威廉希尔app 中考频道提供大量中考资料，在第一时间更新中考资讯。以下是2012中考数学压轴题及答案40例：

32.已知：Rt△ABC的斜边长为5，斜边上的高为2，将这个直角三角形放置在平面直角坐标系中，使其斜边AB与x轴重合(其中OA

(1)求线段OA、OB的长和经过点A、B、C的抛物线的关系式.

(2)如图2，点D的坐标为(2，0)，点P(m，n)是该抛物线上的一个动点(其中m>0，n>0)，连接DP交BC于点E.

①当△BDE是等腰三角形时，直接写出此时点E的坐标.

②又连接CD、CP(如图3)，△CDP是否有最大面积?若有，求出△CDP的最大面积和此时点P的坐标;若没有，请说明理由.

解：(1)由题意知Rt△△AOC∽Rt△COB，∴ = .

∴OC 2=OA·OB=OA(AB-OA)，即22=OA(5-OA).

∴OA 2-5OA+4=0，∵OA

∴A(-1，0)，B(4，0)，C(0，2).

∴可设所求抛物线的关系式为y=a(x+1)(x-4).······················· 3分

将点C(0，2)代入，得2=a(0+1)(0-4)，∴a=- .

∴经过点A、B、C的抛物线的关系式为y=- (x+1)(x-4).·· 4分

即y=- x 2+ x+2.

(2)①E1(3， )，E2( ， )，E3( ， ).······················· 7分

关于点E的坐标求解过程如下(原题不作要求，本人添加，仅供参考)：

设直线BC的解析式为y=kx+b.

则 解得 ∴直线BC的解析式为y=- x+2.

∵点E在直线BC上，∴E(x，- x+2).

若ED=EB，过点E作EH⊥x轴于H，如图2，则DH= DB=1.

∴OH=OD+DH=2+1=3.

∴点E的横坐标为3，代入直线BC的解析式，得y=- ×3+2= .

∴E1(3， ).

若DE=DB，则(x-2)2+(- x+2)2=22.

整理得5x 2-24x+16=0，解得x1=4(舍去)，x2= .

∴y=- × +2= ，∴E2( ， ).

若BE=BD，则(x-4)2+(- x+2)2=22.

整理得5x 2-24x+16=0，解得x1= (此时点P在第四象限，舍去)，x2= .

∴y=- ×( )+2= ，∴E3( ， ).

②△CDP有最大面积.······································································ 8分

过点D作x轴的垂线，交PC于点M，如图3.

设直线PC的解析式为y=px+q，将C(0，2)，P(m，n)代入，

得 解得 ∴直线PC的解析式为y= x+2，∴M(2， +2).

S△CDP=S△CDM+S△PDM= xP·yM

= m( +2)

=m+n-2

=m+(- m2+ m+2)-2

=- m2+ m

=- (m- )2+ ∴当m= 时，△CDP有最大面积，最大面积为 .···················· 9分

此时n=- ×( )2+ × +2= ∴此时点P的坐标为( ， ).··················································· 10分

33.如图，已知抛物线y=x 2+4x+3交x轴于A、B两点，交y轴于点C，抛物线的对称轴交x轴于点E，点B的坐标为(-1，0).

(1)求抛物线的对称轴及点A的坐标;

(2)在平面直角坐标系xOy中是否存在点P，与A、B、C三点构成一个平行四边形?若存在，请写出点P的坐标;若不存在，请说明理由;

(3)连结CA与抛物线的对称轴交于点D，在抛物线上是否存在点M，使得直线CM把四边形DEOC分成面积相等的两部分?若存在，请求出直线CM的解析式;若不存在，请说明理由.

解：(1)对称轴为直线x=- =-2，即x=-2;··············································· 2分

令y=0，得x 2+4x+3=0，解得x1=-1，x2=-3.

∵点B的坐标为(-1，0)，∴点A的坐标为(-3，0).·········································· 4分

(2)存在，点P的坐标为(-2，3)，(2，3)和(-4，-3).································ 7分

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

当x=0时，y=x 2+4x+3=3，∴点C的坐标为(0，3).

