题目内容
P,A,B为双曲线
-
=1上不重合的三点,其中A,B关于原点对称,且直线PA,PB的斜率分别为k1,k2,则k1•k2=______.
| x2 |
| 16 |
| y2 |
| 4 |
∵A,B关于原点对称,∴可设A(x1,y1),则B(-x1,-y1).
设P(x2,y2).
由P,A,B为双曲线
-
=1上不重合的三点,
∴
-
=1,
-
=1,∴
=
.
∴k1•k2=
×
=
=
=
.
故答案为
.
设P(x2,y2).
由P,A,B为双曲线
| x2 |
| 16 |
| y2 |
| 4 |
∴
| ||
| 16 |
| ||
| 4 |
| ||
| 16 |
| ||
| 4 |
| ||||
| 16 |
| ||||
| 4 |
∴k1•k2=
| y2-y1 |
| x2-x1 |
| y2+y1 |
| x2+x1 |
| ||||
|
| 4 |
| 16 |
| 1 |
| 4 |
故答案为
| 1 |
| 4 |
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