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：

21.如图，在平面直角坐标系中，已知矩形ABCD的三个顶点B(4，0)、C(8，0)、D(8，8).抛物线y=ax 2+bx过A、C两点.

(1)直接写出点A的坐标，并求出抛物线的解析式;

(2)动点P从点A出发，沿线段AB向终点B运动，同时点Q从点C出发，沿线段CD向终点D运动，速度均为每秒1个单位长度，运动时间为t秒.过点P作PE⊥AB交AC于点E.

① 过点E作EF⊥AD于点F，交抛物线于点G.当t为何值时，线段EG最长?

② 连接EQ，在点P、Q运动的过程中，判断有几个时刻使得△CEQ是等腰三角形?请直接写出相应的t值.

解：(1)点A的坐标为(4，8).·········································································· 1分

将A(4，8)、C(8，0)两点坐标分别代入y=ax 2+bx，

得 解得a=- ，b=4.

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

(2)①在Rt△APE和Rt△ABC中，tan∠PAE= = ，即 = = .

∴PE= AP= t，PB=8-t.

∴点E的坐标为(4+ t，8-t).2

∴点G的纵坐标为- (4+ t)2+4(4+ t)=- t 2+8.············· 5分

∴EG=- t 2+8-(8-t)=- t 2+t

∵- <0，∴当 =4时，线段EG最长为2.···································· 7分

②共有三个时刻.················································································· 8分

t1= ，t2= ，t3=40- .··················································· 11分

22.如图，抛物线y=-x 2+2x+3与x轴相交于A、B两点(点A在点B的左侧)，与y轴相交于点C，顶点为D.

(1)直接写出A、B、C三点的坐标和抛物线的对称轴;

(2)连结BC，与抛物线的对称轴交于点E，点P为线段BC上的一个动点，过点P作PF∥DE交抛物线于点F，设点P的横坐标为m.

①用含m的代数式表示线段PF的长，并求出当m为何值时，四边形PEDF为平行四边形?

②设△BCF的面积为S，求S与m的函数关系式.

解：(1)A(-1，0)，B(3，0)，C(0，3).················································· 2分

抛物线的对称轴是：x=1.·································································· 3分

(2)①设直线BC的解析式为：y=kx+b.

将B(3，0)，C(0，3)分别代入得：

解得 ∴直线BC的解析式为y=-x+3.

当x=1时，y=-1+3=2，∴E(1，2).

当x=m时，y=-m+3，∴P(m，-m+3).······························· 4分

将x=1代入y=-x 2+2x+3，得y=4，∴D(1，4).

将x=m代入y=-x 2+2x+3，得y=-m 2+2m+3.

∴F(m，-m 2+2m+3).································································ 5分

∴线段DE=4-2=2，线段PF=-m 2+2m+3-(-m+3)=-m 2+3m 6分

∵PF∥DE，∴当PF=DE时，四边形PEDF为平行四边形.

由-m 2+3m=2，解得：m1=2，m2=1(不合题意，舍去).

∴当m=2时，四边形PEDF为平行四边形.······································ 7分

②设直线PF与x轴交于点M.

由B(3，0)，O(0，0)，可得：OB=OM+MB=3.

则S=S△BPF +S△CPF··············································································· 8分

= PF·BM+ PF·OM

= PF·OB

= (-m 2+3m)×3

=- m 2+ m(0≤m≤3)

即S与m的函数关系式为：S=- m 2+ m(0≤m≤3).··········· 9分

23.如图，在矩形OABC中，已知A、C两点的坐标分别为A(4，0)、C(0，2)，D为OA的中点.设点P是∠AOC平分线上的一个动点(不与点O重合).

(1)试证明：无论点P运动到何处，PC总与PD相等;

(2)当点P运动到与点B的距离最小时，试确定过O、P、D三点的抛物线的解析式;

(3)设点E是(2)中所确定抛物线的顶点，当点P运动到何处时，△PDE的周长最小?求出此时点P的坐标和△PDE的周长;

(4)设点N是矩形OABC的对称中心，是否存在点P，使∠CPN=90°?若存在，请直接写出点P的坐标.