AO=3，EO=2，AE=1，CO=3.

∵DE∥CO，

∴△AED∽△AOC.∴ = ，即 = .

∴DE=1.··················································································································· 9分

∵DE∥CO，且DE≠CO，∴四边形DEOC为梯形.

S梯形DEOC= (1+3)×2=4.

设直线CM交x轴于点F，如图.

若直线CM把梯形DEOC分成面积相等的两部分，则S△COF=2

即 CO·FO=2.∴ ×3FO=2，∴FO= .

∴点F的坐标为(- ，0).···················································· 10分

∵直线CM经过点C(0，3)，∴设直线CM的解析式为y=kx+3.

把F(- ，0)代入，得- k+3=0.··································································· 11分

∴k= .

∴直线CM的解析式为y= x+3.········································································· 12分

34.在平面直角坐标系中，现将一块等腰直角三角板ABC放在第二象限，斜靠在两坐标轴上，且点A(0，2)，点C(-1，0)，如图所示;抛物线y=ax 2+ax-2经过点B.

(1)求点B的坐标;

(2)求抛物线的解析式;

(3)在抛物线上是否还存在点P(点B除外)，使△ACP仍然是以AC为直角边的等腰直角三角形?若存在，求所有点P的坐标;若不存在，请说明理由.

解：(1)过点B作BD⊥x轴于D.

∵∠BCD+∠ACO=90°，∠ACO+∠CAO=90°.

∴∠BCD=∠CAO.····································································································· 1分

又∵∠BDC=∠COA=90°，BC=CA.

∴Rt△BCD≌Rt△CAO，······························································································ 2分

∴BD=CO=1，CD=AO=2.····················································································· 3分

∴点B的坐标为(-3，1);······················································································· 4分

(2)把B(-3，1)代入y=ax 2+ax-2，得1=9a-3a-2，解得a= .············ 6分

∴抛物线的解析式为y= x 2+ x-2;··································································· 7分

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

①延长BC至点P1，使CP1=BC，则得到以点C为直角顶点的等腰直角三角形△ACP1.

····································································································································· 9分

过点P1作P1M⊥x轴.

∵CP1=BC，∠P1CM=∠BCD，∠P1MC=∠BDC=90°.

∴Rt△P1CM≌Rt△BCD，········································································ 10分

∴CM=CD=2，P1M=BD=1，可求得点P1(1，-1);······················· 11分

把x=1代入y= x 2+ x-2，得y=-1.

∴点P1(1，-1)在抛物线上.······························································· 12分

②过点A作AP2⊥AC，且使AP2=AC，则得到以点A为直角顶点的等腰直角三角形△ACP2.

··········································································································· 13分

过点P2作P2N⊥y轴，同理可证Rt△P2NA≌Rt△AOC.·········································· 14分

P2N=AO=2，AN=CO=1.可求得点P2(2，1).·················································· 15分

把x=2代入y= x 2+ x-2，得y=1.

∴点P2(2，1)在抛物线上.····················································································· 16分

综上所述，在抛物线上还存在点P1(1，-1)和P2(2，1)，使△ACP仍然是以AC为直角边的等腰直角三角形.

35.如图，在平面直角坐标中，二次函数图象的顶点坐标为C(4，- )，且在x轴上截得的线段AB的长为6.

(1)求二次函数的解析式;

(2)点P在y轴上，且使得△PAC的周长最小，求：

①点P的坐标;

②△PAC的周长和面积;

(3)在x轴上方的抛物线上，是否存在点Q，使得以Q、A、B三点为顶点的三角形与△ABC相似?如果存在，求出点Q的坐标;如果不存在，请说明理由.

解：(1)设二次函数的解析式为y=a(x -4)2- (a≠0)，且A(x1，0)，B(x2，0).

∵y=a(x -4)2- =ax 2-8ax+16a- ∴x1+x2=8，x1x2=16- .

∴AB 2=(x1-x2)2=(x1+x2)2-4x1x2=82-4(16- )=36，∴a= .

