由定义得 | AF | = x1 + 1,| BF | = x2 + 1.
从而有 | AF | 2 + | BF | 2-| AB | 2
= (x1 + 1) 2 + (x2 + 1) 2-(x1-x2) 2-(y1-y2) 2
= -2 (x1 + x2)-6 .
- 答案
由定义得 | AF | = x1 + 1,| BF | = x2 + 1.
从而有 | AF | 2 + | BF | 2-| AB | 2
= (x1 + 1) 2 + (x2 + 1) 2-(x1-x2) 2-(y1-y2) 2
= -2 (x1 + x2)-6 .