88.(08山东济宁26题)(12分)

中，，，cm．长为1cm的线段在的边上沿方向以1cm/s的速度向点运动(运动前点与点重合)．过分别作的垂线交直角边于两点，线段运动的时间为s．

(1)若的面积为，写出与的函数关系式(写出自变量的取值范围)；

(2)线段运动过程中，四边形有可能成为矩形吗？若有可能，求出此时的值；若不可能，说明理由；

(3)为何值时，以为顶点的三角形与相似？

(08山东济宁26题解析)解：(1)当点在上时，，．

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

当点在上时，．

．·················································· 4分

(2)，．．

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

由条件知，若四边形为矩形，需，即，

．

当s时，四边形为矩形．································································· 8分

(3)由(2)知，当s时，四边形为矩形，此时，

．··························································································· 9分

除此之外，当时，，此时．

，．．····························· 10分

，．

又，．········································ 11分

，．

当s或s时，以为顶点的三角形与相似．··················· 12分

87.(08青海省卷28题)王亮同学善于改进学习方法，他发现对解题过程进行回顾反思，效果会更好．某一天他利用30分钟时间进行自主学习．假设他用于解题的时间(单位：分钟)与学习收益量的关系如图甲所示，用于回顾反思的时间(单位：分钟)与学习收益量的关系如图乙所示(其中是抛物线的一部分，为抛物线的顶点)，且用于回顾反思的时间不超过用于解题的时间．

(1)求王亮解题的学习收益量与用于解题的时间之间的函数关系式，并写出自变量的取值范围；

(2)求王亮回顾反思的学习收益量与用于回顾反思的时间之间的函数关系式；

(3)王亮如何分配解题和回顾反思的时间，才能使这30分钟的学习收益总量最大？

(学习收益总量解题的学习收益量回顾反思的学习收益量)

(08青海省卷28题解析)解：(1)设，

把代入，得．

．······································································································ (1分)

自变量的取值范围是：．···························································· (2分)

(2)当时，

设，···················································································· (3分)

把代入，得，．

．································································· (5分)

当时，

············································································································· (6分)

即．

(3)设王亮用于回顾反思的时间为分钟，学习效益总量为，

则他用于解题的时间为分钟．

当时，

．························· (7分)

当时，．············································································ (8分)

当时，

．····································································· (9分)

随的增大而减小，

当时，．

综合所述，当时，，此时．································ (10分)

即王亮用于解题的时间为26分钟，用于回顾反思的时间为4分钟时，学习收益总量最大．

······················································································································· (11分)

86.(08青海西宁28题)如图14，已知半径为1的与轴交于两点，为的切线，切点为，圆心的坐标为，二次函数的图象经过两点．

(1)求二次函数的解析式；

(2)求切线的函数解析式；

(3)线段上是否存在一点，使得以为顶点的三角形与相似．若存在，请求出所有符合条件的点的坐标；若不存在，请说明理由．

(08青海西宁28题解析)解：(1)圆心的坐标为，半径为1，，……1分

二次函数的图象经过点，

可得方程组················································································ 2分

解得：二次函数解析式为············································· 3分

(2)过点作轴，垂足为．······························································· 4分

是的切线，为切点，(圆的切线垂直于经过切点的半径)．

在中，

为锐角，···························· 5分

，

在中，．

．

点坐标为························································································· 6分

设切线的函数解析式为，由题意可知，······ 7分

切线的函数解析式为···································································· 8分

(3)存在．············································································································ 9分

①过点作轴，与交于点．可得(两角对应相等两三角形相似)

，············································ 10分

②过点作，垂足为，过点作，垂足为．

可得(两角对应相等两三角开相似)

在中，，，

在中，，

，······································· 11分

符合条件的点坐标有，······················································ 12分

85.(08内蒙古赤峰25题)(本题满分14分)

