（圓錐曲線）廣州八月調(diào)研考試

2022-08-18 18:08 作者:因戀相思久 0人讀過 | 我要投稿

22.(12分）

已知雙曲線 $%5CGamma%20%EF%BC%9A%5Cfrac%7Bx%5E2%20%7D%7Ba%5E2%20%7D%20-%5Cfrac%7By%5E2%20%7D%7Bb%5E2%20%7D%20%3D1%EF%BC%88a%EF%BC%8Cb%EF%BC%9E0%EF%BC%89$ 經(jīng)過雙曲線 $%5CGamma%20$ 上的點(diǎn) $A%EF%BC%882%2C1%EF%BC%89$ 作互相垂直的直線AM，AN分別交雙曲線 $%5CGamma%20$ 于M，N兩點(diǎn)，設(shè)線段AM，AN的中點(diǎn)分別為B，C，直線OB，OC（O為坐標(biāo)原點(diǎn)）的斜率都存在且它們的乘積為 $-%5Cfrac%7B1%7D%7B4%7D%20$

（1）求雙曲線 $%5CGamma%20$ 的方程

（2）過點(diǎn)A作 $AD%E2%8A%A5MN$ （D為垂足），請(qǐng)問：是否存在定點(diǎn)E，使得 $%5Cvert%20DE%20%5Cvert%20$ 為定值，若存在，求出點(diǎn)E的坐標(biāo)；若不存在，請(qǐng)說明理由

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

因?yàn)榫€段AM，AN的中點(diǎn)分別為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 $k_%7BOB%7D%20k_%7BOC%7D%20%3D%5Cfrac%7B1%2By_%7B1%7D%20%7D%7B2%2Bx_%7B1%7D%20%7D%20%C2%B7%5Cfrac%7B1%2By_%7B2%7D%20%7D%7B2%2Bx_%7B2%7D%20%7D%20%3D-%5Cfrac%7B1%7D%7B4%7D%20$

下面利用中點(diǎn)弦

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%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" alt="%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?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Cfrac%7B4%7D%7Ba%5E2%20%7D%20-%5Cfrac%7B1%7D%7Bb%5E2%20%7D%20%3D1" alt="%5Cfrac%7B4%7D%7Ba%5E2%20%7D%20-%5Cfrac%7B1%7D%7Bb%5E2%20%7D%20%3D1"/>

解得 $b%5E2%3D1%2Ca%5E2%3D2$

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

（2）

當(dāng)直線MN斜率存在時(shí)

設(shè)直線MN： $n%EF%BC%88x-2%EF%BC%89%2Bm%EF%BC%88y-1%EF%BC%89%3D1$

下面進(jìn)行齊次化

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

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

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

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

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

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

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

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

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

斜率不存在時(shí)，設(shè)直線MN為x=t，此時(shí)不妨令 $M%EF%BC%88t%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CN%EF%BC%88t%20%EF%BC%8C-y_%7B1%7D%20%EF%BC%89%EF%BC%88y_%7B1%7D%20%EF%BC%9E1%EF%BC%89$

由幾何關(guān)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

聯(lián)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 $D%EF%BC%886%EF%BC%8C1%EF%BC%89$

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