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28.如图，Rt△ABC的顶点坐标分别为A(0， )，B(- ， )，C(1，0)，∠ABC=90°，BC与y轴的交点为D，D点坐标为(0， )，以点D为顶点、y轴为对称轴的抛物线过点B.

(1)求该抛物线的解析式;

(2)将△ABC沿AC折叠后得到点B的对应点B′，求证：四边形AOCB′是矩形，并判断点B′是否在(1)的抛物线上;

(3)延长BA交抛物线于点E，在线段BE上取一点P，过P点作x轴的垂线，交抛物线于点F，是否存在这样的点P，使四边形PADF是平行四边形?若存在，求出点P的坐标，若不存在，说明理由.

解：(1)∵抛物线的顶点为D(0， )

∴可设抛物线的解析式为y=ax 2+ .············································ 1分

∵B(- ， )在抛物线上

∴a(- )2+ = ，∴a= .····················· 3分

∴抛物线的解析式为y= x 2+ .··················· 5分

(2)∵B(- ， )，C(1，0)

∴BC= = 又B′C=BC，OA= ，∴B′C=OA.············································· 6分

∵AC= = =2

∴AB= = =1

又AB′=AB，OC=1，∴AB′=OC.················································ 7分

∴四边形AOCB′是矩形.··································································· 8分

∵B′C= ，OC=1

∴点B′ 的坐标为(1， )····························································· 9分

将x=1代入y= x 2+ 得y= ∴点B′ 在抛物线上.······································································ 10分

(3)存在····································································································· 11分

理由如下：

设直线AB的解析式为y=kx+b，则

解得 ∴直线AB的解析式为y= ···················································· 12分

∵P、F分别在直线AB和抛物线上，且PF∥AD

∴设P(m， )，F(m， m 2+ )

∴PF=( )-( m 2+ )=- m 2+ + AD= = 若四边形PADF是平行四边形，则有PF=AD.

即- m 2+ + = 解得m1=0(不合题意，舍去)，m2= .····································· 13分

当m= 时， = × + = .

∴存在点P( ， )，使四边形PADF是平行四边形.············· 14分

29.如图1，平移抛物线F1：y=x 2后得到抛物线F2.已知抛物线F2经过抛物线F1的顶点M和点A(2，0)，且对称轴与抛物线F1交于点B，设抛物线F2的顶点为N.

(1)探究四边形ABMN的形状及面积(直接写出结论);

(2)若将已知条件中的“抛物线F1：y=x 2”改为“抛物线F1：y=ax 2”(如图2)，“点A(2，0)”改为“点A(m，0)”，其它条件不变，探究四边形ABMN的形状及其面积，并说明理由;

(3)若将已知条件中的“抛物线F1：y=x 2”改为“抛物线F1：y=ax 2+c”(如图3)，“点A(2，0)”改为“点A(m，c)”其它条件不变，求直线AB与 轴的交点C的坐标(直接写出结论).

解：(1)四边形ABMN是正方形，其面积为2.····················································· 1分

(2)四边形ABMN是菱形.当m>0时，四边形ABMN的面积为 ;当 <0时，四边形ABMN的面积为- .······························································································· 2分

(说明：如果没有说理过程，探究的结论正确的得2分)

理由如下：

∵平移抛物线F1后得到抛物线F2，且抛物线F2经过原点O.

∴设抛物线F2的解析式为y=ax 2+bx.

∵抛物线F2经过点A(m，0)，∴am 2+bm=0.

由题意可知m≠0，∴b=-am.

∴抛物线F2的解析式为y=ax 2-amx.·············································· 3分

∴y=a(x- )2- ∴抛物线F2的对称轴为直线x= ，顶点N( ，- ).········ 4分

∵抛物线F2的对称轴与抛物线F1的交点为B，∴点B的横坐标为 .

∵点B在抛物线F1：y=ax 2上

∴yB=a( )2= ············································································ 5分

设抛物线F2的对称轴与x轴交于点P，如图1.

∵a>0，∴BP= .

∵顶点N( ，- )，∴NP=|- |= .

∴BP=NP.······························································ 6分

∵抛物线是轴对称图形，∴OP=AP.

∴四边形ABMN是平行四边形.····························· 7分

∵BN是抛物线F2的对称轴，∴BN⊥OA.

∴四边形ABMN是菱形.····································································· 8分

∵BN=BP+NP，∴BN= .

∵四边形ABMN的面积为 ×OA·BN= ×|m|× ∴当m>0时，四边形ABMN的面积为 ×m× = .·········· 9分

当m<0时，四边形ABMN的面积为 ×(-m)× =- . 10分

(3)点C的坐标为(0， +c)(参考图2).