在平面直角坐标系中给定以下五个点．

(1)请从五点中任选三点，求一条以平行于轴的直线为对称轴的抛物线的解析式；

(2)求该抛物线的顶点坐标和对称轴，并画出草图；

(3)已知点在抛物线的对称轴上，直线过点且垂直于对称轴．验证：以为圆心，为半径的圆与直线相切．请你进一步验证，以抛物线上的点为圆心为半径的圆也与直线相切．由此你能猜想到怎样的结论．

(08内蒙古赤峰25题解析)25．解：(1)设抛物线的解析式为，

且过点，

由在H ．

则．········································································································ (2分)

得方程组，

解得．

抛物线的解析式为················ (4分)

(2)由············· (6分)

得顶点坐标为，对称轴为．·········· (8分)

(3)①连结，过点作直线的垂线，垂足为，

则．

在中，，，

，

，

以点为圆心，为半径的与直线相切．····························· (10分)

②连结过点作直线的垂线，垂足为．过点作垂足为，

则．

在中，，．

．

以点为圆心为半径的与直线相切．································ (12分)

③以抛物线上任意一点为圆心，以为半径的圆与直线相切．····· (14分)

84.(08辽宁12市26题)(本题14分)26．如图16，在平面直角坐标系中，直线与轴交于点，与轴交于点，抛物线经过三点．

(1)求过三点抛物线的解析式并求出顶点的坐标；

(2)在抛物线上是否存在点，使为直角三角形，若存在，直接写出点坐标；若不存在，请说明理由；

(3)试探究在直线上是否存在一点，使得的周长最小，若存在，求出点的坐标；若不存在，请说明理由．

(08辽宁12市26题解析)

解：(1)直线与轴交于点，与轴交于点．

，························································································· 1分

点都在抛物线上，

抛物线的解析式为························································ 3分

顶点······························································································· 4分

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

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

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

(3)存在·············································································································· 10分

理由：

解法一：

延长到点，使，连接交直线于点，则点就是所求的点．

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

过点作于点．

点在抛物线上，

在中，，

，，

在中，，

，，··············································· 12分

设直线的解析式为

　 解得

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

　 解得　

在直线上存在点，使得的周长最小，此时．··· 14分

解法二：

过点作的垂线交轴于点，则点为点关于直线的对称点．连接交于点，则点即为所求．················································································ 11分

过点作轴于点，则，．

，

同方法一可求得．

在中，，，可求得，

为线段的垂直平分线，可证得为等边三角形，

垂直平分．

即点为点关于的对称点．············································· 12分

设直线的解析式为，由题意得

　 解得

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

　 解得　

在直线上存在点，使得的周长最小，此时．··· 14分

83.(08辽宁沈阳26题)(本题14分)26．如图所示，在平面直角坐标系中，矩形的边在轴的负半轴上，边在轴的正半轴上，且，，矩形绕点按顺时针方向旋转后得到矩形．点的对应点为点，点的对应点为点，点的对应点为点，抛物线过点．

(1)判断点是否在轴上，并说明理由；

(2)求抛物线的函数表达式；

(3)在轴的上方是否存在点，点，使以点为顶点的平行四边形的面积是矩形面积的2倍，且点在抛物线上，若存在，请求出点，点的坐标；若不存在，请说明理由．

(08辽宁沈阳26题解析)解：(1)点在轴上················································· 1分

理由如下：

连接，如图所示，在中，，，

，

由题意可知：

点在轴上，点在轴上．········································································· 3分

(2)过点作轴于点

，

在中，，

点在第一象限，

点的坐标为·························································································· 5分

由(1)知，点在轴的正半轴上

点的坐标为

点的坐标为···························································································· 6分

抛物线经过点，

由题意，将，代入中得

　 解得

所求抛物线表达式为：······················································· 9分

(3)存在符合条件的点，点．········································································· 10分

理由如下：矩形的面积

以为顶点的平行四边形面积为．

由题意可知为此平行四边形一边，

又

边上的高为2··································································································· 11分

依题意设点的坐标为

点在抛物线上

解得，，

，

以为顶点的四边形是平行四边形，

，，

当点的坐标为时，

点的坐标分别为，；

当点的坐标为时，

点的坐标分别为，．··············································· 14分

82.(08广东肇庆25题)(本小题满分10分)

已知点A(a，)、B(2a，y)、C(3a，y)都在抛物线上.

