關于極點極線的一對基礎命題

2022-10-19 16:46 作者:數(shù)學老頑童 0人讀過 | 我要投稿

已知橢圓 $%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1$ （ $a%3Eb%3E0$ ）及其所在平面內(nèi)一點 $P%5Cleft(%20x_0%2Cy_0%20%5Cright)%20$ （不在橢圓上，也不在橢圓中心），過 $P$ 的直線 $l_1$ 與橢圓交于 $A$ 、 $B$ 兩點，與直線 $l_2$ ： $%5Cfrac%7Bx_0x%7D%7Ba%5E2%7D%2B%5Cfrac%7By_0y%7D%7Bb%5E2%7D%3D1$ 交于點 $Q$ ，證明： $%5Cfrac%7B1%7D%7B%5Cleft%7C%20PA%20%5Cright%7C%7D%2B%5Cfrac%7B1%7D%7B%5Cleft%7C%20PB%20%5Cright%7C%7D%3D%5Cfrac%7B2%7D%7B%5Cleft%7C%20PQ%20%5Cright%7C%7D$ .

設 $l_1$ 的參數(shù)方程為 $%5Cbegin%7Bcases%7D%09x%3Dx_0%2Bt%5Ccos%20%20%5Calpha%20%2C%5C%5C%09y%3Dy_0%2Bt%5Csin%20%20%5Calpha%20%2C%5C%5C%5Cend%7Bcases%7D$

（ $t$ 為參數(shù)），

與橢圓聯(lián)立得

$%5Cleft(%20%5Cfrac%7B%5Ccos%20%5E2%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B%5Csin%20%5E2%5Calpha%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20t%5E2%2B%5Cleft(%20%5Cfrac%7B2x_0%5Ccos%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B2y_0%5Csin%20%5Calpha%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20t%2B%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D-1%3D0$

各項同除以 $t%5E2$ ，得

$%5Cleft(%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D-1%20%5Cright)%20%5Cleft(%20%5Cfrac%7B1%7D%7Bt%7D%20%5Cright)%20%5E2%2B%5Cleft(%20%5Cfrac%7B2x_0%5Ccos%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B2y_0%5Csin%20%5Calpha%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20%5Cfrac%7B1%7D%7Bt%7D%2B%5Cleft(%20%5Cfrac%7B%5Ccos%20%5E2%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B%5Csin%20%5E2%5Calpha%7D%7Bb%5E2%7D%20%5Cright)%20%3D0$

依韋達定理可得

$%5Cfrac%7B1%7D%7Bt_1%7D%2B%5Cfrac%7B1%7D%7Bt_2%7D%3D%5Cfrac%7B2%5Cleft(%20%5Cfrac%7Bx_0%5Ccos%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7By_0%5Csin%20%5Calpha%7D%7Bb%5E2%7D%20%5Cright)%7D%7B1-%5Cleft(%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D%5Cright)%7D$ .

聯(lián)立 $l_1$ 與 $l_2$ ，解得

$%5Cfrac%7B1%7D%7Bt_0%7D%3D%5Cfrac%7B%5Cfrac%7Bx_0%5Ccos%20%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7By_0%5Csin%20%20%5Calpha%7D%7Bb%5E2%7D%7D%7B1-%5Cleft(%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D%20%5Cright)%7D$ .

所以

$%5Cfrac%7B1%7D%7B%5Cleft%7C%20PA%20%5Cright%7C%7D%2B%5Cfrac%7B1%7D%7B%5Cleft%7C%20PB%20%5Cright%7C%7D%3D%5Cfrac%7B1%7D%7Bt_1%7D%2B%5Cfrac%7B1%7D%7Bt_2%7D%3D%5Cfrac%7B2%7D%7Bt_0%7D%3D%5Cfrac%7B2%7D%7B%5Cleft%7C%20PQ%20%5Cright%7C%7D$

已知橢圓 $%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1$ （ $a%3Eb%3E0$ ）及其所在平面內(nèi)一點 $P%5Cleft(%20x_0%2Cy_0%20%5Cright)%20$ （不在橢圓上，也不在橢圓中心），過 $P$ 的直線 $l$ 與橢圓交于 $A$ 、 $B$ 兩點，點 $Q$ 滿足： $%5Cfrac%7B1%7D%7B%5Cleft%7C%20PA%20%5Cright%7C%7D%2B%5Cfrac%7B1%7D%7B%5Cleft%7C%20PB%20%5Cright%7C%7D%3D%5Cfrac%7B2%7D%7B%5Cleft%7C%20PQ%20%5Cright%7C%7D$ ，求 $Q$ 的軌跡方程.

設 $l$ 的參數(shù)方程為 $%5Cbegin%7Bcases%7D%09x%3Dx_0%2Bt%5Ccos%20%20%5Calpha%20%2C%5C%5C%09y%3Dy_0%2Bt%5Csin%20%20%5Calpha%20%2C%5C%5C%5Cend%7Bcases%7D$

（ $t$ 為參數(shù)），

與橢圓聯(lián)立得

各項同除以 $t%5E2$ ，得

依韋達定理可得

所以

$%5Cbegin%7Baligned%7D%0A%09%5Cfrac%7B2%7D%7Bt_Q%7D%26%3D%5Cfrac%7B2%7D%7B%5Cleft%7C%20PQ%20%5Cright%7C%7D%3D%5Cfrac%7B1%7D%7B%5Cleft%7C%20PA%20%5Cright%7C%7D%2B%5Cfrac%7B1%7D%7B%5Cleft%7C%20PB%20%5Cright%7C%7D%3D%5Cfrac%7B1%7D%7Bt_1%7D%2B%5Cfrac%7B1%7D%7Bt_2%7D%5C%5C%0A%09%26%3D%5Cfrac%7B2%5Cleft(%20%5Cfrac%7Bx_0%5Ccos%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7By_0%5Csin%20%5Calpha%7D%7Bb%5E2%7D%20%5Cright)%7D%7B1-%5Cleft(%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D%20%5Cright)%7D%5C%5C%0A%5Cend%7Baligned%7D$

所以

$t_Q%3D%5Cfrac%7B1-%5Cleft(%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D%20%5Cright)%7D%7B%5Cfrac%7Bx_0%5Ccos%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7By_0%5Csin%20%5Calpha%7D%7Bb%5E2%7D%7D$ .

所以

$%5Cbegin%7Bcases%7D%09x_Q%3Dx_0%2B%5Cfrac%7B1-%5Cleft(%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D%20%5Cright)%7D%7B%5Cfrac%7Bx_0%5Ccos%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7By_0%5Csin%20%5Calpha%7D%7Bb%5E2%7D%7D%5Ccdot%20%5Ccos%20%20%5Calpha%20%2C%5C%5C%09y_Q%3Dy_0%2B%5Cfrac%7B1-%5Cleft(%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D%20%5Cright)%7D%7B%5Cfrac%7Bx_0%5Ccos%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7By_0%5Csin%20%5Calpha%7D%7Bb%5E2%7D%7D%5Ccdot%20%5Csin%20%20%5Calpha%20%2C%5C%5C%5Cend%7Bcases%7D$