【答案】
分析:(1)過(guò)D作x軸垂線,由拋物線的對(duì)稱性可知拋物線與x軸另一交點(diǎn)為(-1,0).再根據(jù)交點(diǎn)式即可求出過(guò)A、D、C三點(diǎn)的拋物線的解析式;
(2)由外接圓知識(shí)知M為對(duì)稱軸與AC中垂線的交點(diǎn).由等腰直角三角形性質(zhì)可得M點(diǎn)的坐標(biāo),連MC得MC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/0.png)
,即為半徑;
(3)由對(duì)稱性可知:當(dāng)ED+EC+FD+FC最小時(shí),E為對(duì)稱軸與AC交點(diǎn),F(xiàn)為BD與y軸交點(diǎn),再根據(jù)待定系數(shù)法求出BD直線解析式,從而得到E,F(xiàn)的坐標(biāo),再根據(jù)兩點(diǎn)坐標(biāo)公式即可求得EF的長(zhǎng);
(4)先求出直線CP的解析式為y=x-3或y=-x+3,再分情況討論求得以P、Q、C為頂點(diǎn)的三角形與△ADC相似時(shí)點(diǎn)P、Q的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/images1.png)
解:(1)由題意知C(3,0)、A(0,3).
如圖1,過(guò)D作x軸垂線,由矩形性質(zhì)得D(2,3).
由拋物線的對(duì)稱性可知拋物線與x軸另一交點(diǎn)為(-1,0).
設(shè)拋物線的解析式為y=a(x+1)(x-3).
將(0,3)代入得a=-1,所以y=-x
2+2x+3.
(2)由外接圓知識(shí)知M為對(duì)稱軸與AC中垂線的交點(diǎn).
由等腰直角三角形性質(zhì)得OM平分∠AOC,即y
OM=x,
∴M(1,1).
連MC得MC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/1.png)
,即半徑為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/2.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/images4.png)
(3)如圖2,由對(duì)稱性可知:當(dāng)ED+EC+FD+FC最小時(shí),E為對(duì)稱軸與AC交點(diǎn),F(xiàn)為BD與y軸交點(diǎn),
∵∠B=45°,∠AOB=90°,
∴AO=BO=3,故B點(diǎn)坐標(biāo)為:(-3,0),
再利用D(2,3),代入y=ax+b,得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/3.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/4.png)
,
故BD直線解析式為:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/5.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/6.png)
,
當(dāng)x=0,y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/7.png)
,根據(jù)對(duì)稱軸為直線x=1,則y=2,
故F(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/8.png)
)、E(1,2),
EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/11.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/images14.png)
(4)可得△ADC中,AD=2,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/12.png)
,DC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/13.png)
.
假設(shè)存在,顯然∠QCP<90°,則∠QCP=45°或∠QCP=∠CAD.
如圖3,當(dāng)∠QCP=45°時(shí),OR=OC=3,
則R點(diǎn)坐標(biāo)為(0,-3),將C,R代入y=ax+b得出:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/14.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/15.png)
,
這時(shí)直線CP的解析式為y=x-3,同理可得另一解析式為:y=-x+3.
當(dāng)直線CP的解析式為y=x-3時(shí),
則x-3=-x
2+2x+3,
解得:x
1=-2,x
2=3,
可求得P(-2,-5),
故PC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/16.png)
=5
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/17.png)
.
設(shè)CQ=x,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/18.png)
,
解得:x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/19.png)
或x=15.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/images23.png)
∴Q (-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/20.png)
,0)或(-12,0).
當(dāng)y=-x+3即P與A重合時(shí),CQ=y,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/22.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/23.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/24.png)
,或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/25.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/26.png)
,
解得CQ=2或9,
故Q (1,0)或(-6,0).
如圖4,當(dāng)∠QCP=∠ACD時(shí),設(shè)CP交y軸于H,連接ED,則ED⊥AC,
∴DE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/27.png)
,EC=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/28.png)
,
易證:△CDE∽△CHQ,
所以
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/29.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/30.png)
,
∴HO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/31.png)
.
可求HC的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/32.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/33.png)
.
聯(lián)解
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/34.png)
,
得P(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/35.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/36.png)
),PC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/37.png)
.
設(shè)CQ=x,知
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/38.png)
,
∴x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/39.png)
或x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/40.png)
,
∴Q(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/41.png)
,0)或(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/42.png)
,0).
同理當(dāng)H在y軸正半軸上時(shí),HC的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/43.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/44.png)
.
∴P’(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/45.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/46.png)
),
∴PC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/47.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/48.png)
,
∴CQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/49.png)
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/50.png)
,所以Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/51.png)
,0)或(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/52.png)
,0).
綜上所述,P
1(-2,-5)、Q
1(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/53.png)
,0)或(-12,0);P
2(0,3)、Q
2(1,0)或(-6,0);P
3(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/54.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/55.png)
)、Q
3(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/56.png)
,0)或(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/57.png)
,0);P
4(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/58.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/59.png)
)、Q
4(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/60.png)
,0)或(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190906720404875/SYS201311011909067204048024_DA/61.png)
,0).
點(diǎn)評(píng):此題主要考查了二次函數(shù)的綜合題型,其中涉及到的知識(shí)點(diǎn)有拋物線的對(duì)稱軸公式和三角函數(shù)關(guān)系等知識(shí),利用三角形三邊關(guān)系得出|TM-TF|是解題關(guān)鍵.