(1)求抛物线与x轴的交点坐标；

(2)当a=1时，求△ABC的面积；

(3)是否存在含有、y、y，且与a无关的等式？如果存在，试给出一个，并加以证明；如果不存在，说明理由.

(08广东肇庆25题解析)(本小题满分10分)

解：(1)由5=0，·············································································· (1分)

得，．·················································································· (2分)

∴抛物线与x轴的交点坐标为(0，0)、(，0)．······································· (3分)

(2)当a=1时，得A(1，17)、B(2，44)、C(3，81)，······························ (4分)

分别过点A、B、C作x轴的垂线，垂足分别为D、E、F，则有

=S - - ···················································· (5分)

　　　　 =--···································· (6分)

=5(个单位面积)········································································ (7分)

(3)如：． ········································································· (8分)

事实上， =45a²+36a．　　　　　　　　　　　　　　　

　　　　 3()=3[5×(2a)²+12×2a-(5a²+12a)] =45a²+36a．············ (9分)

∴． ···················································································· (10分)

81.(08广东茂名25题)(本题满分10分)

如图，在平面直角坐标系中，抛物线=－++经过A(0，－4)、B(，0)、 C(，0)三点，且-=5．

(1)求、的值；(4分)

(2)在抛物线上求一点D，使得四边形BDCE是以BC为对　　 角线的菱形；(3分)

(3)在抛物线上是否存在一点P，使得四边形BPOH是以OB为对角线的菱形？若存在，求出点P的坐标，并判断这个菱形是否为正方形？若不存在，请说明理由．(3分)

解：　　　　　　　　　　　

(08广东茂名25题解析)解：(1)解法一：

∵抛物线=－++经过点A(0，－4)，

　 ∴=－4 ……1分

又由题意可知，、是方程－++=0的两个根，

∴+=，　 =－=6··································································· 2分

由已知得(-)=25

又(-)=(+)－4=－24

∴ －24=25　　　　　　　　　　　 　　　　　　　

解得=± ··········································································································· 3分

当=时，抛物线与轴的交点在轴的正半轴上，不合题意，舍去．

∴=－． ·········································································································· 4分

解法二：∵、是方程－++c=0的两个根，

　即方程2－3+12=0的两个根．

∴=，··········································································· 2分

∴－==5，

　　　　 解得 =±······························································································· 3分

　　　　 (以下与解法一相同．)　　

　　 (2)∵四边形BDCE是以BC为对角线的菱形，根据菱形的性质，点D必在抛物线的对称轴上， 　　 5分

　　　　　 又∵=－－－4=－(+)+　 ································· 6分

　　　　　　 ∴抛物线的顶点(－，)即为所求的点D．······································· 7分

　　 (3)∵四边形BPOH是以OB为对角线的菱形，点B的坐标为(－6，0)，

根据菱形的性质，点P必是直线=-3与

抛物线=－--4的交点， ···························································· 8分

　　　　 ∴当=－3时，=－×(－3)－×(－3)－4=4，

　　　　 ∴在抛物线上存在一点P(－3，4)，使得四边形BPOH为菱形． ·················· 9分

　　　　　 四边形BPOH不能成为正方形，因为如果四边形BPOH为正方形，点P的坐标只能是(－3，3)，但这一点不在抛物线上．······································································································· 10分

76.(08天津市卷26题)(本小题10分)

已知抛物线，

(Ⅰ)若，，求该抛物线与轴公共点的坐标；

(Ⅱ)若，且当时，抛物线与轴有且只有一个公共点，求的取值范围；

(Ⅲ)若，且时，对应的；时，对应的，试判断当时，抛物线与轴是否有公共点？若有，请证明你的结论；若没有，阐述理由．

(08天津市卷26题解析)解(Ⅰ)当，时，抛物线为，

方程的两个根为，．

∴该抛物线与轴公共点的坐标是和．　 ················································ 2分

(Ⅱ)当时，抛物线为，且与轴有公共点．

对于方程，判别式≥0，有≤． ········································ 3分

①当时，由方程，解得．

此时抛物线为与轴只有一个公共点．································· 4分

②当时，

时，，

时，．

由已知时，该抛物线与轴有且只有一个公共点，考虑其对称轴为，

应有　 即

解得．

综上，或．　　 ················································································ 6分

