（圆锥曲线）广州八月调研考试

2022-08-18 18:08 作者:因恋相思久 0人读过 | 我要投稿

22.(12分）

已知双曲线 $%5CGamma%20%EF%BC%9A%5Cfrac%7Bx%5E2%20%7D%7Ba%5E2%20%7D%20-%5Cfrac%7By%5E2%20%7D%7Bb%5E2%20%7D%20%3D1%EF%BC%88a%EF%BC%8Cb%EF%BC%9E0%EF%BC%89$ 经过双曲线 $%5CGamma%20$ 上的点 $A%EF%BC%882%2C1%EF%BC%89$ 作互相垂直的直线AM，AN分别交双曲线 $%5CGamma%20$ 于M，N两点，设线段AM，AN的中点分别为B，C，直线OB，OC（O为坐标原点）的斜率都存在且它们的乘积为 $-%5Cfrac%7B1%7D%7B4%7D%20$

（1）求双曲线 $%5CGamma%20$ 的方程

（2）过点A作 $AD%E2%8A%A5MN$ （D为垂足），请问：是否存在定点E，使得 $%5Cvert%20DE%20%5Cvert%20$ 为定值，若存在，求出点E的坐标；若不存在，请说明理由

（1）设 $M%EF%BC%88x_%7B1%7D%20%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CN%EF%BC%88x_%7B2%7D%20%EF%BC%8Cy_%7B2%7D%20%EF%BC%89$

因为线段AM，AN的中点分别为B，C

所以 $B%EF%BC%88%5Cfrac%7B2%2Bx_%7B1%7D%20%7D%7B2%7D%EF%BC%8C%20%5Cfrac%7B1%2By_%7B1%7D%20%20%7D%7B2%7D%20%EF%BC%89%EF%BC%8CC%EF%BC%88%5Cfrac%7B2%2Bx_%7B2%7D%20%7D%7B2%7D%EF%BC%8C%20%5Cfrac%7B1%2By_%7B2%7D%20%20%7D%7B2%7D%EF%BC%89$

因此 $k_%7BOB%7D%20k_%7BOC%7D%20%3D%5Cfrac%7B1%2By_%7B1%7D%20%7D%7B2%2Bx_%7B1%7D%20%7D%20%C2%B7%5Cfrac%7B1%2By_%7B2%7D%20%7D%7B2%2Bx_%7B2%7D%20%7D%20%3D-%5Cfrac%7B1%7D%7B4%7D%20$

下面利用中点弦

因为 $%5Cfrac%7Bx_%7B1%7D%5E2%20%20%7D%7Ba%5E2%20%7D%20-%5Cfrac%7By_%7B1%7D%5E2%20%20%7D%7Bb%5E2%20%7D%20%3D1%EF%BC%8C%5Cfrac%7B4%7D%7Ba%5E2%20%7D%20-%5Cfrac%7B1%7D%7Bb%5E2%20%7D%20%3D1%EF%BC%8C%E4%B8%A4%E5%BC%8F%E7%9B%B8%E5%87%8F%E6%9C%89$

$%5Cfrac%7Bb%5E2%7D%7Ba%5E2%7D%3D%20%5Cfrac%7B1%2By_%7B1%7D%20%7D%7B2%2Bx_%7B1%7D%20%7D%20.%5Cfrac%7B1-y_%7B1%7D%20%7D%7B2-x_%7B1%7D%20%7D%3Dk_%7BOB%0A%7D%20k_%7BAM%7D%20%20$

同理可得 $%5Cfrac%7Bb%5E2%7D%7Ba%5E2%7D%3D%20%5Cfrac%7B1%2By_%7B2%7D%20%7D%7B2%2Bx_%7B2%7D%20%20%7D%20.%5Cfrac%7B1-y_%7B2%7D%20%7D%7B2-x_%7B2%7D%20%7D%3Dk_%7BOC%0A%7D%20k_%7BAN%7D%20$

所以 $%5Cfrac%7Bb%5E4%7D%7Ba%5E4%7D%20%3Dk_%7BOB%7D%20k_%7BAM%7D%20%20k_%7BOC%7D%20k_%7BAN%7D%20%20%3D%5Cfrac%7B1%7D%7B4%7D%20%EF%BC%8C%EF%BC%88AM%E2%8A%A5AN%EF%BC%89$

所以 $2b%5E2%3Da%5E2$ 因为 $%5Cfrac%7B4%7D%7Ba%5E2%20%7D%20-%5Cfrac%7B1%7D%7Bb%5E2%20%7D%20%3D1$