30.如图，抛物线的顶点为A(2，1)，且经过原点O，与x轴的另一个交点为B.

(1)求抛物线的解析式;

(2)在抛物线上求点M，使△MOB的面积是△AOB面积的3倍;

(3)连结OA，AB，在x轴下方的抛物线上是否存在点N，使△OBN与△OAB相似?若存在，求出N点的坐标;若不存在，说明理由.

解：(1)由题意，可设抛物线的解析式为y=a(x-2)2+1.

∵抛物线经过原点，∴a(0-2)2+1=0，∴a=- .

∴抛物线的解析式为y=- (x-2)2+1=- x 2+x.···················· 3分

(2)△AOB和所求△MOB同底不等高，若S△MOB =3S△AOB ，则△MOB的高是△AOB高的3倍，

即M点的纵坐标是-3.······································································· 5分

∴- x 2+x=-3，整理得x 2-4x-12=0，解得x1=6，x2=-2.

∴满足条件的点有两个：M1(6，-3)，M2(-2，-3)························ 7分

(3)不存在.································································································ 8分

理由如下：

由抛物线的对称性，知AO=AB，∠AOB=∠ABO.

若△OBN∽△OAB，则∠BON=∠BOA=∠BNO.

设ON交抛物线的对称轴于A′ 点，则A′ (2，-1).

∴直线ON的解析式为y=- x.

由 x=- x 2+x，得x1=0，x2=6.

∴N(6，-3).

过点N作NC⊥x轴于C.

在Rt△BCN中，BC=6-4=2，NC=3

∴NB= = .

∵OB=4，∴NB≠OB，∴∠BON≠∠BNO，∴△OBN与△OAB不相似.

同理，在对称轴左边的抛物线上也不存在符合条件的 点.

∴在x轴下方的抛物线上不存在点N，使△OBN与△OAB相似.····· 10分

31.如图，在直角坐标系中，点A的坐标为(-2，0)，连结OA，将线段OA绕原点O顺时针旋转120°，得到线段OB.

(1)求点B的坐标;

(2)求经过A、O、B三点的抛物线的解析式;

(3)在(2)中抛物线的对称轴上是否存在点C，使△BOC的周长最小?若存在，求出点C的坐标;若不存在，请说明理由.

(4)如果点P是(2)中的抛物线上的动点，且在x轴的下方，那么△PAB是否有最大面积?若有，求出此时P点的坐标及△PAB的最大面积;若没有，请说明理由.

(1)如图1，过点B作BM⊥x轴于M.

由旋转性质知OB=OA=2.

∵∠AOB=120°，∴∠BOM=60°.

∴OM=OB·cos60°=2× =1，BM=OB·sin60°=2× = .

∴点B的坐标为(1， ).······································· 1分

(2)设经过A、O、B三点的抛物线的解析式为y=ax 2+bx+c

∵抛物线过原点，∴c=0.

∴ 解得 ∴所求抛物线的解析式为y= x 2+ x.·································· 3分

(3)存在.································································································· 4分

如图2，连接AB，交抛物线的对称轴于点C，连接OC.

∵OB的长为定值，∴要使△BOC的周长最小，必须BC+OC的长最小.

∵点A与点O关于抛物线的对称轴对称，∴OC=AC.

∴BC+OC=BC+AC=AB.

由“两点之间，线段最短”的原理可知：此时BC+OC最小，点C的位置即为所求.

设直线AB的解析式为y=kx+m，将A(-2，0)，B(1， )代入，得

解得 ∴直线AB的解析式为y= x+ .

抛物线的对称轴为直线x= =-1，即x=-1.

将x=-1代入直线AB的解析式，得y= ×(-1)+ = .

∴点C的坐标为(-1， ).···························································· 6分

(4)△PAB有最大面积.··········································································· 7分

如图3，过点P作y轴的平行线交AB于点D.

∵S△PAB =S△PAD+S△PBD

= (yD-yP)(xB-xA)

= [( x+ )-( x 2+ x)](1+2)

=- x 2- x+ =- (x+ )2+ ∴当x=- 时，△PAB的面积有最大值，最大值为 .············· 8分

此时yP= ×(- )2+ ×(- )=- .

∴此时P点的坐标为(- ，- ).················································ 9分

相关链接：

2012中考数学压轴题及答案40例（8）

2012年上海普陀区中考二模数学试题

2012中考数学最新模拟题（一）