解：(1)∵点D是OA的中点，∴OD=2，∴OD=OC.

又∵OP是∠COD的角平分线，∴∠POC=∠POD=45°.

∴△POC≌∠POD，∴PC=PD;·························································· 3分

(2)如图，过点B作∠AOC的平分线的垂线，垂足为P，点P即为所求.

易知点F的坐标为(2，2)，故BF=2，作PM⊥BF.

∵△PBF是等腰直角三角形，∴PM= BF=1.

∴点P的坐标为(3，3).

∵抛物线经过原点

∴可设抛物线的解析式为y=ax 2+bx.

又∵抛物线经过点P(3，3)和点D(2，0)

∴ 解得 ∴过O、P、D三点的抛物线的解析式为y=x 2-2x;························· 7分

(3)由等腰直角三角形的对称性知D点关于∠AOC的平分线的对称点即为C点.

连接EC，它与∠AOC的平分线的交点即为所求的P点(因为PE+PD=EC，而两点之间线段最短)，此时△PED的周长最小.

∵抛物线y=x 2-2x的顶点E的坐标(1，-1)，C点的坐标(0，2)

设CE所在直线的解析式为y=kx+b

则 解得 ∴CE所在直线的解析式为y=-3x+2.

联立 ，解得 ，故点P的坐标为( ， ).

△PED的周长即是CE+DE= ;··········································· 11分

(4)存在点P，使∠CPN=90°，其坐标为( ， )或(2，2).······· 14分

24.如图1，已知抛物线经过坐标原点O和x轴上另一点E，顶点M的坐标为(2，4);矩形ABCD的顶点A与点O重合，AD、AB分别在x轴、y轴上，且AD=2，AB=3.

(1)求该抛物线所对应的函数关系式;

(2)将矩形ABCD以每秒1个单位长度的速度从图1所示的位置沿x轴的正方向匀速平行移动，同时一动点P也以相同的速度从点A出发向B匀速移动，设它们运动的时间为t秒(0≤t≤3)，直线AB与该抛物线的交点为N(如图2所示).

①当t= 时，判断点P是否在直线ME上，并说明理由;

②设以P、N、C、D为顶点的多边形面积为S，试问S是否存在最大值?若存在，求出这个最大值;若不存在，请说明理由.

解：(1)∵因所求抛物线的顶点M的坐标为(2，4)

∴可设其对应的函数关系式为y=a(x -2)2+4.········································ 1分

又抛物线经过坐标原点O(0，0)，∴a(0-2)2+4=0.················· 2分

解得a=-1.······················································································· 3分

∴所求函数关系式为y=-(x -2)2+4，即y=-x 2+4x.·············· 4分

(2)①点P不在直线ME上，理由如下：··················································· 5分

根据抛物线的对称性可知E点的坐标为(4，0).

设直线ME的解析式为y=kx+b，将M(2，4)，E(4，0)代入，得

解得 .

∴直线ME的解析式为y=-2x+8.··················································· 6分

当t= 时，OA=AP= ，∴P( ， ).······································· 7分

∵点P的坐标不满足直线ME的解析式y=-2x+8

∴当t= 时，点P不在直线ME上.·················································· 8分

②S存在最大值，理由如下：······························································· 9分

∵点A在x轴的非负半轴上，且N在抛物线上，∴OA=AP=t.

∴P(t，t)，N(t，-t 2+4t)，∴AN=-t 2+4t(0≤ ≤3)

∴PN=AN-AP=-t 2+4t-t=-t 2+3t=t(3-t)≥0······················ 10分

(ⅰ)当PN=0，即t=0或t=3时，以点P，N，C，D为顶点的多边形是三角形，此三角形的高为AD.

