【答案】
分析:(1)可在直角三角形BOA中,根據(jù)AB的長和∠AOB的度數(shù),求出OA的長.根據(jù)折疊的性質(zhì)可知:OC=OA,∠COA=60°,過C作x軸的垂線,即可用三角形函數(shù)求出C點(diǎn)的坐標(biāo);
(2)根據(jù)(1)求出的A,C點(diǎn)的坐標(biāo),用待定系數(shù)法即可求出拋物線的解析式;
(3)根據(jù)等腰梯形的性質(zhì),如果過M,P兩點(diǎn)分別作底的垂線ME和PQ,那么CE=PQ,可先設(shè)出此時(shí)P點(diǎn)的坐標(biāo),然后表示出M點(diǎn)的坐標(biāo),CE就是C點(diǎn)縱坐標(biāo)與M點(diǎn)縱坐標(biāo)的差,QD就是P點(diǎn)縱坐標(biāo)和D點(diǎn)縱坐標(biāo)的差.由此可得出關(guān)于P點(diǎn)橫坐標(biāo)的方程,可求出P點(diǎn)的橫坐標(biāo),進(jìn)而可求出P點(diǎn)的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/images0.png)
解:(1)過點(diǎn)C作CH⊥x軸,垂足為H
∵在Rt△OAB中,∠OAB=90°,∠BOA=30°,AB=2
∴OB=4,OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/0.png)
由折疊知,∠COB=30°,OC=OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/1.png)
∴∠COH=60°,OH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/2.png)
,CH=3
∴C點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/3.png)
,3);
(2)∵拋物線y=ax
2+bx(a≠0)經(jīng)過C(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/4.png)
,3)、A(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/5.png)
,0)兩點(diǎn),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/6.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/7.png)
,
∴此拋物線的解析式為:y=-x
2+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/8.png)
x.
解法一:(3)存在.
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/9.png">的頂點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/10.png)
,3)
所以頂點(diǎn)坐標(biāo)為點(diǎn)C(8分)
作MP⊥x軸,垂足為N,
設(shè)PN=t,因?yàn)椤螧OA=30°,
所以O(shè)N=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/11.png)
t
∴P(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/12.png)
t,t)(9分)
作PQ⊥CD,垂足為Q,ME⊥CD,垂足為E
把
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/13.png)
t代入
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/14.png)
得:y=-3t
2+6t
∴M(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/15.png)
t,-3t
2+6t),E(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/16.png)
,-3t
2+6t)(10分)
同理:Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/17.png)
,t),D(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/18.png)
,1)
要使四邊形CDPM為等腰梯形,只需CE=QD(這時(shí)△PQD≌△MEC)
即3-(-3t
2+6t)=t-1,解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/19.png)
,t
2=1(不合題意,舍去)(11分)
∴P點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/20.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/21.png)
)(12分)
∴存在滿足條件的點(diǎn)P,使得四邊形CDPM為等腰梯形,此時(shí)P點(diǎn)的坐為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/23.png)
);
解法二:
(3)存在.
由(2)可得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/24.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/25.png)
得頂點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/26.png)
,3),
即點(diǎn)C恰好為頂點(diǎn);(8分)
設(shè)MP交x軸于點(diǎn)N,
∵M(jìn)P∥y軸,CH為拋物線的對(duì)稱軸
∴MP∥CD且CM與DP不平行
∴四邊形CDPM為梯形
若要使四邊形CDPM為等腰梯形,只需∠MCD=∠PDC
由∠PDC=∠ODH=90°-∠DOA=60°,則∠MCD=60°
又∵∠BCD=90°-∠OCH=60°,
∴∠MCD=∠BCD,
∴此時(shí)點(diǎn)M為拋物線與線段CB所在直線的交點(diǎn)(9分)
設(shè)BC的解析式為y=mx+n
由(2)得C(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/27.png)
,3)、B(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/28.png)
,2)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/29.png)
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/30.png)
∴直線BC的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/31.png)
(10分)
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/32.png)
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/33.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/34.png)
∴ON=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/35.png)
(11分)
在Rt△OPN中,tan∠PON=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/36.png)
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/37.png)
∴P點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/38.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/39.png)
)(12分)
∴存在滿足條件的點(diǎn)P,使得四邊形CDPM為等腰梯形,此時(shí)P點(diǎn)的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/40.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211103849369143088/SYS201312111038493691430024_DA/41.png)
).
點(diǎn)評(píng):本題著重考查了待定系數(shù)法求二次函數(shù)解析式、圖形翻折變換、三角形全等、等腰梯形的性質(zhì)等重要知識(shí)點(diǎn),綜合性強(qiáng),考查學(xué)生數(shù)形結(jié)合的數(shù)學(xué)思想方法.