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：

16.如图，已知与 轴交于点 和 的抛物线 的顶点为 ，抛物线 与 关于 轴对称，顶点为 .

(1)求抛物线 的函数关系式;

(2)已知原点 ，定点 ， 上的点 与 上的点 始终关于 轴对称，则当点 运动到何处时，以点 为顶点的四边形是平行四边形?

(3)在 上是否存在点 ，使 是以 为斜边且一个角为 的直角三角形?若存，求出点 的坐标;若不存在，说明理由.

解：(1)由题意知点 的坐标为 .

设 的函数关系式为 .

又 点 在抛物线 上，

，解得 .

抛物线 的函数关系式为 (或 ).

(2) 与 始终关于 轴对称，

与 轴平行.

设点 的横坐标为 ，则其纵坐标为 ，

， ，即 .

当 时，解得 .

当 时，解得 .

当点 运动到 或 或 或 时，

，以点 为顶点的四边形是平行四边形.

(3)满足条件的点 不存在.理由如下：若存在满足条件的点 在 上，则

， (或 )，

.

过点 作 于点 ，可得 .

， ， .

点 的坐标为 .

但是，当 时， .

不存在这样的点 构成满足条件的直角三角形.

17.如图，抛物线y=-x 2+bx+c与x轴交于A(1，0)，B(-3，0)两点.

(1)求该抛物线的解析式;

(2)设(1)中的抛物线交y轴于C点，在该抛物线的对称轴上是否存在点Q，使得△QAC的周长最小?若存在，求出点Q的坐标;若不存在，请说明理由;

(3)在(1)中的抛物线上的第二象限内是否存在一点P，使△PBC的面积最大?，若存在，求出点P的坐标及△PBC的面积最大值;若不存在，请说明理由.

解：(1)将A(1，0)，B(-3，0)代入y=-x 2+bx+c得

················································································ 2分

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

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

(2)存在.····································································································· 5分

该抛物线的对称轴为x=- =-1

∵抛物线交x轴于A、B两点，∴A、B两点关于抛物线的对称轴x=-1对称.

由轴对称的性质可知，直线BC与x=-1的交点即为所求的Q点，此时△QAC的周长最小，如图1.

将x=0代入y=-x 2-2x+3，得y=3.

∴点C的坐标为(0，3).

设直线BC的解析式为y=kx+b1，

将B(-3，0)，C(0，3)代入，得

解得 ∴直线BC的解析式为y=x+3.··············· 6分

联立 解得 ∴点Q的坐标为(-1，2).··································································· 7分

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

设P点的坐标为(x，-x 2-2x+3)(-3

∵S△PBC =S四边形PBOC -S△BOC =S四边形PBOC - ×3×3=S四边形PBOC - 当S四边形PBOC有最大值时，S△PBC就最大.

∵S四边形PBOC =SRt△PBE+S直角梯形PEOC ························································· 9分

= BE·PE+ (PE+OC)·OE

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

=- (x+ )2+ + 当x=- 时，S四边形PBOC最大值为 + .

∴S△PBC最大值= + - = .············ 10分

当x=- 时，-x 2-2x+3=-(- )2-2×(- )+3= .

∴点P的坐标为(- ， ).···························································· 11分

18.如图，已知抛物线y=a(x-1)2+ (a≠0)经过点A(-2，0)，抛物线的顶点为D，过O作射线OM∥AD.过顶点D平行于 轴的直线交射线OM于点C，B在 轴正半轴上，连结BC.

(1)求该抛物线的解析式;

(2)若动点P从点O出发，以每秒1个长度单位的速度沿射线OM运动，设点P运动的时间为t(s).问：当t为何值时，四边形DAOP分别为平行四边形?直角梯形?等腰梯形?

(3)若OC=OB，动点P和动点Q分别从点O和点B同时出发，分别以每秒1个长度单位和2个长度单位的速度沿OC和BO运动，当其中一个点停止运动时另一个点也随之停止运动.设它们的运动的时间为t(s)，连接PQ，当t为何值时，四边形BCPQ的面积最小?并求出最小值及此时PQ的长.

解：(1)把A(-2，0)代入y=a(x-1)2+ ，得0=a(-2-1)2+ .

∴a=- ····························································································· 1分

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

(2)设点D的坐标为(xD，yD)，由于D为抛物线的顶点

∴xD=- =1，yD=- ×1 2+ ×1+ = .

∴点D的坐标为(1， ).

如图，过点D作DN⊥x轴于N，则DN= ，AN=3，∴AD= =6.

∴∠DAO=60°······················································································· 4分

∵OM∥AD

①当AD=OP时，四边形DAOP为平行四边形.

∴OP=6

∴t=6(s)································································ 5分

②当DP⊥OM时，四边形DAOP为直角梯形.

过点O作OE⊥AD轴于E.

在Rt△AOE中，∵AO=2，∠EAO=60°，∴AE=1.

(注：也可通过Rt△AOE∽Rt△AND求出AE=1)

∵四边形DEOP为矩形，∴OP=DE=6-1=5.

∴t=5(s)····························································································· 6分

③当PD=OA时，四边形DAOP为等腰梯形，此时OP=AD-2AE=6-2=4.

∴t=4(s)

综上所述，当t=6s、5s、4s时，四边形DAOP分别为平行四边形、直角梯形、等腰梯形.

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

(3)∵∠DAO=60°，OM∥AD，∴∠COB=60°.

又∵OC=OB，∴△COB是等边三角形，∴OB=OC=AD=6.