∴二次函数的解析式为y= (x -4)2- .······················································· 2分

(2)①如图1，作点A关于y轴的对称点A′，连结A′C交y轴于点P，连结PA，则点P为所求.

令y=0，得 (x -4)2- =0，解得x1=1，x2=7.

∴A(1，0)，B(7，0).∴OA=1，∴OA′=1.

设抛物线的对称轴与x轴交于点D，则AD=3，A′D=5，DC= .

∵△A′OP∽△ADC，∴ = ，即 = ，∴OP= .

∴P(0，- ).········································································································ 4分

②∵A′C= = = AC= = = ∴△PAC的周长=PA+PC+AC=A′C+AC= + .····································· 5分

S△PAC=S△A′AC - S△A′AP= A′A(DC-OP)= ×2×( - )= .

························································································· 7分

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

∵tan∠BAC= = ，∴∠BAC=30°.

同理，∠ABC=30°，∴∠ACB=120°，AC=BC.

①若以AB为腰，∠BAQ1为顶角，使△ABQ1∽△CBA，则AQ1=AB=6，∠BAQ1=120°.

如图2，过点Q1作Q1H⊥x轴于H，则

Q1H=AQ1·sin60°=6× = ，HA=AQ1·cos60°=6× =3.

HO=HA-OA=3-1=2.

∴点Q1的坐标为(-2， ).

把x=-2代入y= (x -4)2- ，得y= (-2-4)2- = .

∴点Q1在抛物线上.·································································································· 9分

②若以BA为腰，∠ABQ2为顶角，使△ABQ2∽△ACB，由对称性可求得点Q1的坐标为(10， ).

同样，点Q2也在抛物线上.····················································································· 10分

③若以AB为底，AQ，BQ为腰，点Q在抛物线的对称轴上，不合题意，舍去.

····························································································· 11分

综上所述，在x轴上方的抛物线上存在点Q1(-2， )和Q2(10， )，使得以Q、A、B三点为顶点的三角形与△ABC相似.············································································································· 12分

36.如图，抛物线y=ax 2+bx+c(a≠0)与x轴交于A(-3，0)、B两点，与y轴相交于点C(0， ).当x=-4和x=2时，二次函数y=ax 2+bx+c(a≠0)的函数值y相等，连结AC、BC.

(1)求实数a，b，c的值;

(2)若点M、N同时从B点出发，均以每秒1个单位长度的速度分别沿BA、BC边运动，其中一个点到达终点时，另一点也随之停止运动.当运动时间为t秒时，连结MN，将△BMN沿MN翻折，B点恰好落在AC边上的P处，求t的值及点P的坐标;

y

O

x

C

N

B

P

M

A

(3)在(2)的条件下，抛物线的对称轴上是否存在点Q，使得以B，N，Q为顶点的三角形与△ABC相似?若存在，请求出点Q的坐标;若不存在，请说明理由.

解：(1)由题意得

解得a=- ，b=- ，c= .

······················································ 3分

(2)由(1)知y=- x 2- x+ ，令y=0，得- x 2- x+ =0.

解得x1=-3，x2=1.

∵A(-3，0)，∴B(1，0).

又∵C(0， )，∴OA=3，OB=1，OC= ，∴AB=4，BC=2.

∴tan∠ACO= = ，∴∠ACO=60°，∴∠CAO=30°.

同理，可求得∠CBO=60°，∠BCO=30°，∴∠ACB=90°.

∴△ABC是直角三角形.

又∵BM=BN=t，∴△BMN是等边三角形.

∴∠BNM=60°，∴∠PNM=60°，∴∠PNC=60°.

∴Rt△PNC∽Rt△ABC，∴ = .

由题意知PN=BN=t，NC=BC-BN=2-t，∴ = .

∴t= .······················································· 4分

∴OM=BM-OB= -1= .

如图1，过点P作PH⊥x轴于H，则PH=PM·sin60°= × = .

MH=PM·cos60°= × = .

∴OH=OM+MH= + =1.

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

(3)存在.

由(2)知△ABC是直角三角形，若△BNQ与△ABC相似，则△BNQ也是直角三角形.

∵二次函数y=- x 2- x+ 的图象的对称轴为x=-1.

