24．(10分)解：(1)设抛物线的表达式为 1分

点在抛物线的图象上．

∴

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

∴抛物线的表达式为············································································· 4分

(2)设窗户上边所在直线交抛物线于C、D两点，D点坐标为(k，t)

已知窗户高1.6m，∴··························································· 5分

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

∴(m)·············································································· 7分

又设最多可安装n扇窗户

∴····················································································· 9分

．

答：最多可安装4扇窗户．···················································································· 10分

(本题不要求学生画出4个表示窗户的小矩形)

91.(10分)某工厂要赶制一批抗震救灾用的大型活动板房．如图，板房一面的形状是由矩形和抛物线的一部分组成，矩形长为12m，抛物线拱高为5.6m．

(1)在如图所示的平面直角坐标系中，求抛物线的表达式．

(2)现需在抛物线AOB的区域内安装几扇窗户，窗户的底边在AB上，每扇窗户宽1.5m，高1.6m，相邻窗户之间的间距均为0.8m，左右两边窗户的窗角所在的点到抛物线的水平距离至少为0.8m．请计算最多可安装几扇这样的窗户？

106.(08四川乐山28题)28．如图(18)，在平面直角坐标系中，的边在轴上，且，以为直径的圆过点．若点的坐标为，，A、B两点的横坐标，是关于的方程的两根．

(1)求、的值；

(2)若平分线所在的直线交轴于点，试求直线对应的一次函数解析式；

(3)过点任作一直线分别交射线、(点除外)于点、．则的是否为定值？若是，求出该定值；若不是，请说明理由．

(08四川乐山28题解析)28．解：(1)以为直径的圆过点，

，而点的坐标为，

由易知，

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

即：，解之得：或．

，，

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

由根与系数关系有：

，

解之，．································································································ 4分

(2)如图(3)，过点作，交于点，

易知，且，

在中，易得，··········· 5分

，

，

又，有，

，······································································································· 6分

，

则，即，························································································ 7分

易求得直线对应的一次函数解析式为：．··················································· 8分

解法二：过作于，于，

由，

求得．········································································································ 5分

又，

求得．····························································································· 7分

即，

易求得直线对应的一次函数解析式为：．··················································· 8分

(3)过点作于，于．

为的平分线，．

由，有········································································· 9分

由，有······································································ 10分

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

即．···················································································· 12分

105.(08湖南邵阳25题)25．如图(十七)，将含角的直角三角板()绕其直角顶点逆时针旋转解()，得到，与相交于点，过点作交于点，连结．设，的面积为，的面积为．

(1)求证：是直角三角形；

(2)试求用表示的函数关系式，并写出的取值范围；

(3)以点为圆心，为半径作，

①当直线与相切时，试探求与之间的关系；

②当时，试判断直线与的位置关系，并说明理由．

(08湖南邵阳25题解析)25． (1)，

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

又，···························································· 2分

，

，即是直角三角形；······························································ 3分

(2)在中，，

，

，，

；··············································································· 4分

；··························· 5分

(3)①直线与相切时，则．

，

．

，································································· 6分

又，

是等边三角形，，

，

又；······································································ 7分

②当时，

则有，解之得或；··················································· 8分

(i)当时，，

在中，，，

在中，，········································· 9分

，即，

直线与相离；···························································································· 10分

(ii)当时，

同理可求出：，·············································· 11分

，

直线与相交．···························································································· 12分

104.(08贵州遵义27题)27。(14分)如图(1)所示，一张平行四边形纸片ABCD，AB=10，AD=6，BD=8，沿对角线BD把这张纸片剪成△AB₁D₁和△CB₂D₂两个三角形(如图(2)所示)，将△AB₁D₁沿直线AB₁方向移动(点B₂始终在AB₁上，AB₁与CD₂始终保持平行)，当点A与B₂重合时停止平移，在平移过程中，AD₁与B₂D₂交于点E，B₂C与B₁D₁交于点F，

(1)当△AB₁D₁平移到图(3)的位置时，试判断四边形B₂FD₁E是什么四边形？并证明你的结论；

(2)设平移距离B₂B₁为x，四边形B₂FD₁E的面积为y，求y与x的函数关系式；并求出四边形B₂FD₁E的面积的最大值；

(3)连结B₁C(请在图(3)中画出)。当平移距离B₂B₁的值是多少时，△ B₁B₂F与△ B₁CF相似？

(08贵州遵义27题解析)解：(1) 四边形B₂FD₁E是矩形。

　 因为△AB₁D₁平移到图(3)的，所以四边形B₂FD₁E是一个平行四边形，又因为在平行四边形ABCD中，AB=10，AD=6，BD=8，则有∠ADB是直角。所以四边形B₂FD₁E是矩形。

(2)因为三角形B₁B₂F与三角形AB₁D₁相似，则有B₂F==0.6X,B₁F==0.8x

所以sB₂FD₁E=B₂F×D₁F=0.6X × (8-0.8x)=4.8x-0.48x²

即y=4.8x-0.48x²=12-0.48(x-5)

当x=5时，y=12是最大的值。

(3)要使△ B₁B₂F与△ B₁CF相似，则有　 即

解之得：x=3.6

103.(08云南省卷24题) 24．(本大题满分14分)如图12，已知二次函数图象的顶点坐标为C(1,0)，直线与该二次函数的图象交于A、B两点，其中A点的坐标为(3,4)，B点在轴上.