(Ⅲ)对于二次函数，

由已知时，；时，，

又，∴．

于是．而，∴，即．

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

∵关于的一元二次方程的判别式

，　

∴抛物线与轴有两个公共点，顶点在轴下方．····························· 8分

又该抛物线的对称轴，

由，，，

得，

∴．

又由已知时，；时，，观察图象，

可知在范围内，该抛物线与轴有两个公共点． ············································ 10分

77(08湖北宜昌25题)如图1，已知四边形OABC中的三个顶点坐标为O(0，0)，A(0，n)，C(m，0)．动点P从点O出发依次沿线段OA，AB，BC向点C移动，设移动路程为z，△OPC的面积S随着z的变化而变化的图象如图2所示．m，n是常数， m＞1，n＞0．

(1)请你确定n的值和点B的坐标；

(2)当动点P是经过点O，C的抛物线y＝ax+bx+c的顶点，且在双曲线y＝上时，求这时四边形OABC的面积．

(08湖北宜昌25题解析)解：(1) 从图中可知，当P从O向A运动时，△POC的面积S＝mz， z由0逐步增大到2,则S由0逐步增大到m，故OA＝2，n＝2 . (1分)

同理，AB＝1，故点B的坐标是(1，2).(2分)

(2)解法一：

∵抛物线y＝ax+bx+c经过点O(0，0)，C(m ，0),∴c＝0，b＝－am，(3分)

∴抛物线为y＝ax－amx，顶点坐标为(，－am²).(4分)

如图1，设经过点O，C，P的抛物线为l.

当P在OA上运动时，O,P都在y轴上，

这时P,O,C三点不可能同在一条抛物线上，

∴这时抛物线l不存在, 故不存在m的值..①

当点P与C重合时，双曲线y＝不可能经过P,

故也不存在m的值.②(5分)

(说明：①②任做对一处评1分，两处全对也只评一分)

当P在AB上运动时，即当0<x≤1时，y＝2，

抛物线l的顶点为P(，2).

∵P在双曲线y＝上，可得 m＝，∵>2,与 x＝≤1不合，舍去.(6分)③

容易求得直线BC的解析式是：，(7分)

当P在BC上运动，设P的坐标为　 (x,y)，当P是顶点时 x＝，

故得y＝＝，顶点P为(，)，

∵1< x＝<m，∴m>2，又∵P在双曲线y＝上，

于是，×＝，化简后得5m－22m+22＝0,

解得,,(8分)

与题意2<x＝<m不合,舍去.④(9分)

故由①②③④，满足条件的只有一个值：.

这时四边形OABC的面积＝＝.(10分)

(2)解法二：

∵抛物线y＝ax+bx+c经过点O(0，0)，C(m ，0)

∴c＝0，b＝－am，(3分)

∴抛物线为y＝ax－amx，顶点坐标P为(，－am²). (4分)

∵m＞1，∴＞0，且≠m，

∴P不在边OA上且不与C重合. (5分)

∵P在双曲线y＝上，∴×(－ am²)＝即a＝－ .

.①当1＜m≤2时，＜≤1，如图2,分别过B，P作x轴的垂线，

M，N为垂足，此时点P在线段AB上，且纵坐标为2，

∴－am²＝2，即a＝－.

而a＝－ ，∴－ ＝－，m＝＞2，而1＜m≤2，不合题意，舍去.(6分)

②当m≥2时，＞1，如图3,分别过B，P作x轴的垂线，M，N为垂足，ON＞OM，

此时点P在线段CB上，易证Rt△BMC∽Rt△PNC，

∴BM∶PN＝MC∶NC，即:　 2∶PN＝(m－1)∶，∴PN＝(7分)

而P的纵坐标为－ am²，∴＝－ am²，即a＝

而a＝－，∴－ ＝

化简得：5m²－22m+22＝0.解得：m＝ ，(8分)

但m≥2，所以m＝舍去，(9分)

取m ＝ .

