简介

若圆锥曲线题围绕某特殊点展开，可将坐标系原点平移至此点，减小计算量。

例 1

已知椭圆 $E:{\displaystyle \frac{x^2}{4}}+\phantom{0}$$\phantom{0}{\displaystyle \frac{y^2}{3}}=\phantom{0}$$\phantom{0}1$，过 $P(2,\phantom{0}$$\phantom{0}1)$ 的直线 $l$ 与椭圆 $E$ 交于 $A,\phantom{0}$$\phantom{0}B$ 且 ${\overrightarrow{OP}}^2=\phantom{0}$$\phantom{0}4\overrightarrow{PA}\cdot \overrightarrow{PB}$，求 $l$ 的方程。

标答

联立 $\{\begin{array}{l}y=k(x-2)+1\\ {\displaystyle \frac{x^2}{4}}+{\displaystyle \frac{y^2}{3}}=1\end{array}$ 得

$$(3+4k^2)x^2-8k(2k-1)x+16k^2-16k-8=0$$

$\mathrm{\Delta }=\phantom{0}$$\phantom{0}[-\phantom{0}$$\phantom{0}8k(2k-\phantom{0}$$\phantom{0}1){]}^2-\phantom{0}$$\phantom{0}4(3+\phantom{0}$$\phantom{0}4k^2)(16k^2-\phantom{0}$$\phantom{0}16k-\phantom{0}$$\phantom{0}8)>0$

$\therefore 32(6k+\phantom{0}$$\phantom{0}3)>0,\phantom{0}$$\phantom{0}k>-\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}$

$x_1+\phantom{0}$$\phantom{0}{x}_2=\phantom{0}$$\phantom{0}{\displaystyle \frac{8k(2k-1)}{3+4k^2}},\phantom{0}$$\phantom{0}{x}_1{x}_2=\phantom{0}$$\phantom{0}{\displaystyle \frac{16k^2-16k-8}{3+k^2}}$

$\because {\overrightarrow{OP}}^2=\phantom{0}$$\phantom{0}4\overrightarrow{PA}\cdot \overrightarrow{PB}$

$\therefore (x_1-\phantom{0}$$\phantom{0}2,\phantom{0}$$\phantom{0}{y}_1-\phantom{0}$$\phantom{0}1)\cdot (x_2-\phantom{0}$$\phantom{0}2,\phantom{0}$$\phantom{0}{y}_2-\phantom{0}$$\phantom{0}1)=\phantom{0}$$\phantom{0}{\displaystyle \frac{5}{4}}$

$\therefore [x_1{x}_2-\phantom{0}$$\phantom{0}2(x_1+\phantom{0}$$\phantom{0}{x}_2)+\phantom{0}$$\phantom{0}4](1+\phantom{0}$$\phantom{0}{k}^2)=\phantom{0}$$\phantom{0}{\displaystyle \frac{5}{4}}$

$\therefore [{\displaystyle \frac{16k^2-16k-8}{3+4k^2}}-2{\displaystyle \frac{8k(2k-1)}{3+4k^2}}+4](1+\phantom{0}$$\phantom{0}{k}^2)=\phantom{0}$$\phantom{0}{\displaystyle \frac{4+4k^2}{3+4k^2}}=\phantom{0}$$\phantom{0}{\displaystyle \frac{5}{4}}$

解得 $k=\phantom{0}$$\phantom{0}\pm {\displaystyle \frac{1}{2}},\phantom{0}$$\phantom{0}k=\phantom{0}$$\phantom{0}-\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}$ 不合题意，舍去。

$\therefore l:y=\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}x$。

快速解法

将坐标系原点平移至 $P(2,\phantom{0}$$\phantom{0}1)$。为了区分坐标系，令 $\lambda =\phantom{0}$$\phantom{0}x-\phantom{0}$$\phantom{0}2,\phantom{0}$$\phantom{0}\mu =\phantom{0}$$\phantom{0}y-\phantom{0}$$\phantom{0}1$。新坐标系中 $A({\lambda }_1,\phantom{0}$$\phantom{0}{\mu }_1),\phantom{0}$$\phantom{0}B({\lambda }_2,\phantom{0}$$\phantom{0}{\mu }_2)$。

设 $l:\mu =\phantom{0}$$\phantom{0}k\lambda$。于是 $\overrightarrow{PA}\cdot \overrightarrow{PB}=\phantom{0}$$\phantom{0}(k^2+\phantom{0}$$\phantom{0}1){\lambda }_1{\lambda }_2=\phantom{0}$$\phantom{0}{\displaystyle \frac{5}{4}}$。

