（圆锥曲线）利用两点式的变换来解决常见问题

2022-08-24 22:59 作者:因恋相思久 0人读过 | 我要投稿

先来一道开胃菜

抛物线 $C%EF%BC%9Ay%5E2%3D2px%20%EF%BC%88p%EF%BC%9E0%EF%BC%89$ 上一定点 $P%EF%BC%88x_%7B0%7D%EF%BC%8Cy_%7B0%7D%20%EF%BC%89%EF%BC%88y_%7B0%7D%20%EF%BC%9E0%EF%BC%89$ 作两条直线分别交抛物线于 $A%EF%BC%88x_%7B1%7D%20%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CB%EF%BC%88x_%7B2%7D%20%EF%BC%8Cy_%7B2%7D%20%EF%BC%89$ .

问：当PA，PB的斜率存在且倾斜角互补时，求 $%5Cfrac%7By_%7B1%7D%20%2By_%7B2%7D%20%7D%7By_%7B0%7D%20%7D%20$ 的值，并证明直线AB的斜率是非零常数

解：.

因为直线PA: $%EF%BC%88y_%7B1%7D%20%2By_%7B0%7D%20%EF%BC%89y%3D2px%2By_%7B1%7D%20y_%7B0%7D%20$

直线PB： $%EF%BC%88y_%7B2%7D%20%2By_%7B0%7D%20%EF%BC%89y%3D2px%2By_%7B2%7D%20y_%7B0%7D%20$

直线AB： $%EF%BC%88y_%7B1%7D%20%2By_%7B2%7D%20%EF%BC%89y%3D2px%2By_%7B1%7D%20y_%7B2%7D%20$

当PA，PB的斜率存在且倾斜角互补时，即 $k_%7BAP%7D%2B%20k_%7BBP%7D%20%3D0$

所以 $%5Cfrac%7B2p%7D%7By_%7B1%7D%20%2By_%7B0%7D%20%7D%20%2B%5Cfrac%7B2p%7D%7By_%7B2%7D%20%2By_%7B0%7D%20%7D%20%3D0$

化简有 $y_%7B1%7D%20%2By_%7B2%7D%20%3D-2y_%7B0%7D%20$

所以 $%5Cfrac%7By_%7B1%7D%20%2By_%7B2%7D%20%7D%7By_%7B0%7D%20%7D%20%3D-2$

$k_%7BAB%7D%20%3D%5Cfrac%7B2p%7D%7B%7By_%7B1%7D%20%2By_%7B2%7D%20%7D%7D%20%3D-%5Cfrac%7Bp%7D%7By_%7B0%7D%20%7D%20$

升级版：

已知抛物线 $%5CGamma%20%20%EF%BC%9Ay%5E2%3D2px%20%EF%BC%88p%EF%BC%9E0%EF%BC%89$ 上三点直线 $A%EF%BC%882%2C2%EF%BC%89%EF%BC%8CB%EF%BC%8CC$ ，直线AC，AB是圆 $%EF%BC%88x-2%EF%BC%89%5E2%2By%5E2%3D1$ 的两条切线，求直线BC的方程

解：

把 $A$ 代进 $%5CGamma%20$ 得 $p%3D1$

设 $B%EF%BC%88x_%7B1%7D%20%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CC%EF%BC%88x_%7B2%7D%20%EF%BC%8Cy_%7B2%7D%20%EF%BC%89$

直线AB： $%EF%BC%88y_%7B1%7D%20%2B2%20%EF%BC%89y%3D2x%2B2y_%7B1%7D%20$

直线AC： $%EF%BC%88y_%7B2%7D%20%2B2%20%EF%BC%89y%3D2x%2B2y_%7B2%7D%20$

直线BC： $%EF%BC%88y_%7B1%7D%20%2By_%7B2%7D%20%EF%BC%89y%3D2x%2By_%7B1%7D%20y_%7B2%7D%20$

直线AB与圆相切，由点到直线距离公式有

$%5Cfrac%7B%5Cvert%204%2B2y_%7B1%7D%20%20%5Cvert%20%7D%7B%5Csqrt%7B2%5E2%2B%20%EF%BC%88y_%7B1%7D%20%2B2%EF%BC%89%5E2%20%7D%20%7D%20%3D1$

化简有 $3y_%7B1%7D%20%5E2%2B12y_%7B1%7D%20%2B8%3D0$

同理 $3y_%7B2%7D%20%5E2%2B12y_%7B2%7D%20%2B8%3D0$

所以

$y_%7B1%7D%20y_%7B2%7D%20%3D%5Cfrac%7B8%7D%7B3%7D%20%EF%BC%8Cy_%7B1%7D%2B%20y_%7B2%7D%3D-4%EF%BC%88%E5%90%8C%E6%9E%84%2B%E9%9F%A6%E8%BE%BE%EF%BC%89$

