题目内容
(2010•武汉模拟)过定点P(2,1)的直线l交x轴正半轴于A,交y轴正半轴于B,O为坐标原点,则△OAB周长的最小值为( )
分析:作PM⊥x轴于M,PN⊥y轴于N,则ON=2,ON=1.设∠OAB=∠NPB=α,则NB=2tonα,MA=cotα,AP=cscα,PB=2secα.
于是△OAB的周长L=(2+cotα)+(1+2tanα)+(cscα+secα)=6+(cot
-1)+
,由此能够导出L≥6+2
=10.
于是△OAB的周长L=(2+cotα)+(1+2tanα)+(cscα+secα)=6+(cot
| α |
| 2 |
| 4 | ||
cot
|
| 4 |
解答:解:作PM⊥x轴于M,PN⊥y轴于N,则ON=2,ON=1.
设∠OAB=∠NPB=α,则NB=2tonα,MA=cotα,AP=cscα,PB=2secα.
于是△OAB的周长L=(2+cotα)+(1+2tanα)+(cscα+secα)
=3+
+
=3+
+
=3+cot
+
=3+cot
+
=5+cot
+
=6+(cot
-1)+
,
∵α∈(0,
),∴
∈(0,
),cot
-1>0,
∴L≥6+2
=10.
故选B.
设∠OAB=∠NPB=α,则NB=2tonα,MA=cotα,AP=cscα,PB=2secα.
于是△OAB的周长L=(2+cotα)+(1+2tanα)+(cscα+secα)
=3+
| 1+cosα |
| sinα |
| 2(1+sinα) |
| cosα |
=3+
2cos 2
| ||||
2sin
|
2(cos
| ||||
cos2
|
=3+cot
| α |
| 2 |
2(cos
| ||||
cos
|
=3+cot
| α |
| 2 |
2(cos
| ||||||
cos
|
=5+cot
| α |
| 2 |
4sin
| ||||
cos
|
=6+(cot
| α |
| 2 |
| 4 | ||
cot
|
∵α∈(0,
| π |
| 2 |
| α |
| 2 |
| π |
| 4 |
| α |
| 2 |
∴L≥6+2
| 4 |
故选B.
点评:本题考查基本不等式在最值问题中的应用,解题时要认真审题,仔细解答,注意三角函数的合理运用.
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