联立 $\{\begin{array}{l}\mu =k\lambda \\ {\displaystyle \frac{(\lambda +2{)}^2}{4}}+{\displaystyle \frac{(\mu +1{)}^2}{3}}=1\end{array}$ 得

$$(4k^2+3){\lambda }^2+(12+8k)\lambda +4=0$$

$\mathrm{\Delta }=\phantom{0}$$\phantom{0}192k+\phantom{0}$$\phantom{0}96>0,\phantom{0}$$\phantom{0}\therefore k>-\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}$。

${\lambda }_1{\lambda }_2=\phantom{0}$$\phantom{0}{\displaystyle \frac{4}{4k^2+3}}$

代入得 $(k^2+\phantom{0}$$\phantom{0}1){\displaystyle \frac{4}{4k^2+3}}=\phantom{0}$$\phantom{0}{\displaystyle \frac{5}{4}}$，解得 $k=\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}$。$\therefore y=\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}x$。

例 2

已知双曲线 $C:x^2-\phantom{0}$$\phantom{0}{\displaystyle \frac{y^2}{16}}=\phantom{0}$$\phantom{0}1$。$T$ 在直线 $x=\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}$ 上。过 $T$ 的两条直线分别交 $C$ 的右支于 $A,\phantom{0}$$\phantom{0}B$ 和 $P,\phantom{0}$$\phantom{0}Q$。若 $|TA|\cdot |TB|=\phantom{0}$$\phantom{0}|TP|\cdot |TQ|$，求 $k_{AB}+\phantom{0}$$\phantom{0}{k}_{PQ}$。

快速解法

设 $T({\displaystyle \frac{1}{2}},t)$。将坐标系原点平移至 $T$。令 $\lambda =\phantom{0}$$\phantom{0}x-\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}},\phantom{0}$$\phantom{0}\mu =\phantom{0}$$\phantom{0}y-\phantom{0}$$\phantom{0}t$。

设 $AB:\mu =\phantom{0}$$\phantom{0}{k}_1\lambda ,\phantom{0}$$\phantom{0}PQ:\mu =\phantom{0}$$\phantom{0}{k}_2\lambda$。

$\because |TA|\cdot |TB|=\phantom{0}$$\phantom{0}|TP|\cdot |TQ|$

$\therefore ({\lambda }_A^2+\phantom{0}$$\phantom{0}{\mu }_A^2)({\lambda }_B^2+\phantom{0}$$\phantom{0}{\mu }_B^2)=\phantom{0}$$\phantom{0}({\lambda }_P^2+\phantom{0}$$\phantom{0}{\mu }_P^2)({\lambda }_Q^2+\phantom{0}$$\phantom{0}{\mu }_Q^2)$

$\therefore (k_1^2+\phantom{0}$$\phantom{0}1{)}^2({\lambda }_A{\lambda }_B{)}^2=\phantom{0}$$\phantom{0}(k_2^2+\phantom{0}$$\phantom{0}1{)}^2({\lambda }_P{\lambda }_Q{)}^2$

联立 $\{\begin{array}{l}\mu =k\lambda \\ {(\lambda +{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{(\mu +t{)}^2}{16}}=1\end{array}$ 得

$$(16-k^2){\lambda }^2+(16-2kt)\lambda -t^2-12=0$$

$\mathrm{\Delta }>0$ 解得 $k<4$。

${\lambda }_1{\lambda }_2=\phantom{0}$$\phantom{0}{\displaystyle \frac{-t^2-12}{16-k^2}}$，即 ${\lambda }_A{\lambda }_B=\phantom{0}$$\phantom{0}{\displaystyle \frac{-t^2-12}{16-k_1^2}},\phantom{0}$$\phantom{0}{\lambda }_P{\lambda }_Q=\phantom{0}$$\phantom{0}{\displaystyle \frac{-t^2-12}{16-k_2^2}}$

代入得 ${\displaystyle \frac{k_1^2+1}{16-k_1^2}}=\phantom{0}$$\phantom{0}{\displaystyle \frac{k_2^2+1}{16-k_2^2}}$。设左边表达式 $=s,\phantom{0}$$\phantom{0}\therefore k_1,\phantom{0}$$\phantom{0}{k}_2$ 是 $f(x)=\phantom{0}$$\phantom{0}{\displaystyle \frac{k^2+1}{16-k^2}}=\phantom{0}$$\phantom{0}s$ 的两根。

$\because f(-x)=\phantom{0}$$\phantom{0}f(x)$，且 $f(x)=\phantom{0}$$\phantom{0}s$ 仅有两根，$\therefore (k_1,\phantom{0}$$\phantom{0}s),\phantom{0}$$\phantom{0}(k_2,\phantom{0}$$\phantom{0}s)$ 关于 $y$ 轴对称。

$\therefore k_{AB}+\phantom{0}$$\phantom{0}{k}_{PQ}=\phantom{0}$$\phantom{0}{k}_1+\phantom{0}$$\phantom{0}{k}_2=\phantom{0}$$\phantom{0}0$。