【答案】
分析:(1)將拋物線的解析式進(jìn)行配方,即可得出頂點(diǎn)的坐標(biāo).
(2)由(1)的拋物線解析式不難求出A、B兩點(diǎn)的坐標(biāo),而A、B關(guān)于點(diǎn)G對(duì)稱,由此求得G點(diǎn)的坐標(biāo),進(jìn)而能求出AG、GH、AH的長;然后分兩種情況討論:
①⊙P與⊙G內(nèi)切,且與直線AH相切時(shí);設(shè)⊙P與AH的切點(diǎn)為C,連接CP,由相似三角形:△HPC和△HAG,列出關(guān)于HP、CP、AG、AH的比例關(guān)系式,由此求出點(diǎn)P的坐標(biāo);
②⊙P與⊙G外切,且與直線AH相切時(shí);設(shè)⊙P與AH的切點(diǎn)為D,連接DP,后面的思路和解法同①.
(3)在四邊形GMNA中,只有GA邊是確定的,另外的三邊長都不明確,所以在求四邊形的最小周長時(shí)需要做兩個(gè)對(duì)稱點(diǎn):①作點(diǎn)A關(guān)于直線L的對(duì)稱點(diǎn)A′,②作點(diǎn)G關(guān)于y軸的對(duì)稱點(diǎn)G′;連接A′G′,那么該直線與直線L和y軸的交點(diǎn)即為符合條件的N、M點(diǎn).
解答:解:(1)∵y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/0.png)
(x
2+2x-24)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/1.png)
(x+1)
2+12,
∴頂點(diǎn)H(-1,12).
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/images2.png)
(2)由y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/2.png)
(x
2+2x-24)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/3.png)
(x+6)(x-4)知:A(-6,0)、B(4,0),則:G(-1,0)、E(-1,5);
在Rt△HAG中,AG=GE=5,HG=12,則:AH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/4.png)
=13;
設(shè)PE=x,分兩種情況討論:
①⊙P與⊙G內(nèi)切,且與直線AH相切;
HE=HG-GE=12-5=7,HP1=7+x;
設(shè)⊙P與AH的切點(diǎn)為C,連接CP,如右圖,則有:Rt△HCP
1∽R(shí)t△HGA,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/6.png)
?
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/8.png)
,解得:x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/9.png)
∴GP
1=GE-EP
1=5-x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/10.png)
,P
1(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/11.png)
);
②⊙P與⊙G外切,且與直線AH相切;
設(shè)⊙P與AH的切點(diǎn)為D,同①可知:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/13.png)
?
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/15.png)
,解得:x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/16.png)
∴GP
2=GE+EP
2=5+x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/17.png)
,P
2(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/18.png)
);
綜上,點(diǎn)P的坐標(biāo)為(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/19.png)
)或(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/20.png)
).
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/images22.png)
(3)由題意知,直線L:y=5;
作A(-6,0)關(guān)于直線L的對(duì)稱點(diǎn)A′,則:A′(-6,10);
作G(-1,0)關(guān)于y軸的對(duì)稱點(diǎn)G′,則:G′(1,0);
連接A′G′,則直線A′G′與y軸、直線L的交點(diǎn)為符合條件的M、N點(diǎn);
設(shè)直線A′G′的解析式為:y=kx+b,代入A′、G′兩點(diǎn)的坐標(biāo),有:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/21.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/22.png)
∴直線A′G′:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/23.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/24.png)
;
則:M(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/25.png)
)、N(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/26.png)
,5).
綜上,四邊形GMNA的周長有最小值,此時(shí)M(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/27.png)
)、N(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191950034099810/SYS201311011919500340998021_DA/28.png)
,5).
點(diǎn)評(píng):這道二次函數(shù)綜合題綜合考查了圓與軸對(duì)稱圖形的性質(zhì)等重要知識(shí)點(diǎn);(2)題中,⊙P、⊙G的內(nèi)、外切關(guān)系要分開進(jìn)行討論,連接切點(diǎn)作出相似三角形也是重要的解題思路;最后一題中,根據(jù)軸對(duì)稱圖形的性質(zhì)以及兩點(diǎn)間線段最短作出兩個(gè)對(duì)稱點(diǎn)是解答題目的關(guān)鍵所在.