【答案】
分析:(1)將A、C兩點坐標代入拋物線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/0.png)
x
2+bx+c,運用待定系數(shù)法即可求出b,c的值;
(2)先由兩角對應(yīng)相等的兩三角形相似證明△AOP∽△PEB,再根據(jù)相似三角形對應(yīng)邊的比相等得到
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/2.png)
=2,則PE=2,進而求出點D的坐標,然后將D(t+2,4)代入(1)中求出的拋物線的解析式,即可求出t的值;
(3)由于t=8時,點B與點D重合,△ABD不存在,所以分0<t<8和t>8兩種情況進行討論,在每一種情況下,當(dāng)以A、B、D為頂點的三角形與△AOP相似時,又分兩種情況:△POA∽△ADB與△POA∽△BDA,根據(jù)相似三角形對應(yīng)邊的比相等列出比例式,求解即可;
(4)設(shè)BP的中點為N,由P(t,0),B(t+2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/3.png)
),根據(jù)中點坐標公式得出N(t+1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/4.png)
),由勾股定理求出AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/5.png)
.過點N作FN∥AC交y軸于點F,過點F作FH⊥AC于點H.運用待定系數(shù)法求出AC的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/6.png)
x+4,根據(jù)解析式平移的規(guī)律設(shè)FN的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/7.png)
x+m,將N(t+1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/8.png)
)代入,得出m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/9.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/10.png)
.由△AFH∽△ACO,根據(jù)相似三角形對應(yīng)邊的比相等得出FH=2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/11.png)
,又當(dāng)以PB為直徑的圓與直線AC相切時,F(xiàn)H=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/12.png)
BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/13.png)
AP,列出方程2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/15.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/16.png)
,解方程即可求出t的值.
解答:解:(1)∵拋物線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/17.png)
x
2+bx+c過點A(0,4)和C(8,0),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/18.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/19.png)
.
故所求b,c的值分別為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/20.png)
,4;
(2)∵∠AOP=∠PEB=90°,∠OAP=∠EPB=90°-∠APO,
∴△AOP∽△PEB且相似比為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/22.png)
=2,
∵AO=4,
∴PE=2,OE=OP+PE=t+2,
又∵DE=OA=4,
∴點D的坐標為(t+2,4),
∴點D落在拋物線上時,有-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/23.png)
(t+2)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/24.png)
(t+2)+4=4,
解得t=3或t=-2,
∵t>0,
∴t=3.
故當(dāng)t為3時,點D落在拋物線上;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/images25.png)
(3)存在t,能夠使得以A、B、D為頂點的三角形與△AOP相似,理由如下:
①當(dāng)0<t<8時,如圖1.
若△POA∽△ADB,則PO:AD=AO:BD,
即t:(t+2)=4:(4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/25.png)
t),
整理,得t
2+16=0,
∴t無解;
若△POA∽△BDA,同理,解得t=-2±2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/26.png)
(負值舍去);
②當(dāng)t>8時,如圖3.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/images28.png)
若△POA∽△ADB,則PO:AD=AO:BD,
即t:(t+2)=4:(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/27.png)
t-4),
解得t=8±4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/28.png)
(負值舍去);
若△POA∽△BDA,同理,解得t無解;
綜上可知,當(dāng)t=-2+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/29.png)
或8+4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/30.png)
時,以A、B、D為頂點的三角形與△AOP相似;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/images33.png)
(4)如圖2.∵A(0,4),C(8,0),
∴AC的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/31.png)
x+4.
設(shè)BP的中點為N,由P(t,0),B(t+2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/32.png)
),可得N(t+1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/33.png)
),AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/34.png)
.
過點N作FN∥AC交y軸于點F,過點F作FH⊥AC于點H,
設(shè)直線FN的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/35.png)
x+m,將N(t+1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/36.png)
)代入,
可得-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/37.png)
(t+1)+m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/38.png)
,即m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/39.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/40.png)
.
由△AFH∽△ACO,可得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/41.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/42.png)
,
∵AF=4-m,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/43.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/44.png)
,
∴FH=2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/45.png)
,
當(dāng)以PB為直徑的圓與直線AC相切時,F(xiàn)H=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/46.png)
BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/47.png)
AP,
∴2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/48.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/49.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/50.png)
,
將m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/51.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/52.png)
代入,整理得:31t
2-336t+704=0,
解得:t=8,t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/53.png)
.
故以PB為直徑的圓與直線AC相切時,t的值為8或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/54.png)
.
點評:本題考查了運用待定系數(shù)法求二次函數(shù)、一次函數(shù)的解析式,二次函數(shù)圖象上點的坐標特征,旋轉(zhuǎn)的性質(zhì),相似三角形的判定與性質(zhì),勾股定理,切線的性質(zhì)等知識,綜合性較強,難度較大.由相似三角形的判定與性質(zhì)求出點D的坐標是解決(2)小題的關(guān)鍵;進行分類討論是解決(3)小題的關(guān)鍵;根據(jù)切線及旋轉(zhuǎn)的性質(zhì)得出FH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/55.png)
BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192745294495933/SYS201311011927452944959025_DA/56.png)
AP解決(4)小題的關(guān)鍵.