由以上,这时四边形OABC的面积为：

(AB+OC) ×OA＝(1+m) ×2＝. (10分)

74.(08广东东莞22题)(本题满分9分)将两块大小一样含30°角的直角三角板，叠放在一起，使得它们的斜边

AB重合，直角边不重合，已知AB=8，BC=AD=4，AC与BD相交于点E，连结CD．

(1)填空：如图9，AC=　　　　 ，BD=　　　　 ；四边形ABCD是　　　 梯形.

(2)请写出图9中所有的相似三角形(不含全等三角形).

(3)如图10，若以AB所在直线为轴，过点A垂直于AB的直线为轴建立如图10的平面直角坐标系，保持ΔABD不动，将ΔABC向轴的正方向平移到ΔFGH的位置，FH与BD相交于点P，设AF=t，ΔFBP面积为S，求S与t之间的函数关系式，并写出t的取值值范围.

(08广东东莞22题解析)解：(1)，，…………………………1分

等腰；…………………………2分

　 (2)共有9对相似三角形.(写对3－5对得1分，写对6－8对得2分，写对9对得3分)

　 　①△DCE、△ABE与△ACD或△BDC两两相似，分别是：△DCE∽△ABE，△DCE∽△ACD，△DCE∽△BDC，△ABE∽△ACD，△ABE∽△BDC；(有5对)

②△ABD∽△EAD，△ABD∽△EBC；(有2对)

③△BAC∽△EAD，△BAC∽△EBC；(有2对)

所以，一共有9对相似三角形.…………………………………………5分

(3)由题意知，FP∥AE，

　　 ∴ ∠1＝∠PFB，

又∵ ∠1＝∠2＝30°，

　 ∴ ∠PFB＝∠2＝30°,

∴ FP＝BP.…………………………6分

过点P作PK⊥FB于点K，则.

∵ AF＝t，AB＝8，

∴ FB＝8－t，.

在Rt△BPK中，. ……………………7分

∴ △FBP的面积，

∴ S与t之间的函数关系式为：

　　　 ，或. …………………………………8分

t的取值范围为：. …………………………………………………………9分

75(08甘肃兰州28题)(本题满分12分)如图19-1，是一张放在平面直角坐标系中的矩形纸片，为原点，点在轴的正半轴上，点在轴的正半轴上，，．

(1)在边上取一点，将纸片沿翻折，使点落在边上的点处，求两点的坐标；

(2)如图19-2，若上有一动点(不与重合)自点沿方向向点匀速运动，运动的速度为每秒1个单位长度，设运动的时间为秒()，过点作的平行线交于点，过点作的平行线交于点．求四边形的面积与时间之间的函数关系式；当取何值时，有最大值？最大值是多少？

(3)在(2)的条件下，当为何值时，以为顶点的三角形为等腰三角形，并求出相应的时刻点的坐标．

(08甘肃兰州28题解析)(本题满分12分)

解：(1)依题意可知，折痕是四边形的对称轴，

在中，，．

．．

点坐标为(2，4)．································································································ 2分

在中，，　 又．

　．　 解得：．

点坐标为···································································································· 3分

(2)如图①，．

，又知，，

， 又．

而显然四边形为矩形．

·························································· 5分

，又

当时，有最大值．········································································ 6分

(3)(i)若以为等腰三角形的底，则(如图①)

在中，，，为的中点，

．

又，为的中点．

过点作，垂足为，则是的中位线，

，，

当时，，为等腰三角形．

此时点坐标为．··························································································· 8分

(ii)若以为等腰三角形的腰，则(如图②)

在中，．

过点作，垂足为．

，．

．

，．

，，

当时，()，此时点坐标为．·························· 11分

综合(i)(ii)可知，或时，以为顶点的三角形为等腰三角形，相应点的坐标为或．·········································································································· 12分