frac{\frac{2}{\sqrt{3}}}{\sqrt{2 - \frac{4}{3}}} = \sqrt{2},$ 所以 $k_{F H} \in [- \sqrt{2}, \sqrt{2}]$ .
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或者设 $H (\frac{2}{\sqrt{3}} \cos α, \frac{2}{\sqrt{3}} \sin α)$ , 然后 $k_{F H} = \frac{\frac{2}{\sqrt{3}} \sin α}{\frac{2}{\sqrt{3}} \cos α - \sqrt{2}} \leq \frac{\frac{2}{\sqrt{3}} \sin α}{\sqrt{2} + \sqrt{\frac{2}{3}} \sin α - \sqrt{2}} = \sqrt{2} .$

经典结论的证明

$O A, O B$ 与坐标轴重合的情况容易验证.

证法一化齐次联立 设 $l_{A B} : α x + β y = 1$ , 与 $C$ 联立得 $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - (α x + β y)^{2} = 0.$ 令 $\frac{y}{x} = k$ , 则 $(\frac{1}{b^{2}} - β^{2}) k^{2} - 2 α β k + (\frac{1}{a^{2}} - α^{2}) = 0.$ 由 $O A ⊥ O B$ , 得 $- 1 = k_{1} k_{2} = \frac{\frac{1}{a^{2}} - α^{2}}{\frac{1}{b^{2}} - β^{2}} \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = α^{2} + β^{2} .$ 于是用点到直线距离公式可得 $O H = \frac{1}{\sqrt{α^{2} + β^{2}}} = \frac{a b}{\sqrt{a^{2} + b^{2}}}$ .
证法二极坐标 设 $A (r_{1} \cos θ, r_{1} \sin θ)$ , $B (r_{2} \cos (θ + \frac{π}{2}), r_{2} \sin (θ + \frac{π}{2}))$ , 则 $\frac{r_{1}^{2} \cos^{2} θ}{a^{2}} + \frac{r_{1}^{2} \sin^{2} θ}{b^{2}} = 1, \frac{r_{2}^{2} \cos^{2} (θ + \frac{π}{2})}{a^{2}} + \frac{r_{2}^{2} \sin^{2} (θ + \frac{π}{2})}{b^{2}} = 1.$ 从而 $\frac{1}{O H^{2}} = \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \Rightarrow O H = \frac{a b}{\sqrt{a^{2} + b^{2}}} .$
证法三硬解坐标 设 $l_{O A} : y = k x$ , $l_{O B} : y = - \frac{1}{k} x$ . 容易解得 $x_{A}^{2} = {(\frac{1}{a^{2}} + \frac{k^{2}}{b^{2}})}^{- 1}, x_{B}^{2} = {(\frac{1}{a^{2}} + \frac{1}{k^{2} b^{2}})}^{- 1},$ 于是 $\frac{1}{O H^{2}} = \frac{1}{O A^{2}} + \frac{1}{O B^{2}} = \frac{1}{(1 + k^{2}) x_{A}^{2}} + \frac{1}{(1 + \frac{1}{k^{2}}) x_{B}^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} .$
证法四韦达定理 设 $l_{A B} : y = k x + m$ , 联立得 $(a^{2} k^{2} + b^{2}) x^{2} + 2 k m a^{2} x + a^{2} (m^{2} - b^{2}) = 0.$ 当 $Δ > 0$ , 由韦达定理得 $x_{A} + x_{B} = - \frac{2 k m a^{2}}{a^{2} k^{2} + b^{2}}, x_{A} x_{b} = \frac{a^{2} (m^{2} - b^{2})}{a^{2} k^{2} + b^{2}} .$ 由 $O A ⊥ O B$ 得 $\begin{aligned} 0 & = \vec{O A} \cdot \vec{O B} = (k^{2} + 1) x_{A} x_{B} + k m (x_{A} + x_{B}) + m^{2} \\ = \frac{(a^{2} + b^{2}) m^{2} - a^{2} b^{2} (k^{2} + 1)}{a^{2} k^{2} + b^{2}} \\ \Rightarrow & \frac{m^{2}}{k^{2} + 1} = \frac{a^{2} b^{2}}{a^{2} + b^{2}}, \end{aligned}$ 于是 $O H = \frac{| m |}{\sqrt{k^{2} + 1}} = \frac{a b}{\sqrt{a^{2} + b^{2}}}$ .