∴点P在对称轴上.

∵PN∥x轴，∴PN⊥对称轴.

又∵QN≥PN，PN=BN，∴QN≥BN.

∴△BNQ不存在以点Q为直角顶点的情形.

①如图2，过点N作QN⊥对称轴于Q，连结BQ，则△BNQ是以点N为直角顶点的直角三角形，且QN>PN，∠MNQ=30°.

∴∠PNQ=30°，∴QN= = = .

∴ = = .

∵ =tan60°= ，∴ ≠ .

∴当△BNQ以点N为直角顶点时，△BNQ与△ABC不相似.··········· 7分

②如图3，延长NM交对称轴于点Q，连结BQ，则∠BMQ=120°.

∵∠AMP=60°，∠AMQ=∠BMN=60°，∴∠PMQ=120°.

∴∠BMQ=∠PMQ，又∵PM=BM，QM=QM.

∴△BMQ≌△PMQ，∴∠BQM=∠PQM=30°.

∵∠BNM=60°，∴∠QBN=90°.

∵∠CAO=30°，∠ACB=90°.

∴△BNQ∽△ABC.················································ 8分

∴当△BNQ以点B为直角顶点时，△BNQ∽△ABC.

设对称轴与x轴的交点为D.

∵∠DMQ=∠DMP=60°，DM=DM，∴Rt△DMQ≌Rt△DMP.

∴DQ=PD，∴点Q与点P关于x轴对称.

∴点Q的坐标为(-1，- ).··············································································· 9分

综合①②得，在抛物线的对称轴上存在点Q(-1，- )，使得以B，N，Q为顶点的三角形与△ABC相似. 10分

37.如图①，已知抛物线y=ax 2+bx+3(a≠0)与x轴交于点A(1，0)和点B(-3，0)，与y轴交于点C.

(1)求抛物线的解析式;

(2)设抛物线的对称轴与x轴交于点M，问在对称轴上是否存在点P，使△CMP为等腰三角形?若存在，请直接写出所有符合条件的点P的坐标;若不存在，请说明理由;

(3)如图②，若点E为第二象限抛物线上一动点，连接BE、CE，求四边形BOCE面积的最大值，并求此时E点的坐标.

解：(1)由题意得 .········································································ 1分

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

∴所求抛物线的解析式为y=-x 2-2x+3;··································· 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)解法一：

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

则EF=-m 2-2m+3，BF=m+3，OF=-m.········· 8分

∴S四边形BOCE =S△BEF +S梯形FOCE

= BF·EF + (EF+OC)·OF

= (m+3)(-m 2-2m+3)+ (-m 2-2m+6)(-m).···································· 9分

=- m 2- m+ ································································································· 10分

=- (m+ )2+ ∴当m=- 时，S四边形BOCE 最大，且最大值为 .·············································· 11分

此时y=-(- )2-2×(- )+3= ∴此时E点的坐标为(- ， ).··········································································· 12分

解法二：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 =S△BEF +S梯形FOCE

= BF·EF + (EF+OC)·OF

= (3+x)· y+ (3+y)(-x).·························································· 9分

= (y-x)= (-x 2-3x+3).······················································ 10分

=- (x+ )2+ ∴当x=- 时，S四边形BOCE 最大，且最大值为 .··············································· 11分

此时y=-(- )2-2×(- )+3= ∴此时E点的坐标为(- ， ).··········································································· 12分

38.如图，已知抛物线y=ax 2+bx+c与x轴交于A、B两点，与y轴交于点C.其中点A在x轴的负半轴上，点C在y轴的负半轴上，线段OA、OC的长(OA

(1)求A、B、C三点的坐标;

(2)求此抛物线的解析式;

(3)若点D是线段AB上的一个动点(与点A、B不重合)，过点D作DE∥BC交AC于点E，连结CD，设BD的长为m，△CDE的面积为S，求S与m的函数关系式，并写出自变量m的取值范围.S是否存在最大值?若存在，求出最大值并求此时D点坐标;若不存在，请说明理由.

解：(1)∵OA、OC的长是方程x 2-5x+4=0的两个根，OA

∴OA=1，OC=4.

