题目内容
设A、B为在双曲线
-
=1(b>a>0)上两点,O为坐标原点.若OA丄OB,则△AOB面 积的最小值为______.
| x2 |
| a2 |
| y2 |
| b2 |
设直线OA的方程为y=kx,则直线OB的方程为y=-
x,
则点A(x1,y1)满足
故x12=
,y12=
,
∴|OA|2=x12+y12=
,同理|OB|2=
,
故|OA|2•|OB|2=
•
=
∵
=
≤
(当且仅当k=±1时,取等号)
∴|OA|2•|OB|2≥
,又b>a>0,
故S△AOB=
|OA||OB|的最小值为
.
| 1 |
| k |
则点A(x1,y1)满足
|
| a2b2 |
| b2-a2k2 |
| k2a2b2 |
| b2-a2k2 |
∴|OA|2=x12+y12=
| (1+k2)a2b2 |
| b2-a2k2 |
| (1+k2)a2b2 |
| k2b2-a2 |
故|OA|2•|OB|2=
| (1+k2)a2b2 |
| b2-a2k2 |
| (1+k2)a2b2 |
| k2b2-a2 |
| (1+k2)2(a2b2)2 |
| -a2b2+(a4+b4)k2-k4a2b2 |
∵
| k2 |
| (k2+1)2 |
| 1 | ||
k2+
|
| 1 |
| 4 |
∴|OA|2•|OB|2≥
| 4a4b4 |
| (b2-a2)2 |
故S△AOB=
| 1 |
| 2 |
| a2b2 |
| b2-a2 |
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