【答案】
分析:(1)過點B作BE⊥y軸于點E,作BF⊥x軸于點F.依題意得BF=OE=2,利用勾股定理求出OF,然后可得點B的坐標(biāo).設(shè)直線AB的解析式是y=kx+b,把已知坐標(biāo)代入可求解.
(2)由△ABD由△AOP旋轉(zhuǎn)得到,證明△ABD≌△AOP.AP=AD,∠DAB=∠PAO,∠DAP=∠BAO=60°,△ADP是等邊三角形.利用勾股定理求出DP.在Rt△BDG中,∠BGD=90°,∠DBG=60°.利用三角函數(shù)求出BG=BD•cos60°,DG=BD•sin60°.然后求出OH,DH,然后求出點D的坐標(biāo).
(3)本題分三種情況進(jìn)行討論,設(shè)點P的坐標(biāo)為(t,0):
①當(dāng)P在x軸正半軸上時,即t>0時,關(guān)鍵是求出D點的縱坐標(biāo),方法同(2),在直角三角形DBG中,可根據(jù)BD即OP的長和∠DBG的正弦函數(shù)求出DG的表達(dá)式,即可求出DH的長,根據(jù)已知的△OPD的面積可列出一個關(guān)于t的方程,即可求出t的值.
②當(dāng)P在x軸負(fù)半軸,但D在x軸上方時.即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/0.png)
<t≤0時,方法同①類似,也是在直角三角形DBG用BD的長表示出DG,進(jìn)而求出GF的長,然后同①.
③當(dāng)P在x軸負(fù)半軸,D在x軸下方時,即t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/1.png)
時,方法同②.
綜合上面三種情況即可求出符合條件的t的值.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/images2.png)
解:(1)如圖1,過點B作BE⊥y軸于點E,作BF⊥x軸于點F.由已知得:
BF=OE=2,OF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/3.png)
,
∴點B的坐標(biāo)是(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/4.png)
,2)
設(shè)直線AB的解析式是y=kx+b(k≠0),則有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/5.png)
.
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/6.png)
.
∴直線AB的解析式是y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/7.png)
x+4;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/images9.png)
(2)如圖2,∵△ABD由△AOP旋轉(zhuǎn)得到,
∴△ABD≌△AOP,
∴AP=AD,∠DAB=∠PAO,
∴∠DAP=∠BAO=60°,
∴△ADP是等邊三角形,
∴DP=AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/8.png)
.
如圖2,過點D作DH⊥x軸于點H,延長EB交DH于點G,則BG⊥DH.
方法(一)
在Rt△BDG中,∠BGD=90°,∠DBG=60°.
∴BG=BD•cos60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/9.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/11.png)
.
DG=BD•sin60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/12.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/14.png)
.
∴OH=EG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/15.png)
,DH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/16.png)
∴點D的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/18.png)
)
方法(二)
易得∠AEB=∠BGD=90°,∠ABE=∠BDG,∴△ABE∽△BDG,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/19.png)
;而AE=2,BD=OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/20.png)
,BE=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/21.png)
,AB=4,
則有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/22.png)
,解得BG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/23.png)
,DG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/24.png)
;
∴OH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/25.png)
,DH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/26.png)
;
∴點D的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/27.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/28.png)
).
(3)假設(shè)存在點P,在它的運動過程中,使△OPD的面積等于
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/29.png)
.
設(shè)點P為(t,0),下面分三種情況討論:
①當(dāng)t>0時,如圖,BD=OP=t,DG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/30.png)
t,
∴DH=2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/31.png)
t.
∵△OPD的面積等于
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/32.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/33.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/34.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/35.png)
(舍去)
∴點P
1的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/36.png)
,0).
②∵當(dāng)D在x軸上時,根據(jù)勾股定理求出BD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/37.png)
=OP,
∴當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/38.png)
<t≤0時,如圖,BD=OP=-t,DG=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/39.png)
t,
∴GH=BF=2-(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/40.png)
t)=2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/41.png)
t.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/images44.png)
∵△OPD的面積等于
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/42.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/43.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/44.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/45.png)
,
∴點P
2的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/46.png)
,0),點P
3的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/47.png)
,0).
③當(dāng)t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/48.png)
時,如圖3,BD=OP=-t,DG=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/49.png)
t,
∴DH=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/50.png)
t-2.
∵△OPD的面積等于
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/51.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/52.png)
(-t)【-(2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/53.png)
t)】=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/54.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/55.png)
(舍去),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/56.png)
∴點P
4的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/57.png)
,0),
綜上所述,點P的坐標(biāo)分別為P
1(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/58.png)
,0)、P
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/59.png)
,0)、P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/60.png)
,0)、
P
4(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231449174358673/SYS201310212314491743586026_DA/61.png)
,0).
點評:本題綜合考查的是一次函數(shù)的應(yīng)用,難度較大.