题目内容
梯形ABCD按如图所示放置在直角坐标系中(如图a),AB在x轴上,点D在y轴上,CD∥AB,A(-1,0),C(1,3),抛物线![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_ST/0.png)
(1)求抛物线的解析式与线段BC的长度
(2)当t为何值时,△PHG与△AOD相似(点P与点A对应)?
(3)如图(b),连接AC交y轴于点E,动点Q从点B沿BC以每秒1个单位的速度向终点C运动,设点P、Q同时出发,若其中有一点到达终点,则另一点也立即停止运动.
①请探索:是否存在某一时刻t,使△OPQ是以OP为腰的等腰三角形?若存在,求出此时t的值;若不存在,请说明理由.
②如图(c),连接BD交PQ于F,当t=______
【答案】分析:(1)∵CD∥AB,C(1,3),就可以求出D点的坐标,然后把B、C的坐标代入解析式就可以求出抛物线的解析式.
(2)根据(1)的解析式可以求出顶点G的坐标,从而求出GH,OH进而求出AH的值.利用三角形相似就可以求出PH的值,求出OP的值求出t的值.
(3)①利用等腰三角形的性质,根据3中不同的位置情况,由相似三角形的性质可以求出t的值.
②通过作辅助线证明三角形相似,利用相似三角形的性质对应边成比例可以求出t的值.
解答:![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/images0.png)
解:(1)∵C(1,3),CD∥AB,
∴D(0,3),
∵A(-1,0),
∴![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/0.png)
解得
,
抛物线的解析式为:y=-
.
当y=0时,
,
解得:x1=-1,x2=5.
过点C作CM⊥AB于M,则CM=DO=3,BM=4,在Rt△MCB中,由勾股定理,得
BC=
=5
(2)∵y=-
.
∴y=-![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/7.png)
∴G(2,
)
∴HG=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/9.png)
当△PHG∽△AOD时,
,
∴![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/11.png)
∴PH=1.8
∴OP=0.2或OP=3.8,
∴当t=0.2或3.8时,△PHG∽△AOD.
(3)①存在
过点Q作QN⊥AB于N,
∴△BQN∽△BCM
∴得,QN=
t,BN=
t
OQ=OP时,OQ=OP=BQ=t,
∴BN=ON=
t,
∴OB=
=5,
∴t=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/16.png)
当OP=PQ时,OP=PQ=BQ=t,
∴MN=PN=
t,
∴t+
=5,
∴t=
,
当t=5时,OP=PQ,成立
∴t=
、
或5时△OPQ是以OP为腰的等腰三角形.
②分别过点QN⊥AB、FR⊥AB,垂足为N、R.
∴FR∥QN∥OD
∴
,
∴FR=1,BR=
,PR=
,PN=5-![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/25.png)
∵FR∥QN,
∴△PRF∽△PNQ
∴
,
∴
,
解得:t=
,
∵t=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/29.png)
故答案为:
.
点评:本题是一道二次函数的综合试题,考查了待定系数法求函数的解析式,勾股定理的运用,等腰三角形的性质,相似三角形的判定及性质.
(2)根据(1)的解析式可以求出顶点G的坐标,从而求出GH,OH进而求出AH的值.利用三角形相似就可以求出PH的值,求出OP的值求出t的值.
(3)①利用等腰三角形的性质,根据3中不同的位置情况,由相似三角形的性质可以求出t的值.
②通过作辅助线证明三角形相似,利用相似三角形的性质对应边成比例可以求出t的值.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/images0.png)
解:(1)∵C(1,3),CD∥AB,
∴D(0,3),
∵A(-1,0),
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/0.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/1.png)
抛物线的解析式为:y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/2.png)
当y=0时,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/3.png)
解得:x1=-1,x2=5.
过点C作CM⊥AB于M,则CM=DO=3,BM=4,在Rt△MCB中,由勾股定理,得
BC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/4.png)
(2)∵y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/5.png)
∴y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/7.png)
∴G(2,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/8.png)
∴HG=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/9.png)
当△PHG∽△AOD时,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/10.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/11.png)
∴PH=1.8
∴OP=0.2或OP=3.8,
∴当t=0.2或3.8时,△PHG∽△AOD.
(3)①存在
过点Q作QN⊥AB于N,
∴△BQN∽△BCM
∴得,QN=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/12.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/13.png)
OQ=OP时,OQ=OP=BQ=t,
∴BN=ON=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/14.png)
∴OB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/15.png)
∴t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/16.png)
当OP=PQ时,OP=PQ=BQ=t,
∴MN=PN=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/17.png)
∴t+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/18.png)
∴t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/19.png)
当t=5时,OP=PQ,成立
∴t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/20.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/21.png)
②分别过点QN⊥AB、FR⊥AB,垂足为N、R.
∴FR∥QN∥OD
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/22.png)
∴FR=1,BR=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/23.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/24.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/25.png)
∵FR∥QN,
∴△PRF∽△PNQ
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/26.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/27.png)
解得:t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/28.png)
∵t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/29.png)
故答案为:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022163259912578557/SYS201310221632599125785023_DA/30.png)
点评:本题是一道二次函数的综合试题,考查了待定系数法求函数的解析式,勾股定理的运用,等腰三角形的性质,相似三角形的判定及性质.
![](http://thumb2018.1010pic.com/images/loading.gif)
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