（圓錐曲線）利用兩點式的變換來解決常見問題

2022-08-24 22:59 作者:因戀相思久 0人讀過 | 我要投稿

先來一道開胃菜

拋物線 $C%EF%BC%9Ay%5E2%3D2px%20%EF%BC%88p%EF%BC%9E0%EF%BC%89$ 上一定點 $P%EF%BC%88x_%7B0%7D%EF%BC%8Cy_%7B0%7D%20%EF%BC%89%EF%BC%88y_%7B0%7D%20%EF%BC%9E0%EF%BC%89$ 作兩條直線分別交拋物線于 $A%EF%BC%88x_%7B1%7D%20%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CB%EF%BC%88x_%7B2%7D%20%EF%BC%8Cy_%7B2%7D%20%EF%BC%89$ .

問：當(dāng)PA，PB的斜率存在且傾斜角互補時，求 $%5Cfrac%7By_%7B1%7D%20%2By_%7B2%7D%20%7D%7By_%7B0%7D%20%7D%20$ 的值，并證明直線AB的斜率是非零常數(shù)

解：.

因為直線PA: $%EF%BC%88y_%7B1%7D%20%2By_%7B0%7D%20%EF%BC%89y%3D2px%2By_%7B1%7D%20y_%7B0%7D%20$

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

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

當(dāng)PA，PB的斜率存在且傾斜角互補時，即 $k_%7BAP%7D%2B%20k_%7BBP%7D%20%3D0$

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

化簡有 $y_%7B1%7D%20%2By_%7B2%7D%20%3D-2y_%7B0%7D%20$

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

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

升級版：

已知拋物線 $%5CGamma%20%20%EF%BC%9Ay%5E2%3D2px%20%EF%BC%88p%EF%BC%9E0%EF%BC%89$ 上三點直線 $A%EF%BC%882%2C2%EF%BC%89%EF%BC%8CB%EF%BC%8CC$ ，直線AC，AB是圓 $%EF%BC%88x-2%EF%BC%89%5E2%2By%5E2%3D1$ 的兩條切線，求直線BC的方程

解：

把 $A$ 代進 $%5CGamma%20$ 得 $p%3D1$

設(shè) $B%EF%BC%88x_%7B1%7D%20%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CC%EF%BC%88x_%7B2%7D%20%EF%BC%8Cy_%7B2%7D%20%EF%BC%89$

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

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

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

直線AB與圓相切，由點到直線距離公式有

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

化簡有 $3y_%7B1%7D%20%5E2%2B12y_%7B1%7D%20%2B8%3D0$

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

所以

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

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

即 $3x%2B6y%2B4%3D0$

再來個加強版（2021全國甲卷）

（1）設(shè)拋物線 $C$ 為 $y%5E2%3D2px%EF%BC%88p%EF%BC%9E0%EF%BC%89$ ，不妨令 $P%EF%BC%881%EF%BC%8C%5Csqrt%7B2p%7D%20%EF%BC%89$

由幾何關(guān)系（射影定理）

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

解得 $p%3D%5Cfrac%7B1%7D%7B2%7D%20$

拋物線 $C%E4%B8%BA%EF%BC%9Ay%5E2%3Dx$

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

（2）

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

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

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

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

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

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

化簡有

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

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

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

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

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

設(shè)直線 $A_%7B2%7D%20A_%7B3%7D%20$ 到 $%5Codot%20M$ 距離為 $d$

所以

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