∵BQ=2t，∴OQ=6-2t(0

过点P作PF⊥x轴于F，则PF= t.················································· 8分

∴S四边形BCPQ =S△COB -S△POQ

= ×6× - ×(6-2t)× t

= (t- )2+ ······················································· 9分

∴当t= (s)时，S四边形BCPQ的最小值为 .······························ 10分

此时OQ=6-2t=6-2× =3，OP= ，OF= ，∴QF=3- = ，PF= .

∴PQ= = = ······································· 11分

19.如图，已知直线y=- x+1交坐标轴于A、B两点，以线段AB为边向上作正方形ABCD，过点A，D，C的抛物线与直线另一个交点为E.

(1)请直接写出点C，D的坐标;

(2)求抛物线的解析式;

(3)若正方形以每秒 个单位长度的速度沿射线AB下滑，直至顶点D落在x轴上时停止.设正方形落在x轴下方部分的面积为S，求S关于滑行时间t的函数关系式，并写出相应自变量t的取值范围;

(4)在(3)的条件下，抛物线与正方形一起平移，直至顶点D落在x轴上时停止，求抛物线上C、E两点间的抛物线弧所扫过的面积.

解：(1)C(3，2)，D(1，3);················································································ 2分

(2)设抛物线的解析式为y=ax 2+bx+c，把A(0，1)，D(1，3)，C(3，2)代入

得 解得 ···················································· 4分

∴抛物线的解析式为y=- x 2+ x+1;········································ 5分

(3)①当点A运动到点F(F为原B点的位置)时

∵AF= = ，∴t= =1(秒).

当0< t ≤1时，如图1.

B′F=AA′= t

∵Rt△AOF∽Rt△∠GB ′F，∴ = .

∴B ′G= ·B ′F= × t= t

正方形落在x轴下方部分的面积为S即为△B ′FG的面积S△B′FG

∴S=S△B′FG= B ′F·B ′G= × t× t= t 2······················ 7分

②当点C运动到x轴上时

∵Rt△BCC ′∽Rt△∠AOB，∴ = .

∴CC ′= ·BC= × = ，∴t= =2(秒).

当1< t ≤2时，如图2.

∵A ′B ′=AB= ，∴A ′F= t- .

∴A ′G= ∵B ′H= t

∴S=S梯形A′B′HG= (A ′G+B ′H)·A ′B ′

= ( + t)· = t- ······································································ 9分

③当点D运动到x轴上时

DD′= t= =3(秒)

当2< t ≤3时，如图3.

∵A ′G= ∴GD′= - = ∴D′H= -

∴S△D′GH = ( )( - )=( )2

∴S=S正方形A′B′C′D′ -S△D′GH

=( )2-( )2

=- t 2+ t- ······································································ 11分

(4)如图4，抛物线上C、E两点间的抛物线弧所扫过的面积为图中阴影部分的面积.

∵t=3，BB′=AA′=DD′= ∴S阴影=S矩形BB′C′C ············································································ 13分

=BB′·BC

= × =15··························································································· 14分

20.已知：抛物线y=x 2-2x+a(a <0)与y轴相交于点A，顶点为M.直线y= x-a分别与x轴，y轴相交于B，C两点，并且与直线AM相交于点N.

(1)填空：试用含a的代数式分别表示点M与N的坐标，则M( ， )，N( ， );

(2)如图，将△NAC沿 轴翻折，若点N的对应点N ′恰好落在抛物线上，AN ′与 轴交于点D，连结CD，求a的值和四边形ADCN的面积;

(3)在抛物线y=x 2-2x+a(a <0)上是否存在一点P，使得以P，A，C，N为顶点的四边形是平行四边形?若存在，求出P点的坐标;若不存在，试说明理由.

解：(1)M(1，a-1)，N( a，- a).······························································· 4分

(2)∵点N ′是△NAC沿 轴翻折后点N的对应点

∴点N ′与点N关于y轴对称，∴N ′(- a，- a).

将N ′(- a，- a)代入y=x 2-2x+a，得- a=(- a)2-2×(- a)+a

整理得4a 2+9a=0，解得a1=0(不合题意，舍去)，a2=- .·· 6分

∴N ′(3， )，∴点N到 轴的距离为3.

∵a=- ，抛物线y=x 2-2x+a与y轴相交于点A，∴A(0，- ).

∴直线AN ′的解析式为y=x - ，将y=0代入，得x = .

∴D( ，0)，∴点D到 轴的距离为 .

∴S四边形ADCN =S△ACN +S△ACN = × ×3+ × × = ·········· 8分

(3)如图，当点P在y轴的左侧时，若四边形ACPN是平行四边形，则PN平行且等于AC.

∴将点N向上平移-2a个单位可得到点P，其坐标为( a，- a)，代入抛物线的解析式，得：- a=( a)2-2× a+a，整理得8a 2+3a=0.

解得a1=0(不合题意，舍去)，a2=- .

∴P(- ， )···················································································· 10分

当点P在y轴的右侧时，若四边形APCN是平行四边形，

则AC与PN互相平分.

∴OA=OC，OP=ON，点P与点N关于原点对称.

∴P(- a， a)，代入y=x 2-2x+a，得

a=(- a)2-2×(- a)+a，整理得8a 2+15a=0.

解得a1=0(不合题意，舍去)，a2=- .

∴P( ，- )····················································································· 12分

∴存在这样的点P，使得以P，A，C，N为顶点的四边形是平行四边形，点P的坐标为

(- ， )或( ，- ).

相关链接：

2012中考数学压轴题及答案40例（6）

2012中考数学压轴题及答案40例（7）