【答案】
分析:(1)根據(jù)折疊的條件得到EO=EF,在直角△CEF中,斜邊大于直角邊,因而EF>EC故EO>EC
(2)四邊形CFGH與四邊形CNMO的面積可以用直角△CEF的面積,可以證明四邊形CFGH與四邊形CNMO的面積相等.因而就可以求出m的值.
(3)已知OC=1,可以得到C點(diǎn)的坐標(biāo)是(0,1),易證△EFQ是等邊三角形,已知QF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/0.png)
就可以求出Q點(diǎn)的坐標(biāo),把C,Q點(diǎn)的坐標(biāo)代入函數(shù)y=mx
2+bx+c,就可以求出b,c的值,就可以得到函數(shù)的解析式.
(4)過Q作y軸的垂線,已知E,Q點(diǎn)的坐標(biāo),可以根據(jù)三角形相似,求出OA的長,就可以求出P點(diǎn)的橫坐標(biāo),進(jìn)而求出P點(diǎn)的坐標(biāo).
若△PBK與△AEF相似,根據(jù)相似三角形的對(duì)應(yīng)邊的比相等,可以求出BK的值,即得到K的坐標(biāo).
解答:解:(1)EO>EC,理由如下:
由折疊知,EO=EF,在Rt△EFC中,EF為斜邊,
∴EF>EC,
故EO>EC.
(2)m為定值,理由如下:
∵S
四邊形CFGH=CF
2=EF
2-EC
2=EO
2-EC
2=(EO+EC)(EO-EC)=CO•(EO-EC),
S
四邊形CMNO=CM•CO=|CE-EO|•CO=(EO-EC)•CO,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/1.png)
.
(3)∵CO=1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/2.png)
,
∴EF=EO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/3.png)
,
∴cos∠FEC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/4.png)
,
∴∠FEC=60°,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/5.png)
,
∴△EFQ為等邊三角形,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/6.png)
.
作QI⊥EO于I,EI=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/7.png)
,IQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/8.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/images9.png)
∴IO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/9.png)
,
∴Q點(diǎn)坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/10.png)
.
∵拋物線y=mx
2+bx+c過點(diǎn)C(0,1),Q
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/11.png)
,m=1,
∴可求得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/12.png)
,c=1,
∴拋物線解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/13.png)
.
(4)由(3),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/14.png)
,
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/15.png)
時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/16.png)
<AB,
∴P點(diǎn)坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/images19.png)
∴BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/18.png)
AO.
方法1:若△PBK與△AEF相似,而△AEF≌△AEO,則分情況如下:
①
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/19.png)
時(shí),BK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/20.png)
,
∴K點(diǎn)坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/21.png)
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/22.png)
;
②
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/23.png)
時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/24.png)
,
∴K點(diǎn)坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/25.png)
或(0,1).
故直線KP與y軸交點(diǎn)T的坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/26.png)
.
方法2:若△BPK與△AEF相似,由(3)得:∠BPK=30°或60°.
過P作PR⊥y軸于R,則∠RTP=60°或30°.
①當(dāng)∠RTP=30°時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/27.png)
,
②當(dāng)∠RTP=60°時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/28.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329168014/SYS201311031015583291680019_DA/29.png)
.
點(diǎn)評(píng):本題主要考查了待定系數(shù)法求函數(shù)解析式,以及相似三角形的性質(zhì),相似三角形的對(duì)應(yīng)邊的比相等.