所以直线BC为 $-4y%3D2x%2B%5Cfrac%7B8%7D%7B3%7D%20$

即 $3x%2B6y%2B4%3D0$

再来个加强版（2021全国甲卷）

（1）设抛物线 $C$ 为 $y%5E2%3D2px%EF%BC%88p%EF%BC%9E0%EF%BC%89$ ，不妨令 $P%EF%BC%881%EF%BC%8C%5Csqrt%7B2p%7D%20%EF%BC%89$

由几何关系（射影定理）

$1%3D%EF%BC%88%5Csqrt%7B2p%7D%20%EF%BC%89%5E2$

解得 $p%3D%5Cfrac%7B1%7D%7B2%7D%20$

抛物线 $C%E4%B8%BA%EF%BC%9Ay%5E2%3Dx$

显然 $%5Codot%20M%E7%9A%84%E6%96%B9%E7%A8%8B%E4%B8%BA%EF%BC%88x-2%EF%BC%89%5E2%2By%5E2%3D1%20$

（2）

设 $A_%7B1%7D%20%EF%BC%88x_%7B1%7D%20%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CA_%7B2%7D%20%EF%BC%88x_%7B2%7D%20%EF%BC%8Cy_%7B2%7D%20%EF%BC%89%2CA_%7B3%7D%20%EF%BC%88x_%7B3%7D%EF%BC%8C%20y_%7B3%7D%20%EF%BC%89$

直线 $A_%7B1%7D%20A_%7B2%7D%20%EF%BC%9A%EF%BC%88y_%7B1%7D%20%2By_%7B2%7D%20%EF%BC%89y%3Dx%2By_%7B1%7D%20y_%7B2%7D%20$

直线 $A_%7B1%7D%20A_%7B3%7D%20%EF%BC%9A%EF%BC%88y_%7B1%7D%20%2By_%7B3%7D%20%EF%BC%89y%3Dx%2By_%7B1%7D%20y_%7B3%7D%20$

直线 $A_%7B2%7D%20A_%7B3%7D%20%EF%BC%9A%EF%BC%88y_%7B2%7D%20%2By_%7B3%7D%20%EF%BC%89y%3Dx%2By_%7B2%7D%20y_%7B3%7D%20$

$%E5%9B%A0%E4%B8%BA%E7%9B%B4%E7%BA%BFA_%7B1%7D%20A_%7B2%7D%20%E4%B8%8E%5Codot%20%20M%E7%9B%B8%E5%88%87$

故 $%5Cfrac%7B%5Cvert%202%2By_%7B1%7D%20y_%7B2%7D%20%20%5Cvert%20%7D%7B%5Csqrt%7B1%2B%EF%BC%88y_%7B1%7D%20%2By_%7B2%7D%20%EF%BC%89%5E2%20%7D%20%7D%20%3D1$

化简有

$%EF%BC%88y_%7B1%7D%5E2%20-1%20%EF%BC%89y_%7B2%7D%5E2%2B2y_%7B1%7D%20y_%7B2%7D%20%2B3-y_%7B1%7D%20%5E2%3D0$ （注意化简要的是 $%20y_%7B2%7D%20$ ）

同理

$%EF%BC%88y_%7B1%7D%5E2%20-1%20%EF%BC%89y_%7B3%7D%5E2%2B2y_%7B1%7D%20y_%7B3%7D%20%2B3-y_%7B1%7D%20%5E2%3D0$

所以

$y_%7B2%7D%20%2By_%7B3%7D%20%3D%5Cfrac%7B-2y_%7B1%7D%20%7D%7By_%7B1%7D%5E2-1%20%20%7D%20%EF%BC%8Cy_%7B2%7D%20y_%7B3%7D%3D%5Cfrac%7B3-y_%7B1%7D%5E2%20%7D%7By_%7B1%7D%5E2-1%20%20%7D$

所以直线 $A_%7B2%7D%20A_%7B3%7D%20$ 方程为 $%5Cfrac%7B-2y_%7B1%7D%20%7D%7By_%7B1%7D%5E2-1%20%20%7D%20y%3Dx%2B%5Cfrac%7B3-y_%7B1%7D%5E2%20%7D%7By_%7B1%7D%5E2-1%20%20%7D$

即 $%EF%BC%88y_%7B1%7D%5E2-1%20%EF%BC%89x%2B2y_%7B1%7Dy%2B3-y_%7B1%7D%5E2%3D0%20$

设直线 $A_%7B2%7D%20A_%7B3%7D%20$ 到 $%5Codot%20M$ 距离为 $d$

所以

$d%3D%5Cfrac%7B%5Cvert%20y_%7B1%7D%5E2%2B1%20%20%5Cvert%20%7D%7B%5Csqrt%7B%EF%BC%88y_%7B1%7D%5E2-1%20%EF%BC%89%5E2%2B4y_%7B1%7D%5E2%20%20%7D%20%7D%20%3D%5Cfrac%7B%5Cvert%20y_%7B1%7D%5E2%2B1%20%20%5Cvert%20%7D%7B%5Csqrt%7B%EF%BC%88y_%7B1%7D%5E2%2B1%20%EF%BC%89%5E2%20%7D%20%7D%3D1$