【答案】
分析:(1)直線AC的解析式中,令x=0,即可求出C點(diǎn)的坐標(biāo).
(2)已知了拋物線圖象上的三點(diǎn)坐標(biāo),可利用待定系數(shù)法求得該拋物線的解析式,進(jìn)而可用公式法或配方法求出拋物線的對(duì)稱軸方程.
(3)設(shè)當(dāng)直線與圓開始有交點(diǎn)時(shí),此圓為⊙P
1,直線與與圓開始沒有交點(diǎn)時(shí),圓為⊙P
2,那么欲求時(shí)間就必須求出PP
1、PP
2的值,設(shè)拋物線的對(duì)稱軸與x軸交于點(diǎn)M,與直線AC交于點(diǎn)N;易求得直線AC的解析式,聯(lián)立拋物線的解析式可求得點(diǎn)N的坐標(biāo),即可得MN的長(zhǎng),設(shè)⊙P
1、⊙P
2與直線AC的切點(diǎn)分別為D、E,易證得△NDP
1∽△COA,根據(jù)相似三角形的比例線段即可求得P
1N的值,從而由P
1M=MN-NP
1求得點(diǎn)P
1的坐標(biāo),同理可求得點(diǎn)P
2的坐標(biāo),已知了P點(diǎn)的縱坐標(biāo),即可求得PP
1、PP
2的長(zhǎng),由此得解.
解答:解:(1)令x=0,y=-4,
∴點(diǎn)C的坐標(biāo)為(0,-4).
(2)設(shè)過A(-3,0),B(1,0),C(0,-4)的函數(shù)解析式為:y=a(x+3)(x-1),
則有:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/images0.png)
a(0+3)(0-1)=-4,
即a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/0.png)
,
∴拋物線的解析式為:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/1.png)
(x+3)(x-1)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/2.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/3.png)
x-4,
對(duì)稱軸為x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/4.png)
,即x=-1.
(3)在Rt△MOC中,OA=3,OC=4,
∴CA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/6.png)
=5,
當(dāng)⊙P向上移動(dòng)時(shí),永遠(yuǎn)不會(huì)與直線AC由公共點(diǎn);
當(dāng)⊙P向下移動(dòng)時(shí),設(shè)⊙P與直線AC有一個(gè)公共點(diǎn)的位置如圖中的⊙P
1和⊙P
2;
⊙P
1與直線AC相切于點(diǎn)D,⊙P
2與直線AC相切于點(diǎn)E,連接P
1D;
則∠NDP
1=90°,又∵M(jìn)N∥OC,∴∠DNP
1=∠ACO;
又∵∠NDP
1=∠COA=90°,∴△NDP
1∽△COA,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/7.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/9.png)
,NP
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/10.png)
;
同理NP
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/11.png)
,把A(-3,0)代入y=kx-4中,-3k-4=0得k=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/12.png)
;
∴直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/13.png)
x-4,把x=-1代入上式,得y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/14.png)
;
∴MN=|-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/15.png)
|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/16.png)
,
∴MP
1=MN-NP
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/17.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/18.png)
=1,
∴PP
1=PM+MP
1=5+1=6;
PP
2=PP
1+2NP
1=6+2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/19.png)
=9
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/20.png)
,t
P→P1=6÷1=6(秒),t
P→P2=9
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/21.png)
÷1=9
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/22.png)
(秒);
綜上所述,經(jīng)過6秒⊙P與直線AC開始有公共點(diǎn),經(jīng)過9
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935727457/SYS201311031015529357274006_DA/23.png)
秒后,⊙P與直線AC不再有公共點(diǎn).
點(diǎn)評(píng):此題主要考查了函數(shù)圖象與坐標(biāo)軸交點(diǎn)坐標(biāo)的求法、二次函數(shù)解析式的確定、直線與圓的位置關(guān)系、切線的性質(zhì)等知識(shí),綜合性強(qiáng),難度較大.