22. 解：( 1)由已知得：解得

c=3,b=2

∴抛物线的线的解析式为

(2)由顶点坐标公式得顶点坐标为(1，4)

所以对称轴为x=1,A,E关于x=1对称，所以E(3,0)

设对称轴与x轴的交点为F

所以四边形ABDE的面积=

=

=

=9

(3)相似

如图，BD=

BE=

DE=

所以, 即： ,所以是直角三角形

所以,且,

所以.

21.解：

(1)m=-5,n=-3

　 (2)y=x+2

(3)是定值.

因为点D为∠ACB的平分线，所以可设点D到边AC,BC的距离均为h，

设△ABC AB边上的高为H,

则利用面积法可得：

(CM+CN)h=MN﹒H

又 H=

化简可得　 (CM+CN)﹒

故 　　　　

20. 解：(1)如图，过点B作BD⊥OA于点D.

　　　 在Rt△ABD中，

　　　 ∵∣AB∣=,sin∠OAB=,

　　　 ∴∣BD∣=∣AB∣·sin∠OAB

　　　　　　　 =×=3.

又由勾股定理，得

∴∣OD∣=∣OA∣-∣AD∣=10-6=4.

∵点B在第一象限，∴点B的坐标为(4，3).　　　　　　　　　　　　 ……3分

设经过O(0,0)、C(4，-3)、A(10,0)三点的抛物线的函数表达式为

　　 y=ax²+bx(a≠0).

由

∴经过O、C、A三点的抛物线的函数表达式为　　　　　　 ……2分

(2)假设在(1)中的抛物线上存在点P，使以P、O、C、A为顶点的四边形为梯形

　 ①∵点C(4，-3)不是抛物线的顶点，

∴过点C做直线OA的平行线与抛物线交于点P₁ 　.

则直线CP₁的函数表达式为y=-3.

对于，令y=-3x=4或x=6.

∴

而点C(4，-3)，∴P₁(6,-3).

在四边形P₁AOC中，CP₁∥OA,显然∣CP₁∣≠∣OA∣.

∴点P₁(6，-3)是符合要求的点.　　　　　　　　　　　　　　　　　 ……1分

②若AP₂∥CO.设直线CO的函数表达式为

　 将点C(4，-3)代入，得

∴直线CO的函数表达式为

　 于是可设直线AP₂的函数表达式为

将点A(10，0)代入，得

∴直线AP₂的函数表达式为

由，即(x-10)(x+6)=0.

∴

而点A(10，0)，∴P₂(-6，12).

过点P₂作P₂E⊥x轴于点E，则∣P₂E∣=12.

在Rt△AP₂E中，由勾股定理，得

而∣CO∣=∣OB∣=5.

∴在四边形P₂OCA中，AP₂∥CO,但∣AP₂∣≠∣CO∣.

∴点P₂(-6，12)是符合要求的点.　　　　　　　　　　　　　　　　　　　 ……1分

③若OP₃∥CA,设直线CA的函数表达式为y=k₂x+b₂

　 将点A(10,0)、C(4,-3)代入，得

∴直线CA的函数表达式为

∴直线OP₃的函数表达式为

由即x(x-14)=0.

∴

而点O(0,0),∴P₃(14，7).

过点P₃作P₃E⊥x轴于点E，则∣P₃E∣=7.

在Rt△OP₃E中，由勾股定理，得

而∣CA∣=∣AB∣=.

∴在四边形P₃OCA中，OP₃∥CA,但∣OP₃∣≠∣CA∣.

∴点P₃(14，7)是符合要求的点.　　　　　　　　　　　　　　　　　　　 ……1分

综上可知，在(1)中的抛物线上存在点P₁(6,-3)、P₂(-6,12)、P₃(14,7),

使以P、O、C、A为顶点的四边形为梯形.　　　　　　　　　　　　　　　　 ……1分

(3)由题知，抛物线的开口可能向上，也可能向下.

　①当抛物线开口向上时，则此抛物线与y轴的副半轴交与点N.

可设抛物线的函数表达式为(a＞0).

即

如图，过点M作MG⊥x轴于点G.

∵Q(-2k，0)、R(5k，0)、G(、N(0，-10ak²)、M

∴

∴　　　　　　　　　　　　　　　 ……2分

②当抛物线开口向下时，则此抛物线与y轴的正半轴交于点N，

　 同理，可得　　　　　　　　　　　　　　　　　 　　　……1分

综上所知，的值为3：20.　　　　　　　　　　　　　　　　　　 ……1分

19. 解：(1)在中，令

，

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

又点在上

的解析式为···················································································· 2分

(2)由，得　 ·························································· 4分

，

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

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

(3)过点作于点

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

················································································································ 8分

由直线可得：

在中，，，则

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

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

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

此抛物线开口向下，当时，

当点运动2秒时，的面积达到最大，最大为．　　　

18. 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

(以上答案仅供参考，如有其它做法，可参照给分)

17. 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．　　 1

16.

