3．设是夹角为的单位向量，若是单位向量，则的取值范围是_ ▲__

2．设复数，则等于_ _▲____．

1. 设集合，若，则与的关系是_ _▲___ (填、、、)

090](1)解：把A(，0)，C(3，)代入抛物线 　得

　　 　················································································· 1分

　 整理得　 ··············· 　……………… 2分　 　解得………………3分

　 ∴抛物线的解析式为 ··································································· 4分

　 (2)令　　 解得

　 ∴ B点坐标为(4，0)

　 又∵D点坐标为(0，)　∴AB∥CD　 ∴四边形ABCD是梯形．

　 ∴S_梯形ABCD ＝···························· 5分

设直线与x轴的交点为H，

与CD的交点为T，

则H(，0)，　 T(，)··················· 6分

∵直线将四边形ABCD面积二等分

∴S_梯形AHTD ＝S_梯形ABCD＝4

∴····································· 7分　

∴···························································· 8分

(3)∵MG⊥轴于点G，线段MG︰AG＝1︰2

　　 ∴设M(m，)，········································ 9分　

∵点M在抛物线上　　 ∴　

解得(舍去) ···························· 10分

∴M点坐标为(3，)····················································································· 11分

根据中心对称图形性质知，MQ∥AF，MQ＝AF，NQ＝EF，

∴N点坐标为(1，) ···················································································· 12分

089]解：(1)圆心在坐标原点，圆的半径为1，

点的坐标分别为

抛物线与直线交于点，且分别与圆相切于点和点，

．······························································································· 2分

点在抛物线上，将的坐标代入

，得：　 解之，得：

抛物线的解析式为：．······································································ 4分

(2)

抛物线的对称轴为，

．···················· 6分

连结，

，，

又，

，

．····································································· 8分

(3)点在抛物线上．································································································ 9分

设过点的直线为：，

将点的坐标代入，得：，

直线为：．······················································································ 10分

过点作圆的切线与轴平行，点的纵坐标为，

将代入，得：．

点的坐标为，··························································································· 11分

当时，，

所以，点在抛物线上．···································································· 12分

说明：解答题各小题中只给出了1种解法，其它解法只要步骤合理、解答正确均应得到相应的分数．

088]解：(1)法一：由图象可知：抛物线经过原点，

设抛物线解析式为．

把，代入上式得：···················································································· 1分

解得······················································································· 3分

∴所求抛物线解析式为········································································ 4分

法二：∵，，

∴抛物线的对称轴是直线．

设抛物线解析式为()······························································ 1分

把，代入得

　　 解得··········································································· 3分

∴所求抛物线解析式为．····························································· 4分

(2)分三种情况：

①当，重叠部分的面积是，过点作轴于点，

∵，在中，，，

在中，，，

∴，

∴．··············································· 6分

②当，设交于点，作轴于点，

，则四边形是等腰梯形，

重叠部分的面积是．

∴，

∴．··········· 8分

③当，设与交于点，交于点，重叠部分的面积是．

因为和都是等腰直角三角形，所以重叠部分的面积是．

∵，，

∴，

∴，

∴

　 ．················································ 10分

(3)存在　　 ···································································································· 12分

　　　　　　 ··································································································· 14分

087](天门)略

086]⑴证明：∵BC是⊙O的直径

∴∠BAC=90^o

又∵EM⊥BC，BM平分∠ABC，

∴AM=ME，∠AMN=EMN

又∵MN=MN，

∴△ANM≌△ENM

⑵∵AB²=AF·AC

∴

又∵∠BAC=∠FAB=90^o

∴△ABF∽△ACB

∴∠ABF=∠C

又∵∠FBC=∠ABC+∠FBA=90^o

∴FB是⊙O的切线

⑶由⑴得AN=EN，AM=EM，∠AMN=EMN，

又∵AN∥ME，∴∠ANM=∠EMN，

∴∠AMN=∠ANM，∴AN=AM，

∴AM=ME=EN=AN

∴四边形AMEN是菱形

∵cos∠ABD=，∠ADB=90^o

∴

设BD=3x，则AB=5x，，由勾股定理

而AD=12，∴x=3

∴BD=9，AB=15

∵MB平分∠AME，∴BE=AB=15

∴DE=BE-BD=6

∵ND∥ME，∴∠BND=∠BME，又∵∠NBD=∠MBE

∴△BND∽△BME，则

设ME=x，则ND=12-x，，解得x=

∴S=ME·DE=×6=45

085]解: (1)由题知: ……………………………………1 分　　

　解得: ……………………………………………………………2分

∴ 所求抛物线解析式为: ……………………………3分

(2) 存在符合条件的点P, 其坐标为P (－1, )或P(－1,－ )

或P (－1, 6)　 或P (－1, )………………………………………………………7分

(3)解法①：

过点E 作EF⊥x 轴于点F , 设E ( a ,--2a+3 )( －3< a < 0 )

∴EF=--2a+3，BF=a+3，OF=－a ………………………………………………8 分

∴S_{四边形BOCE} = BF·EF + (OC +EF)·OF　

=( a+3 )·(－－2a+3) + (－－2a+6)·(－a)……………………………9 分

=………………………………………………………………………10 分

=－+

∴ 当a =－时，S_{四边形BOCE} 最大, 且最大值为 ．……………………………11 分　

此时，点E 坐标为 (－，)……………………………………………………12分

解法②：

过点E 作EF⊥x 轴于点F， 设E ( x ,　 y ) ( －3< x < 0 ) …………………………8分

则S_{四边形BOCE} = (3 + y )·(－x) +　 ( 3 + x )·y ………………………………………9分

　　　　　　 = ( y－x)= ( )　 …………………………………10 分

　　　　　　 = － + 　

∴ 当x =－时，S_{四边形BOCE} 最大，且最大值为 ． …………………………11分

此时，点E 坐标为 (－，) ……………………………………………………12分

084]解：(1)⊙P与x轴相切.

　　 　∵直线y=－2x－8与x轴交于A(4，0)，与y轴交于B(0，－8)，

∴OA=4，OB=8.由题意，OP=－k，∴PB=PA=8+k.

在Rt△AOP中，k²+4²=(8+k)²，

∴k=－3，∴OP等于⊙P的半径，

∴⊙P与x轴相切.

(2)设⊙P与直线l交于C，D两点，连结PC，PD当圆心P在线段OB上时,作PE⊥CD于E.

∵△PCD为正三角形，∴DE=CD=，PD=3，

　∴PE=.

∵∠AOB=∠PEB=90°， ∠ABO=∠PBE，

∴△AOB∽△PEB，

∴，

∴∴，

∴，∴.

当圆心P在线段OB延长线上时,同理可得P(0,－－8)，

∴k=－－8，∴当k=－8或k=－－8时，以⊙P与直线l的两个交点和圆心P为顶点的三角形是正三角形.