题目内容
![精英家教网](http://thumb.1010pic.com/pic3/upload/images/201202/3/ac1f3116.png)
4 | x |
分析:作P1B⊥y轴,P1A⊥x轴,根据等腰直角三角形的性质解答即可.
解答:
解:作P1B⊥y轴,P1A⊥x轴,
∵△P1OA1,△P2A1A2是等腰直角三角形,
∴AP1=BP1,A1D=DA2=DP2,
则OA•OB=4,
∴OA=OB=AA1=2,OA1=4,
设A1D=x,则有(4+x)x=4,
解得x=-2+2
,或x=-2-2
(舍去),
则OA2=4+2x=4-4+4
=4
,A2坐标为(4
,0).
![精英家教网](http://thumb.1010pic.com/pic3/upload/images/201202/3/c9a4b940.png)
∵△P1OA1,△P2A1A2是等腰直角三角形,
∴AP1=BP1,A1D=DA2=DP2,
则OA•OB=4,
∴OA=OB=AA1=2,OA1=4,
设A1D=x,则有(4+x)x=4,
解得x=-2+2
2 |
2 |
则OA2=4+2x=4-4+4
2 |
2 |
2 |
点评:本题考查一定经过某点的函数应适合这个点的横纵坐标.
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