【答案】
分析:(1)因?yàn)辄c(diǎn)B(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/0.png)
,2)在直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/1.png)
x+b上,所以把B點(diǎn)坐標(biāo)代入解析式即可求出未知數(shù)的值,進(jìn)而求出其解析式.根據(jù)直線解析式可求出A點(diǎn)的坐標(biāo)及直線與y軸交點(diǎn)的坐標(biāo),根據(jù)銳角三角函數(shù)的定義即可求出∠BAO的度數(shù).
(2)根據(jù)拋物線平移的性質(zhì)可設(shè)出拋物線平移后的解析式,由拋物線上點(diǎn)的坐標(biāo)特點(diǎn)求出E點(diǎn)坐標(biāo)及對稱軸直線,根據(jù)EF∥x軸可知E,F(xiàn),兩點(diǎn)關(guān)于對稱軸直線對稱,可求出F點(diǎn)的坐標(biāo),把此坐標(biāo)代入(1)所求的直線解析式就可求出未知數(shù)的值,進(jìn)而求出拋物線C的解析式.
(3)根據(jù)特殊角求出D點(diǎn)的坐標(biāo)表達(dá)式,將表達(dá)式代入(2)所求解析式,看能否計(jì)算出P點(diǎn)坐標(biāo),若能,則D點(diǎn)在拋物線C上.反之,不在拋物線上.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/images2.png)
解:(1)設(shè)直線與y軸交于點(diǎn)N,
將x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/2.png)
,y=2代入y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/3.png)
x+b得b=3,
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/4.png)
x+3,
當(dāng)x=0時(shí),y=3,當(dāng)y=0時(shí)x=-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/5.png)
∴A(-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/6.png)
,0),N(0,3);
∴OA=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/7.png)
,ON=3,
∴tan∠BAO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/9.png)
∴∠BAO=30°,
(2)設(shè)拋物線C的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/10.png)
(x-t)
2,則P(t,0),E(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/11.png)
t
2),
∵EF∥x軸且F在拋物線C上,根據(jù)拋物線的對稱性可知F(2t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/12.png)
t
2),
把x=2t,y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/13.png)
t
2代入y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/14.png)
x+3
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/15.png)
t+3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/16.png)
t
2解得t
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/17.png)
,t
2=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/18.png)
(1分)
∴拋物線C的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/19.png)
(x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/20.png)
)
2或y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/21.png)
(x-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/22.png)
)
2;
(3)假設(shè)點(diǎn)D落在拋物線C上,
不妨設(shè)此時(shí)拋物線頂點(diǎn)P(m,0),則拋物線C:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/23.png)
(x-m)
2,AP=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/24.png)
+m,
連接DP,作DM⊥x軸,垂足為M.由已知,得△PAB≌△DAB,
又∵∠BAO=30°,
∴△PAD為等邊三角形,
PM=AM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/25.png)
(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/26.png)
+m),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/images28.png)
∴tan∠DAM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/28.png)
,
∴DM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/29.png)
(9+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/30.png)
m),
OM=PM-OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/31.png)
(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/32.png)
+m)-t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/33.png)
(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/34.png)
-m),
∴M=[-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/35.png)
(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/36.png)
-m),0],
∴D[-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/37.png)
(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/38.png)
-m),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/39.png)
(9+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/40.png)
m)],
∵點(diǎn)D落在拋物線C上,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/41.png)
(9+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/42.png)
m)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/43.png)
[-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/44.png)
(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/45.png)
-m)-m
2,即m
2=27,m=±3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/46.png)
;
當(dāng)m=-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/47.png)
時(shí),此時(shí)點(diǎn)P(-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/48.png)
,0),點(diǎn)P與點(diǎn)A重合,不能構(gòu)成三角形,不符合題意,舍去.
當(dāng)m=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/49.png)
時(shí)P為(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/50.png)
,0)此時(shí)可以構(gòu)成△DAB,
所以點(diǎn)P為(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/51.png)
,0),
∴當(dāng)點(diǎn)D落在拋物線C上,頂點(diǎn)P為(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162156161400063/SYS201310221621561614000024_DA/52.png)
,0).
點(diǎn)評:此題將拋物線與直線相結(jié)合,涉及到動(dòng)點(diǎn)問題,翻折變換問題,有一定的難度.
尤其(3)題是一道開放性問題,需要進(jìn)行探索.要求同學(xué)們有一定的創(chuàng)新能力.