【答案】
分析:(1)根據(jù)點(diǎn)C的坐標(biāo)可求出點(diǎn)F的縱坐標(biāo),結(jié)合題意可得出點(diǎn)F的坐標(biāo),過(guò)點(diǎn)E作EH⊥x軸于點(diǎn)H,利用△AHE∽△AOD,可求出點(diǎn)E的坐標(biāo),從而利用待定系數(shù)法可確定直線EF的解析式,令x=0,可得出點(diǎn)G的坐標(biāo).
(2)延長(zhǎng)HE交CD的延長(zhǎng)線于點(diǎn)M,討論點(diǎn)P的位置,①當(dāng)點(diǎn)P在AB上運(yùn)動(dòng)時(shí),②當(dāng)點(diǎn)P在BC邊上運(yùn)動(dòng)時(shí),③當(dāng)點(diǎn)P在CF上運(yùn)動(dòng)時(shí),分別利用面積相減法可求出答案.
(3)很明顯在BC上存在兩個(gè)點(diǎn)使△PGF為直角三角形,這兩點(diǎn)是通過(guò)①過(guò)點(diǎn)G作GP⊥EF,②過(guò)點(diǎn)F作FP⊥EF得出來(lái)的.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/images0.png)
解:(1)∵C(8,8),DC∥x軸,點(diǎn)F的橫坐標(biāo)為3,
∴OD=CD=8.
∴點(diǎn)F的坐標(biāo)為(3,8),
∵A(-6,0),
∴OA=6,
∴AD=10,
過(guò)點(diǎn)E作EH⊥x軸于點(diǎn)H,
則△AHE∽△AOD.
又∵E為AD的中點(diǎn),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/3.png)
.
∴AH=3,EH=4.
∴OH=3.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/images5.png)
∴點(diǎn)E的坐標(biāo)為(-3,4),
設(shè)過(guò)E、F的直線為y=kx+b,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/4.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/5.png)
∴直線EF為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/6.png)
x+6,
令x=0,則y=6,即點(diǎn)G的坐標(biāo)為(0,6).
(2)延長(zhǎng)HE交CD的延長(zhǎng)線于點(diǎn)M,
則EM=EH=4.
∵DF=3,
∴S
△DEF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/7.png)
×3×4=6,
且S
平行四邊形ABCD=CD•OD=8×8=64.
①當(dāng)點(diǎn)P在AB上運(yùn)動(dòng)時(shí),如圖3,
S=S
平行四邊形ABCD-S
△DEF-S
△APE-S
四邊形PBCF.
∵AP=t,EH=4,
∴S
△APE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/8.png)
×4t=2t,
S
四邊形PBCF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/9.png)
(5+8-t)×8=52-4t.
∴S=64-6-2t-(52-4t),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/images12.png)
即:S=2t+6.
②當(dāng)點(diǎn)P在BC邊上運(yùn)動(dòng)時(shí),
S=S
平行四邊形ABCD-S
△DEF-S
△PCF-S
四邊形ABPE.
過(guò)點(diǎn)P作PN⊥CD于點(diǎn)N.
∵∠C=∠A,sin∠A=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/11.png)
,
∴sin∠C=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/12.png)
.
∵PC=18-t,
∴PN=PC•sin∠C=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/13.png)
(18-t).
∵CF=5,
∴S
△PCF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/14.png)
×5×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/15.png)
(18-t)=36-2t.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/images19.png)
過(guò)點(diǎn)B作BK⊥AD于點(diǎn)K.
∵AB=CD=8,
∴BK=AB•sin∠A=8×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/17.png)
.
∵PB=t-8,
∴S
四邊形ABPE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/18.png)
(t-8+5)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/20.png)
t-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/21.png)
.
∴S=64-6-(36-2t)-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/22.png)
t-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/23.png)
),
即 S=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/24.png)
t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/25.png)
.(8分)
③當(dāng)點(diǎn)P在CF上運(yùn)動(dòng)時(shí),
∵PC=t-18,
∴PF=5-(t-18)=23-t.
∵EM=4,
∴S
△PEF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/26.png)
×4×(23-t)=46-2t.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/images31.png)
綜上:S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/27.png)
(3)存在.
P
1(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/28.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/29.png)
).
P
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/30.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163506721112158/SYS201310221635067211121025_DA/31.png)
).
點(diǎn)評(píng):此題考查了一次函數(shù)的綜合應(yīng)用,綜合了平行四邊形、待定系數(shù)法及直角三角形的性質(zhì),難度較大,關(guān)鍵是仔細(xì)審題,理解每一問(wèn)要求的問(wèn)題,對(duì)于第二問(wèn)要分類(lèi)討論點(diǎn)P的位置,不要遺漏.