【答案】
分析:(1)根據(jù)OC、OA的長,可求得∠OCA=∠ACP=60°(折疊的性質(zhì)),∠BCA=∠OAC=30°,由此可判斷出∠PCB的度數(shù).
(2)過P作PQ⊥OA于Q,在Rt△PAQ中,易知PA=OA=3,而∠PAO=2∠PAC=60°,即可求出AQ、PQ的長,進而可得到點P的坐標,將P、A坐標代入拋物線的解析式中,即可得到b、c的值,從而確定拋物線的解析式,然后將C點坐標代入拋物線的解析式中進行驗證即可.
(3)根據(jù)拋物線的解析式易求得C、D、E點的坐標,然后分兩種情況考慮:
①DE是平行四邊形的對角線,由于CD∥x軸,且C在y軸上,若過D作直線CE的平行線,那么此直線與x軸的交點即為M點,而N點即為C點,D、E的坐標已經(jīng)求得,結(jié)合平行四邊形的性質(zhì)即可得到點M的坐標,而C點坐標已知,即可得到N點的坐標;
②DE是平行四邊形的邊,由于A在x軸上,過A作DE的平行線,與y軸的交點即為N點,而M點即為A點;易求得∠DEA的度數(shù),即可得到∠NAO的度數(shù),已知OA的長,通過解直角三角形可求得ON的值,從而確定N點的坐標,而M點與A點重合,其坐標已知;
同理,由于C在y軸上,且CD∥x軸,過C作DE的平行線,也可找到符合條件的M、N點,解法同上.
解答:解:(1)在Rt△OAC中,OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/0.png)
,OC=1,則∠OAC=30°,∠OCA=60°;
根據(jù)折疊的性質(zhì)知:OA=AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/1.png)
,∠ACO=∠ACP=60°;
∵∠BCA=∠OAC=30°,且∠ACP=60°,
∴∠PCB=30°.
(2)過P作PQ⊥OA于Q;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/images2.png)
Rt△PAQ中,∠PAQ=60°,AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/2.png)
;
∴OQ=AQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/3.png)
,PQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/4.png)
,
所以P(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/5.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/6.png)
);
將P、A代入拋物線的解析式中,得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/7.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/8.png)
;
即y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/9.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/10.png)
x+1;
當x=0時,y=1,故C(0,1)在拋物線的圖象上.
(3)①若DE是平行四邊形的對角線,點C在y軸上,CD平行x軸,
∴過點D作DM∥CE交x軸于M,則四邊形EMDC為平行四邊形,
把y=1代入拋物線解析式得點D的坐標為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/11.png)
,1)
把y=0代入拋物線解析式得點E的坐標為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/12.png)
,0)
∴M(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/13.png)
,0);N點即為C點,坐標是(0,1);
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/images15.png)
②若DE是平行四邊形的邊,
過點A作AN∥DE交y軸于N,四邊形DANE是平行四邊形,
∴DE=AN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/15.png)
=2,
∵tan∠EAN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/16.png)
,
∴∠EAN=30°,
∵∠DEA=∠EAN,
∴∠DEA=30°,
∴M(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/17.png)
,0),N(0,-1);
同理過點C作CM∥DE交y軸于N,四邊形CMDE是平行四邊形,
∴M(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233025648648337/SYS201310212330256486483023_DA/18.png)
,0),N(0,1).
點評:此題考查了矩形的性質(zhì)、圖形的翻折變換、二次函數(shù)解析式的確定、平行四邊形的判定和性質(zhì)等知識,同時考查了分類討論的數(shù)學思想,難度較大.