　 (1)求的值及这个二次函数的关系式；

(2)P为线段AB上的一个动点(点P与A、B不重合)，过P作轴的垂线与这个二次函数的图象交于点E点，设线段PE的长为，点P的横坐标为，求与之间的函数关系式，并写出自变量的取值范围；

(3)D为直线AB与这个二次函数图象对称轴的交点，在线段AB上是否存在一点P，使得四边形DCEP是平行四形？若存在，请求出此时P点的坐标；若不存在，请说明理由.

(08云南省卷24题解析) (1) ∵ 点A(3,4)在直线y=x+m上，

∴ 4=3+m.　　　　　　　　　　　　　　　　 ………………………………(1分)

∴ m=1.　　　　　　　　 　　　　　　　　　………………………………(2分)

　　　　 设所求二次函数的关系式为y=a(x-1)².　　　 ………………………………(3分)

　　　　 ∵ 点A(3,4)在二次函数y=a(x-1)²的图象上，

　　　　 ∴ 4=a(3-1)²,

　　　　 ∴ a=1.　　　　　　　　　　　　　　　　　 ………………………………(4分)

∴ 所求二次函数的关系式为y=(x-1)².

　 即y=x²-2x+1.　　　 　　　　　　　　　　………………………………(5分)

(2) 设P、E两点的纵坐标分别为y_P和y_E .

∴ PE=h=y_P-y_E　　　　　　　　　　　　　　 ………………………………(6分)

　　　 =(x+1)-(x²-2x+1)　　　　　　　　　　 ………………………………(7分)

　　　 =-x²+3x.　　　　　　　　　　　　　　 ………………………………(8分)

　 即h=-x²+3x (0＜x＜3).　　　　　　　　　 ………………………………(9分)

(3) 存在.　　　　　　　　　　　　　　　　　 ………………………………(10分)

解法1：要使四边形DCEP是平行四边形，必需有PE=DC. …………………(11分)

∵ 点D在直线y=x+1上,

∴ 点D的坐标为(1,2),

∴ -x²+3x=2 .

即x²-3x+2=0 .　　　　　　　　　　　　　 ………………………………(12分)

解之，得　 x₁=2，x₂=1 (不合题意，舍去)　　 ………………………………(13分)

∴ 当P点的坐标为(2,3)时，四边形DCEP是平行四边形.　　 ……………(14分)

解法2：要使四边形DCEP是平行四边形，必需有BP∥CE.　 ………………(11分)

设直线CE的函数关系式为y=x+b.

∵ 直线CE 经过点C(1,0),

∴ 0=1+b,

∴ b=-1 .

∴ 直线CE的函数关系式为y=x-1 .

∴ 　　得x²-3x+2=0.　　　　 ………………………………(12分)

解之，得　 x₁=2，x₂=1 (不合题意，舍去)　 ………………………………(13分)

∴ 当P点的坐标为(2,3)时，四边形DCEP是平行四边形.　 ……………(14分)

102.23．如图9，在平面直角坐标系中，以点为圆心，2为半径作圆，交轴于两点，开口向下的抛物线经过点，且其顶点在上．

(1)求的大小；

(2)写出两点的坐标；

(3)试确定此抛物线的解析式；

(4)在该抛物线上是否存在一点，使线段与互相平分？若存在，求出点的坐标；若不存在，请说明理由．

(08新疆乌鲁木齐23题解答)23．解：(1)作轴，为垂足，

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

，······································ 3分

(2)，半径

，故，············································ 5分

········································································· 6分

(3)由圆与抛物线的对称性可知抛物线的顶点的坐标为··································· 7分

设抛物线解析式·················································································· 8分

把点代入上式，解得······································································· 9分

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

(4)假设存在点使线段与互相平分，则四边形是平行四边形········· 11分

且．

轴，点在轴上．·············································································· 12分

又，，即．

又满足，

点在抛物线上······································································································ 13分

所以存在使线段与互相平分．···························································· 14分

21．解：(1)过作轴于点，如图(第21题图)

在中，，

································· 1分

由对称性可知：

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

点的坐标为····························································································· 3分

(2)设经过的抛物线的解析式为，则

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

解之得

抛物线的解析式为：································································· 5分

(3)与两坐标轴相切

圆心应在第一、三象限或第二、四象限的角平分线上．

即在直线或上·························································································· 6分

若点在直线上，根据题意有

解之得

，

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

若点在直线上，根据题意有

解之得，

的半径为或．······································································ 8分

101. 21．如图，已知平面直角坐标系中，有一矩形纸片，为坐标原点，轴，，现将纸片按如图折叠，为折痕，．折叠后，点落在点，点落在线段上的处，并且与在同一直线上．

(1)求的坐标；

(2)求经过三点的抛物线的解析式；

(3)若的半径为，圆心在(2)的抛物线上运动，

与两坐标轴都相切时，求半径的值．

21、 (2008台湾) 　如图，圆O₁、圆O₂、圆O₃三圆两两相切，为圆O₁、圆O₂的公切线，为半圆，且分别与三圆各切于一点。若圆O₁、圆O₂的半径均为1，则圆O₃的半径为何？(　 )　

A. 1 　　B. 　　　C. －1 　　　　D. +1

答案：C