题目内容
(2004•山西)已知二次函数y=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_ST/0.png)
(1)求这个二次函数的解析式,并在下面的坐标系中画出该二次函数的图象;
(2)设D为线段OC上的一点,满足∠DPC=∠BAC,求点D的坐标;
(3)在x轴上是否存在一点M,使以M为圆心的圆与AC、PC所在的直线及y轴都相切?如果存在,请求出点M的坐标;若不存在,请说明理由.
【答案】分析:(1)利用待定系数法求出b,c的值后可求出该函数的解析式;
(2)证明△DPC∽△BAC,利用线段比求出各相关线段的值后易求点C的坐标;
(3)过M作MH⊥AC,MG⊥PC,推出△SCT是等腰直角三角形,M是△SCT的内切圆圆心,根据直线与圆的关系进行解答.
解答:
解:(1)∵二次函数y=
x2+bx+c的图象过点A(-3,6),B(-1,0),
得
,
解得
.
∴这个二次函数的解析式为:
y=
x2-x-
.(4分)
由解析式可求P(1,-2),C(3,0),(5分)
画出二次函数的图象;(6分)
(2)解法一:
易证:∠ACB=∠PCD=45°,
又已知:∠DPC=∠BAC,
∴△DPC∽△BAC,(8分)
∴
,
易求AC=6
,PC=2
,BC=4,
∴DC=
,
∴OD=3-
,
∴D(
,0).(10分)
解法二:过A作AE⊥x轴,垂足为E,
设抛物线的对称轴交x轴于F,
亦可证△AEB∽△PFD,(8分)
∴
,
易求:AE=6,EB=2,PF=2,
∴FD=
,
∴OD=
+1=
,
∴D(
,0);(10分)
(3)存在.
①过M作MH⊥AC,MG⊥PC垂足分别为H、G,设AC交y轴于S,CP的延长线交y轴于T,
∵△SCT是等腰直角三角形,M是△SCT的内切圆圆心,
∴MG=MH=OM,(11分)
又∵MC=
OM且OM+MC=OC,
∴
OM+OM=3,
得OM=3
-3,
∴M(3
-3,0)(12分)
②在x轴的负半轴上,存在一点M′,
同理OM′+OC=M′C,OM′+OC=
OM′
得OM′=3
+3
∴M′
(14分)
即在x轴上存在满足条件的两个点.
点评:本题综合考查的是二次函数的有关知识以及直线与圆的关系,难度较大.
(2)证明△DPC∽△BAC,利用线段比求出各相关线段的值后易求点C的坐标;
(3)过M作MH⊥AC,MG⊥PC,推出△SCT是等腰直角三角形,M是△SCT的内切圆圆心,根据直线与圆的关系进行解答.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/images0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/0.png)
得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/1.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/2.png)
∴这个二次函数的解析式为:
y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/4.png)
由解析式可求P(1,-2),C(3,0),(5分)
画出二次函数的图象;(6分)
(2)解法一:
易证:∠ACB=∠PCD=45°,
又已知:∠DPC=∠BAC,
∴△DPC∽△BAC,(8分)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/5.png)
易求AC=6
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/7.png)
∴DC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/8.png)
∴OD=3-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/9.png)
∴D(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/10.png)
解法二:过A作AE⊥x轴,垂足为E,
设抛物线的对称轴交x轴于F,
亦可证△AEB∽△PFD,(8分)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/11.png)
易求:AE=6,EB=2,PF=2,
∴FD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/12.png)
∴OD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/14.png)
∴D(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/15.png)
(3)存在.
①过M作MH⊥AC,MG⊥PC垂足分别为H、G,设AC交y轴于S,CP的延长线交y轴于T,
∵△SCT是等腰直角三角形,M是△SCT的内切圆圆心,
∴MG=MH=OM,(11分)
又∵MC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/16.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/17.png)
得OM=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/18.png)
∴M(3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/19.png)
②在x轴的负半轴上,存在一点M′,
同理OM′+OC=M′C,OM′+OC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/20.png)
得OM′=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/21.png)
∴M′
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105811561528069/SYS201310191058115615280027_DA/22.png)
即在x轴上存在满足条件的两个点.
点评:本题综合考查的是二次函数的有关知识以及直线与圆的关系,难度较大.
![](http://thumb2018.1010pic.com/images/loading.gif)
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