∵点A在x轴的负半轴，点C在y轴的负半轴

∴A(-1，0)，C(0，-4).

∵抛物线y=ax 2+bx+c的对称轴为x=1

∴由对称性可得B点坐标为(3，0).

∴A、B、C三点的坐标分别是：A(-1，0)，B(3，0)，C(0，-4).

············································································································· 3分

(2)∵点C(0，-4)在抛物线y=ax 2+bx+c图象上，∴c=-4.···················· 4分

将A(-1，0)，B(3，0)代入y=ax 2+bx-4得

解得 ·············································································· 6分

∴此抛物线的解析式为y= x 2- x-4.······························································· 7分

(3)∵BD=m，∴AD=4-m.

在Rt△BOC中，BC 2=OB 2+OC 2=3 2+4 2=25，∴BC=5.

∵DE∥BC，∴△ADE∽△ABC.

∴ = ，即 = .

∴DE= .

过点E作EF⊥AB于点F，则sin∠EDF=sin∠CBA= = .

∴ = ，∴EF= DE= × =4-m.·················································· 9分

∴S =S△CDE =S△ADC -S△ADE

= (4-m)×4- (4-m)(4-m)

=- m 2+2m

=- (m-2)2+2(0

∵- <0

∴当m=2时，S有最大值2.··············································· 11分

此时OD=OB-BD=3-2=1.

∴此时D点坐标为(1，0).·················································································· 12分

39.如图，抛物线y=a(x+3)(x-1)与x轴相交于A、B两点(点A在点B右侧)，过点A的直线交抛物线于另一点C，点C的坐标为(-2，6).

(1)求a的值及直线AC的函数关系式;

(2)P是线段AC上一动点，过点P作y轴的平行线，交抛物线于点M，交x轴于点N.

①求线段PM长度的最大值;

②在抛物线上是否存在这样的点M，使得△CMP与△APN相似?如果存在，请直接写出所有满足条件的点M的坐标(不必写解答过程);如果不存在，请说明理由.

解：(1)由题意得6=a(-2+3)(-2-1)，∴a=-2.······································· 1分

∴抛物线的解析式为y=-2(x+3)(x-1)，即y=-2x 2-4x+6

令-2(x+3)(x-1)=0，得x1=-3，x2=1

∵点A在点B右侧，∴A(1，0)，B(-3，0)

设直线AC的函数关系式为y=kx+b，把A(1，0)、C(-2，6)代入，得

解得 ∴直线AC的函数关系式为y=-2x+2.··········································· 3分

(2)①设P点的横坐标为m(-2≤ m ≤1)，

则P(m，-2m+2)，M(m，-2m 2-4m+6).··································· 4分

∴PM=-2m 2-4m+6-(-2m+2)

=-2m 2-2m+4

=-2(m+ )2+ ∴当m=- 时，线段PM长度的最大值为 .································ 6分

②存在

M1(0，6).················································································································· 7分

M2(- ， ).·········································································································· 9分

点M的坐标的求解过程如下(原题不作要求，本人添加，仅供参考)

ⅰ)如图1，当M为直角顶点时，连结CM，则CM⊥PM，△CMP∽△ANP

∵点C(-2，6)，∴点M的纵坐标为6，代入y=-2x 2-4x+6

得-2x 2-4x+6=6，∴x=-2(舍去)或x=0

∴M1(0，6)

(此时点M在y轴上，即抛物线与y轴的交点，此时直线MN与y轴

重合，点N与原点O重合)

ⅱ)如图2，当C为直角顶点时，设M(m，-2m 2-4m+6)(-2≤ m ≤1)

过C作CH⊥MN于H，连结CM，设直线AC与y轴相交于点D

则△CMP∽△NAP

又∵△HMC∽△CMP，△NAP∽△OAD，∴△HMC∽△OAD

∴ = ∵C(-2，6)，∴CH=m+2，MH=-2m 2-4m+6-6=-2m 2-4m

在y=-2x+2中，令x=0，得y=2

∴D(0，2)，∴OD=2

∴ = 整理得4m 2+9m+2=0，解得m=-2(舍去)或m=- 当m=- 时，-2m 2-4m+6=(- )2-4×(- )+6= ∴M2(- ， )

如图，二次函数的图象经过点D(0， )，且顶点C的横坐标为4，该图象在x轴上截得的线段AB的长为6.