解得 $b%5E2%3D1%2Ca%5E2%3D2$

双曲线 $%5CGamma%20%EF%BC%9A%5Cfrac%7Bx%5E2%20%7D%7B2%20%7D%20-%7By%5E2%20%7D%3D1$

（2）

当直线MN斜率存在时

设直线MN： $n%EF%BC%88x-2%EF%BC%89%2Bm%EF%BC%88y-1%EF%BC%89%3D1$

下面进行齐次化

$x%5E2%20-2y%5E2%3D2$

$%E5%8D%B3%EF%BC%88x-2%EF%BC%89%5E2-2%EF%BC%88y-1%EF%BC%89%5E2-4%EF%BC%88y-1%EF%BC%89%2B4%EF%BC%88x-2%EF%BC%89%20%3D0$

$%EF%BC%88x-2%EF%BC%89%5E2-2%EF%BC%88y-1%EF%BC%89%5E2%20-%5B4%EF%BC%88y-1%EF%BC%89%2B4%EF%BC%88x-2%EF%BC%89%5D%5Bn%EF%BC%88x-2%EF%BC%89%2Bm%EF%BC%88y-1%EF%BC%89%5D%3D0$

$%EF%BC%884m%2B2%EF%BC%89%EF%BC%88y-1%EF%BC%89%5E2%2B%EF%BC%884n-4m%EF%BC%89%EF%BC%88x-2%EF%BC%89%EF%BC%88y-1%EF%BC%89-%EF%BC%884n%2B1%EF%BC%89%EF%BC%88x-2%EF%BC%89%5E2%3D0$

$%EF%BC%884m%2B2%EF%BC%89%EF%BC%88%5Cfrac%7By-1%7D%7Bx-2%7D%20%EF%BC%89%5E2%2B%EF%BC%884n-4m%EF%BC%89%EF%BC%88%5Cfrac%7By-1%7D%7Bx-2%7D%20%EF%BC%89-%EF%BC%884n%2B1%EF%BC%89%3D0$

$%E6%89%80%E4%BB%A5k_%7BAM%7D%20k_%7BAN%7D%20%3D-%5Cfrac%7B4n%2B1%7D%7B4m%2B2%7D%20%3D-1$

$%E6%89%80%E4%BB%A5n%3Dm%2B%5Cfrac%7B1%7D%7B4%7D%20$ 带入直线MN并整理有

$m%EF%BC%88x-3%2By%EF%BC%89%2B%5Cfrac%7B1%7D%7B4%7D%20%EF%BC%88x-6%EF%BC%89%3D0$ 过定点 $P(6%2C-3)$

所以AP中点 $E%EF%BC%884%EF%BC%8C-1%EF%BC%89$ ，由于 $AD%E2%8A%A5MN%EF%BC%88D%E4%B8%BA%E5%9E%82%E8%B6%B3%EF%BC%89$ ，故 $%5Cvert%20DE%20%5Cvert%20%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cvert%20AP%20%5Cvert%20%3D2%5Csqrt%7B2%7D%20$ （圆的性质）

斜率不存在时，设直线MN为x=t，此时不妨令 $M%EF%BC%88t%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CN%EF%BC%88t%20%EF%BC%8C-y_%7B1%7D%20%EF%BC%89%EF%BC%88y_%7B1%7D%20%EF%BC%9E1%EF%BC%89$

由几何关系（射影定理）

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

且 $%5Cfrac%7Bt%5E2%7D%7B2%7D%20-1%3Dy_%7B1%7D%20%5E2$ ②

联立①②得到 $4t%5E2-8t%2B10%3D0$

解得 $t_%7B1%7D%20%3D2%EF%BC%88%E8%88%8D%EF%BC%89%EF%BC%8Ct_%7B2%7D%3D6$

$%E5%9B%A0%E4%B8%BAAD%E2%8A%A5MN%EF%BC%88D%E4%B8%BA%E5%9E%82%E8%B6%B3%EF%BC%89$ 故 $D%EF%BC%886%EF%BC%8C1%EF%BC%89$

此时存在 $E%EF%BC%884%EF%BC%8C-1%EF%BC%89$ 使得 $%5Cvert%20DE%20%5Cvert%20%3D%5Csqrt%7B%EF%BC%886-4%EF%BC%89%5E2%2B%EF%BC%881%2B1%EF%BC%89%5E2%7D%20%3D2%5Csqrt%7B2%7D%20$