∴S= DC·AD= ×3×2=3.······················································ 11分

(ⅱ)当PN≠0时，以点P，N，C，D为顶点的多边形是四边形.

∵PN∥CD，AD⊥CD.

∴S= (CD+PN)·AD= (3-t 2+3t)×2=-t 2+3t+3=-(t- )2+ (0

当t= 时，S最大= .·································································· 12分

综上所述，当t= 时，以点P，N，C，D为顶点的多边形面积S有最大值，最大值为 . 13分

说明：(ⅱ)中的关系式，当t=0和t=3时也适合.

25.如图1，已知抛物线y=ax 2-2ax-3与x轴交于A、B两点，其顶点为C，过点A的直线交抛物线于另一点D(2，-3)，且tan∠BAD=1.

(1)求抛物线的解析式;

(2)连结CD，求证：AD⊥CD;

(3)如图2，P是线段AD上的动点，过点P作y轴的平行线交抛物线于点E，求线段PE长度的最大值;

(4)点Q是抛物线上的动点，在x轴上是否存在点F，使以A，D，F，Q为顶点的四边形是平行四边形?若存在，直接写出点F的坐标;若不存在，请说明理由.

解：(1)如图1，过点D作DH⊥x轴于H，则OH=2，DH=3.

∵tan∠BAD=1，∴AH=DH=3，∴AO=3-2=1.·························· 1分

∴A(-1，0).······················································································ 2分

把A(-1，0)代入y=ax 2-2ax-3，得a+2a-3=0.

∴a=1.································································································ 3分

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

(2)∵y=x 2-2x-3=(x-1)2-4

∴C(1，-4).······················································································ 5分

连结AC，则AD 2=3 2+3 2=18，CD 2=(2-1)2+(-3+4)2=2，AC 2=(1+1)2+4 2=20.

∴AD 2+CD 2=AC 2，∴△ACD是直角三角形，且∠ADC=90°.········ 7分

∴AD⊥CD.·························································································· 8分

(3)设直线AD的解析式为y=kx+b，把A(-1，0)，D(2，-3)代入

求得直线BC的解析式为y=-x-1.·················································· 9分

设点P的横坐标为x，则P(x，-x-1)，E(x，x 2-2x-3).

∵点P在点E的上方

∴EP=(-x-1)-(x 2-2x-3)=-x 2+x+2=-(x- )2+ ····· 10分

∴当x= 时，线段PE长度的最大值= .····································· 12分

(4)存在，点F的坐标分别为F1(-3，0)，F2(1，0)，F3( ，0)，F4( ，0).

······················································································· 16分

关于点F坐标的求解过程(原题不作要求，本人添加，仅供参考)

如图3

①若四边形ADQ1F1为平行四边形，则AF1=DQ1，DQ1∥AF1.

∴点Q1的纵坐标为-3，代入y=x 2-2x-3，得x 2-2x-3=-3，∴x1=0，x2=2.

∵D(2，-3)，∴Q1(0，-3)，∴DQ1=2，∴AF1=2.

∴F1(-3，0).

②若四边形AF2DQ2为平行四边形，同理可得F2(1，0).

③若四边形AQ3F3D为平行四边形，则AQ3=DF3.

∴点Q3的纵坐标为3，代入y=x 2-2x-3，得x 2-2x-3=3，∴x3= ，x4= .

-1-( )= ，OF3=2-( )= .

∴F3( ，0).

④若四边形AQ4F4D为平行四边形，则

OF4=( )-( )+( )= ∴F4( ，0).

26.已知二次函数y=ax 2+bx+c(a≠0)的图象经过点A(1，0)，B(2，0)，C(0，-2)，直线x=m(m>2)与x轴交于点D.