(1)求该二次函数的解析式;

(2)在该抛物线的对称轴上找一点P，使PA+PD最小，求出点P的坐标;

(3)在抛物线上是否存在点Q，使△QAB与△ABC相似?如果存在，求出点Q的坐标;如果不存在，请说明理由.

解：(1)设该二次函数的解析式为y=a(x-h)2+k

∵顶点C的横坐标为4，且过点D(0， )

∴ =16a+k ①

又∵对称轴为直线x=4，图象在x轴上截得的线段AB的长为6

∴A(1，0)，B(7，0)

∴0=9a+k ②

由①②解得a= ，k= ∴该二次函数的解析式为y= (x-4)2 (2)∵点A、B关于直线x=4对称，∴PA=PB

∴PA+PD=PB+PD≥DB

∴当点P在线段DB上时，PA+PD取得最小值

∴DB与对称轴的交点即为所求的点P，如图1

设直线x=4与x轴交于点M

∵PM∥OD，∴∠BPM=∠BDO

又∠PBM=∠DBO，∴△BPM∽△BDO

∴ = ，即 = ，∴PM= ’

∴点P的坐标为(4， )

(3)由(1)知点C(4， )，

又∵AM=3，∴在Rt△ACM中，tan∠ACM= ，∴∠ACM=60°

∵AC=BC，∴∠ACB=120°

①如图2，当点Q在x轴上方时，过Q作QN⊥x轴于N

如果AB=BQ，由△ABC∽△ABQ，得BQ=6，∠ABQ=120°

∴∠QBN=60°

∴QN= ，BN=3，ON=10

∴此时点Q的坐标为(10， )

∵ (10-4)2 = ，∴点Q在抛物线上

如果AB=AQ，由对称性知Q(-2， )，且也在抛物线上

②当点Q在x轴下方时，△QAB就是△ACB

∴此时点Q的坐标为(4， )

综上所述，在抛物线上存在点Q，使△QAB与△ABC相似

点Q的坐标为(10， )或(-2， )或(4， ).

41.已知，如图，抛物线y=ax 2+3ax+c(a>0)与y轴交于C点，与x轴交于A、B两点，A点在B点左侧，点B的坐标为(1，0)，OC=3OB.

(1)求抛物线的解析式;

(2)若点D是线段AC下方抛物线上的动点，求四边形ABCD面积的最大值;

(3)若点E在x轴上，点P在抛物线上，是否存在以A、C、E、P为顶点且以AC为一边的平行四边形?若存在，求点P的坐标;若不存在，请说明理由.

解：(1)∵对称轴x=- =- .······································································ 1分

又∵OC=3OB=3，a>0

∴C(0，-3).······················································································ 2分

方法一：把B(1，0)、C(0，-3)代入y=ax 2+3ax+c得：

解得

∴抛物线的解析式为y= x 2+ x-3.··································································· 4分

方法二：令ax 2+3ax+c=0，则xA+xB=-3

∵B(1，0)，∴xA+1=-3，∴xA=-4

∴A(-4，0)

∴可设抛物线的解析式为y=a(x+4)(x-1)，把C(0，-3)代入

得-3=a(0+4)(0-1)，∴a= ∴抛物线的解析式为y= (x+4)(x-1)

即y= x 2+ x-3.·········································································· 4分

(2)方法一：如图1，过点D作DN⊥x轴，垂足为N，交线段AC于点M

∵S四边形ABCD =S△ABC +S△ACD

= AB·OC+ DM·(AN+ON)

= (4+1)×3+ DM·4

= +2DM.······················································································ 5分

设直线AC的解析式为y=kx+b，把A(-4，0)、C(0，-3)代入

得 解得

∴直线AC的解析式为y=- x-3.················································· 6分

设D(x， x 2+ x-3)，则M(x，- x-3)

