题目内容
(2010•宣武区一模)如图,在第一象限内作与x轴的夹角为30°的射线OC,在射线OC上取一点A,过点A作AH⊥x轴于点H.在抛物线y=x2(x>0)上取一点P,在y轴上取一点Q,使得以P,O,Q为顶点的三角形与△AOH全等,则符合条件的点A的坐标是 .![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_ST/images0.png)
【答案】分析:此题应分四种情况考虑:
①∠POQ=∠OAH=60°,此时A、P重合,可联立直线OA和抛物线的解析式,即可得A点坐标;
②∠POQ=∠AOH=30°,此时∠POH=60°,即直线OP:y=
x,联立抛物线的解析式可得P点坐标,进而可求出OQ、PQ的长,由于△POQ≌△AOH,那么OH=OQ、AH=PQ,由此得到点A的坐标.
③当∠OPQ=90°,∠POQ=∠AOH=30°时,此时△QOP≌△AOH;
④当∠OPQ=90°,∠POQ=∠OAH=60°,此时△OQP≌△AOH;
解答:
解:①当∠POQ=∠OAH=60°,若以P,O,Q为顶点的三角形与△AOH全等,那么A、P重合;
由于∠AOH=30°,
所以直线OA:y=
x,联立抛物线的解析式,
得:
,
解得
,
;
故A(
,
);
②当∠POQ=∠AOH=30°,此时△POQ≌△AOH;
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/images8.png)
易知∠POH=60°,则直线OP:y=
x,联立抛物线的解析式,
得:
,
解得
,
;
故P(
,3),那么A(3,
);
③当∠OPQ=90°,∠POQ=∠AOH=30°时,此时△QOP≌△AOH;
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/images15.png)
易知∠POH=60°,则直线OP:y=
x,联立抛物线的解析式,
得:
,
解得
、
,
故P(
,3),
∴OP=2
,QP=2,
∴OH=OP=2
,AH=QP=2,
故A(2
,2);
④当∠OPQ=90°,∠POQ=∠OAH=60°,此时△OQP≌△AOH;
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/images24.png)
此时直线OP:y=
x,联立抛物线的解析式,
得:
,
解得
、
,
∴P(
,
),
∴QP=
,OP=
,
∴OH=QP,QP=
,AH=OP=
,
故A(
,
).
综上可知:符合条件的点A有四个,且坐标为:则符合条件的点A的坐标是 (
,
)或(3,
)或(2
,2)或(
,
).
点评:此题主要考查的是全等三角形的判定和性质以及函数图象交点坐标的求法;由于全等三角形的对应顶点不明确,因此要注意分类讨论思想的运用.
①∠POQ=∠OAH=60°,此时A、P重合,可联立直线OA和抛物线的解析式,即可得A点坐标;
②∠POQ=∠AOH=30°,此时∠POH=60°,即直线OP:y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/0.png)
③当∠OPQ=90°,∠POQ=∠AOH=30°时,此时△QOP≌△AOH;
④当∠OPQ=90°,∠POQ=∠OAH=60°,此时△OQP≌△AOH;
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/images1.png)
由于∠AOH=30°,
所以直线OA:y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/1.png)
得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/2.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/4.png)
故A(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/6.png)
②当∠POQ=∠AOH=30°,此时△POQ≌△AOH;
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/images8.png)
易知∠POH=60°,则直线OP:y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/7.png)
得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/8.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/10.png)
故P(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/12.png)
③当∠OPQ=90°,∠POQ=∠AOH=30°时,此时△QOP≌△AOH;
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/images15.png)
易知∠POH=60°,则直线OP:y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/13.png)
得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/14.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/15.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/16.png)
故P(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/17.png)
∴OP=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/18.png)
∴OH=OP=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/19.png)
故A(2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/20.png)
④当∠OPQ=90°,∠POQ=∠OAH=60°,此时△OQP≌△AOH;
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/images24.png)
此时直线OP:y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/21.png)
得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/22.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/23.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/24.png)
∴P(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/25.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/26.png)
∴QP=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/27.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/28.png)
∴OH=QP,QP=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/29.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/30.png)
故A(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/31.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/32.png)
综上可知:符合条件的点A有四个,且坐标为:则符合条件的点A的坐标是 (
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/33.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/34.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/35.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/36.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/37.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859210094965/SYS201310212328592100949017_DA/38.png)
点评:此题主要考查的是全等三角形的判定和性质以及函数图象交点坐标的求法;由于全等三角形的对应顶点不明确,因此要注意分类讨论思想的运用.
![](http://thumb2018.1010pic.com/images/loading.gif)
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