解：(1)，．

(2)当时，过点作，交于，如图1，

则，，

，．

(3)①能与平行．

若，如图2，则，

即，，而，

．

②不能与垂直．

若，延长交于，如图3，

则．

．

．

又，，

，

，而，

不存在．

15. 解：(1)解法1：根据题意可得：A(-1,0)，B(3,0)；

则设抛物线的解析式为(a≠0)

又点D(0，-3)在抛物线上，∴a(0+1)(0-3)=-3，解之得：a=1

　∴y=x²-2x-3···································································································· 3分

自变量范围：-1≤x≤3···················································································· 4分

　　　　　 解法2：设抛物线的解析式为(a≠0)

　　　　　　　　　 根据题意可知，A(-1,0)，B(3,0)，D(0，-3)三点都在抛物线上

　　　　　　　　　 ∴，解之得：

∴y=x²-2x-3··············································································· 3分

自变量范围：-1≤x≤3······························································ 4分

　　　　　 (2)设经过点C“蛋圆”的切线CE交x轴于点E，连结CM，

　　　　　　 在Rt△MOC中，∵OM=1，CM=2，∴∠CMO=60°,OC=

　　　　　　 在Rt△MCE中，∵OC=2，∠CMO=60°，∴ME=4

　∴点C、E的坐标分别为(0，)，(-3,0) ·················································· 6分

∴切线CE的解析式为··························································· 8分

(3)设过点D(0，-3)，“蛋圆”切线的解析式为：y=kx-3(k≠0) ·························· 9分

　　　　　　　 由题意可知方程组只有一组解

　 即有两个相等实根，∴k=-2············································· 11分

　 ∴过点D“蛋圆”切线的解析式y=-2x-3····················································· 12分

14. 解：(1)由题意可知，．

解，得 m＝3．　　　　 ………………………………3分

∴ A(3，4)，B(6，2)；

∴ k＝4×3=12．　 　　……………………………4分

(2)存在两种情况，如图：　

①当M点在x轴的正半轴上，N点在y轴的正半轴

上时，设M₁点坐标为(x₁，0)，N₁点坐标为(0，y₁)．

∵ 四边形AN₁M₁B为平行四边形，

∴ 线段N₁M₁可看作由线段AB向左平移3个单位，

再向下平移2个单位得到的(也可看作向下平移2个单位，再向左平移3个单位得到的)．

由(1)知A点坐标为(3，4)，B点坐标为(6，2)，

∴ N₁点坐标为(0，4－2)，即N₁(0，2)；　　　 ………………………………5分

M₁点坐标为(6－3，0)，即M₁(3，0)．　　　 ………………………………6分

设直线M₁N₁的函数表达式为，把x＝3，y＝0代入，解得．

∴ 直线M₁N₁的函数表达式为． ……………………………………8分

②当M点在x轴的负半轴上，N点在y轴的负半轴上时，设M₂点坐标为(x₂，0)，N₂点坐标为(0，y₂)．　

∵ AB∥N₁M₁，AB∥M₂N₂，AB＝N₁M₁，AB＝M₂N₂，

∴ N₁M₁∥M₂N₂，N₁M₁＝M₂N₂．　　

∴ 线段M₂N₂与线段N₁M₁关于原点O成中心对称．　　　

∴ M₂点坐标为(-3，0)，N₂点坐标为(0，-2)．　　 ………………………9分

设直线M₂N₂的函数表达式为，把x＝-3，y＝0代入，解得，

∴ 直线M₂N₂的函数表达式为． 　　

所以，直线MN的函数表达式为或．　 ………………11分

(3)选做题：(9，2)，(4，5)．　 ………………………………………………2分

13. 解：(1)分别过D，C两点作DG⊥AB于点G，CH⊥AB于点H． ……………1分

∵ AB∥CD，　

∴ DG＝CH，DG∥CH．　

∴ 四边形DGHC为矩形，GH＝CD＝1．　

∵ DG＝CH，AD＝BC，∠AGD＝∠BHC＝90°，

∴ △AGD≌△BHC(HL)．　

∴ AG＝BH＝＝3．　 ………2分

∵ 在Rt△AGD中，AG＝3，AD＝5，　

∴ DG＝4． 　　　　　　　　　　　　　　　

∴ ．　　　 ………………………………………………3分

(2)∵ MN∥AB，ME⊥AB，NF⊥AB，　

∴ ME＝NF，ME∥NF．　

∴ 四边形MEFN为矩形．　

∵ AB∥CD，AD＝BC，　　

∴ ∠A＝∠B．　

∵ ME＝NF，∠MEA＝∠NFB＝90°，　　

∴ △MEA≌△NFB(AAS)．

∴ AE＝BF． 　　　　……………………4分　

设AE＝x，则EF＝7－2x．　 ……………5分　

∵ ∠A＝∠A，∠MEA＝∠DGA＝90°，　　

∴ △MEA∽△DGA．

∴ ．

∴ ME＝．　　　　 …………………………………………………………6分

∴ ．　 ……………………8分

当x＝时，ME＝＜4，∴四边形MEFN面积的最大值为．……………9分

(3)能．　　 ……………………………………………………………………10分

由(2)可知，设AE＝x，则EF＝7－2x，ME＝．　

若四边形MEFN为正方形，则ME＝EF．　

　　 即 7－2x．解，得 ．　 ……………………………………………11分

∴ EF＝＜4．　

∴ 四边形MEFN能为正方形，其面积为．