【答案】
分析:(1)根據(jù)拋物線的解析式,易求得A、B的坐標,利用勾股定理即可求得AB的長.
(2)首先根據(jù)A、B的坐標,求出直線AB的解析式,設(shè)出點P的橫坐標,利用直線AB的解析式,即可表示出P點的縱坐標,由此可得到MP、OM、PN的長,從而證得OM=PN,而∠OPC=90°,則∠OPM、∠PCN同為∠CPN的余角,再加上一組直角,即可由AAS判定△OPM≌△PCN,由此得證.
(3)由(2)的全等三角形知PM=CN,由此可求得BC的表達式,OB的長易求得,根據(jù)三角形的面積公式即可得到S、m的函數(shù)關(guān)系式.(需注意的是,自變量的取值范圍會影響到PM的表達式,因此要分類討論)
(4)此題應(yīng)分三種情況討論:
①P為等腰三角形的頂角頂點,由于∠PBN=45°,若PC=PB,那么CP⊥PB,顯然不符合題意;
②C為等腰三角形的頂角頂點,此時PC=BC,由于△OAB是等腰直角三角形,因此P、A重合時,△PCB也是等腰直角三角形,故A點符合點P的要求;
③B為等腰三角形的頂角頂點,此時PB=BC;當C點在第一象限時,顯然不存在這樣的P點,故此時C點必在第四象限,首先設(shè)出點P的坐標,表示出AP、PB、BC的長,根據(jù)所得等量關(guān)系,即可得到點P的坐標.
解答:解:(1)拋物線y=-2x
2+x+1中,令x=0,得y=1,令y=0,得x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/0.png)
,x=1;
故A(0,1),B(1,0);
∴AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/1.png)
.(2分)
(2)∵A(0,1),B(1,0),
∴直線AB:y=-x+1;
設(shè)P(a,-a+1),則有:
PM=a,OM=1-a,PN=MN-PM=1-a,
故OM=PN;
∵∠OPC=90°,則∠OPM+∠CPN=∠CPN+∠PCN=90°,
∴∠OPM=∠PCN;
又∵∠OMP=∠CPN=90°,OM=PN,
∴△OPM≌△PCN,
∴OP=CP.(5分)
(3)易知OA=OB=1,則∠OBA=∠OAB=45°;
若AP=m,則PM=AM=CN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/2.png)
m,OM=BN=1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/3.png)
m,
①當0<m<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/4.png)
時,BC=BN-NC=1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/5.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/6.png)
=1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/7.png)
m,
故S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/8.png)
;(7分)
②當
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/9.png)
<m<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/10.png)
時,BC=CN-BN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/11.png)
-(1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/12.png)
)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/13.png)
m-1,
故S=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/14.png)
.(9分)
(4)假設(shè)存在符合條件的P點;
①△PCB以P為頂角頂點,此時點C位于第一象限;
由于∠PBN=45°,若PC=PB,則∠CPB=90°,顯然不合題意;
②△PCB以C為頂角頂點;
由于△OAB是等腰直角三角形,當P、A重合時,△PCB也是等腰直角三角形,
故A點符合P點的要求,即P(0,1);
③△PBC以B為頂角頂點;
當C點在第一象限時,PB>BC,若PB=BC,則C點必在第四象限;
設(shè)P(a,1-a),則AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/15.png)
a,PB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/16.png)
(1-a),BC=CN-BN=a-(1-a)=2a-1;
若PB=BC,則2a-1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/17.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/18.png)
a,
解得a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/19.png)
,
故P(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/20.png)
,1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/21.png)
);
綜上所述,存在符合條件的P點,且坐標為P(0,1)或(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/22.png)
,1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163747696065226/SYS201310221637476960652024_DA/23.png)
).(12分)
點評:此題主要考查了拋物線與坐標軸交點坐標的求法、全等三角形的判定、等腰直角三角形的性質(zhì)、等腰三角形的構(gòu)成條件等知識.(4)題中,由于等腰三角形的腰和底不確定,一定要分類討論,以免漏解.