(1)求二次函数的解析式;

(2)在直线x=m(m>2)上有一点E(点E在第四象限)，使得E、D、B为顶点的三角形与以A、O、C为顶点的三角形相似，求E点坐标(用含m的代数式表示);

(3)在(2)成立的条件下，抛物线上是否存在一点F，使得四边形ABEF为平行四边形?若存在，请求出m的值及四边形ABEF的面积;若不存在，请说明理由.

解：(1)∵二次函数y=ax 2+bx+c的图象经过点A(1，0)，B(2，0)，C(0，-2)

∴ 解得 ∴二次函数的解析式y=-x 2+3x-2.·············································· 2分

(2)当△EDB∽△AOC时，有 = 或 = ∵AO=1，CO=2，BD=m-2.

当 = 时，得 = ，∴ED= .

∵点E在第四象限，∴E1(m， ).············· 4分

当 = 时，得 = ，∴ED=2m-4.

∵点E在第四象限，∴E2(m，4-2m).············ 6分

(3)假设抛物线上存在一点F，使得四边形ABEF为平行四边形，则

EF=AB=1，点F的横坐标为m-1.

当点E1的坐标为(m， )时，点F1的坐标为(m-1， ).

∵点F1在抛物线的图象上，∴ =-(m-1)2+3(m-1)-2.

∴2m 2-11m+14=0，解得m1= ，m2=2(不合题意，舍去).

∴F1( ，- ).

∴S□ABEF =1× = .········································································ 9分

当点E2的坐标为(m，4-2m)时，点F2的坐标为(m-1，4-2m).

∵点F2在抛物线的图象上，∴4-2m=-(m-1)2+3(m-1)-2.

∴m 2-7m+10=0，解得m1=5，m2=2(不合题意，舍去).

∴F2(4，-6).

∴S□ABEF =1×6=6.········································································· 12分

注：其它解法可参照评分标准给分.

27.已知：t1，t2是方程t 2+2t-24=0，的两个实数根，且t1

(1)求这个抛物线的解析式;

(2)设点P(x，y)是抛物线上一动点，且位于第三象限，四边形OPAQ是以OA为对角线的平行四边形，求□OPAQ的面积S与 之间的函数关系式，并写出自变量 的取值范围;

(3)在(2)的条件下，当□OPAQ的面积为24时，是否存在这样的点P，使□OPAQ为正方形?若存在，求出P点的坐标;若不存在，说明理由.

解：(1)由t 2+2t-24=0，解得t1=-6，t2=4.··············································· 1分

∵t1

∵抛物线y= x 2+bx+c的图象经过点A，B两点

∴ 解得 ∴这个抛物线的解析式为y= x 2+ x+4.

···························································································· 4分

(2)∵点P(x，y)在抛物线上，且位于第三象限，∴y<0，即-y>0.

又∵S=2S△APO=2× ×| OA|·| y |=| OA|·| y |=6| y |

∴S=-6y.·························································································· 6分

=-6( x 2+ x+4)

=-4(x 2+7x+6)

=-4(x+ )2+25.······································································ 7分

令y=0，则 x 2+ x+4=0，解得x1=-6，x2=-1.

∴抛物线与x轴的交点坐标为(-6，0)、(-1，0)

∴x的取值范围为-6

(3)当S=24时，得-4(x+ )2+25=24，解得：x1=-4，x2=-3.· 9分

代入抛物线的解析式得：y1=y2=-4.

∴点P的坐标为(-3，-4)、(-4，-4).

当点P为(-3，-4)时，满足PO=PA，此时，□OPAQ是菱形.

当点P为(-4，-4)时，不满足PO=PA，此时，□OPAQ不是菱形.

·························································································· 10分

要使□OPAQ为正方形，那么，一定有OA⊥PQ，OA=PQ，此时，点的坐标为(-3，-3)，而(-3，-3)不在抛物线y= x 2+ x+4上，故不存在这样的点P，使□OPAQ为正方形. 12分

相关链接：

2012中考数学压轴题及答案40例（7）

2012中考数学压轴题及答案40例（8）