∴DM=- x-3-( x 2+ x-3)=- (x+2)2+3.········································· 7分

当x=-2时，DM有最大值3

此时四边形ABCD面积有最大值，最大值为： +2×3= .····························· 8分

方法二：如图2，过点D作DQ⊥y轴于Q，过点C作CC1∥x轴交抛物线于C1

设D(x， x 2+ x-3)，则DQ=-x，OQ=- x 2- x+3

从图象可判断当点D在CC1下方的抛物线上运动时，四边形ABCD面积才有最大值

则S四边形ABCD =S△BOC +S梯形AOQD -S△CDQ

= OB·OC+ (AO+DQ)·OQ- DQ·CQ

= ×1×3+ (4+DQ)·OQ- DQ·(OQ-3)

= +2OQ+ DQ.······················································ 5分

= -2( x 2+ x-3)- x

=- x 2-6x+ =- (x+2)2+ .··················································· 7分

当x=-2时，四边形ABCD面积有最大值 ···························································································· 8分

(3)如图3

①过点C作CP1∥x轴交抛物线于点P1，过点P1作P1E1∥AC交x轴于点E1，则四边形ACP1E1为平行四边形. 9分

∵C(0，-3)，令 x 2+ x-3=-3

解得x1=0，x2=3，∴CP1=3

∴P1(-3，-3).······································································································ 11分

②平移直线AC交x轴于点E，交x轴上方的抛物线于点P，当AC=PE时，四边形ACEP为平行四边形. 12分

∵C(0，-3)，∴设P(x，3)

由 x 2+ x-3=3，解得x= 或x= ∴P2( ，3)，P3( ，3).································································ 14分

综上所述，存在以A、C、E、P为顶点且以AC为一边的平行四边形，点P的坐标分别为：

P1(-3，-3)，P2( ，3)，P3( ，3)

42.如图，在平面直角坐标系xOy中，抛物线y=- x 2+bx+c与x轴交于A(1，0)、B(5，0)两点.

(1)求抛物线的解析式和顶点C的坐标;

(2)设抛物线的对称轴与x轴交于点D，将∠DCB绕点C按顺时针方向旋转，角的两边CD和CB与x轴分别交于点P、Q，设旋转角为α(0°<α≤90°).

①当α等于多少度时，△CPQ是等腰三角形?

②设BP=t，AQ=s，求s与t之间的函数关系式.

解：(1)根据题意，得..······················································· 1分

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

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

即y=- (x-3)2+2.

∴顶点C的坐标为(3，2)..··················· 4分

(2)①∵CD=DB=AD=2，CD⊥AB，

∴∠DCB=∠CBD=45°.······························ 5分

ⅰ)若CQ=CP，则∠PCD= ∠PCQ=22.5°.

∴当α=22.5°时，△CPQ是等腰三角形.········································ 6分

ⅱ)若CQ=PQ，则∠CPQ=∠PCQ=45°，

此时点Q与D重合，点P与A重合.

∴当α=45°时，△CPQ是等腰三角形.············································ 7分

ⅲ)若PC=PQ，则∠PCQ=∠PQC=45°，此时点Q与B重合，点P与D重合.

∴α=0°，不合题意.································································································ 8分

∴当α=22.5°或45°时，△CPQ是等腰三角形.······················································ 9分

②连接AC，∵AD=CD=2，CD⊥AB，

∴∠ACD=∠CAD=45°，AC=BC= = .············································ 10分

ⅰ)当0°<α≤45°时，

∵∠ACQ=∠ACP+∠PCQ=∠ACP+45°.

∠BPC=∠ACP+∠CAD=∠ACP+45°.

∴∠ACQ=∠BPC.··································································································· 11分

又∵∠CAQ=∠PBC=45°，∴△ACQ∽△BPC.

∴ = .

∴AQ·BP=AC·BC= × =8.·································································· 12分

ⅱ)当45°<α<90°时，同理可得AQ·BP=AC·BC=8.··································· 13分

∴s= .··················································································································· 14分

相关链接：

2012年上海静安区中考二模数学试题

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）