18. 解：(1)點(diǎn)在軸上························································································· 1分

理由如下：

連接，如圖所示，在中，，，

，

由題意可知：

點(diǎn)在軸上，點(diǎn)在軸上．············································································ 3分

(2)過點(diǎn)作軸于點(diǎn)

，

在中，，

點(diǎn)在第一象限，

點(diǎn)的坐標(biāo)為····························································································· 5分

由(1)知，點(diǎn)在軸的正半軸上

點(diǎn)的坐標(biāo)為

點(diǎn)的坐標(biāo)為······························································································· 6分

拋物線經(jīng)過點(diǎn)，

由題意，將，代入中得

　 解得

所求拋物線表達(dá)式為：·························································· 9分

(3)存在符合條件的點(diǎn)，點(diǎn)．············································································ 10分

理由如下：矩形的面積

以為頂點(diǎn)的平行四邊形面積為．

由題意可知為此平行四邊形一邊，

又

邊上的高為2······································································································ 11分

依題意設(shè)點(diǎn)的坐標(biāo)為

點(diǎn)在拋物線上

解得，，

，

以為頂點(diǎn)的四邊形是平行四邊形，

，，

當(dāng)點(diǎn)的坐標(biāo)為時(shí)，

點(diǎn)的坐標(biāo)分別為，；

當(dāng)點(diǎn)的坐標(biāo)為時(shí)，

點(diǎn)的坐標(biāo)分別為，．·················································· 14分

(以上答案僅供參考，如有其它做法，可參照給分)

17. 解：(1)直線與軸交于點(diǎn)，與軸交于點(diǎn)．

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

點(diǎn)都在拋物線上，

拋物線的解析式為······························································ 3分

頂點(diǎn)····································································································· 4分

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

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

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

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

理由：

解法一：

延長到點(diǎn)，使，連接交直線于點(diǎn)，則點(diǎn)就是所求的點(diǎn)．

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

過點(diǎn)作于點(diǎn)．

點(diǎn)在拋物線上，

在中，，

，，

在中，，

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

設(shè)直線的解析式為

　 解得

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

　 解得　

在直線上存在點(diǎn)，使得的周長最小，此時(shí)．········· 14分

解法二：

過點(diǎn)作的垂線交軸于點(diǎn)，則點(diǎn)為點(diǎn)關(guān)于直線的對(duì)稱點(diǎn)．連接交于點(diǎn)，則點(diǎn)即為所求．···················································································· 11分

過點(diǎn)作軸于點(diǎn)，則，．

，

同方法一可求得．

在中，，，可求得，

為線段的垂直平分線，可證得為等邊三角形，

垂直平分．

即點(diǎn)為點(diǎn)關(guān)于的對(duì)稱點(diǎn)．··················································· 12分

設(shè)直線的解析式為，由題意得

　 解得

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

　 解得　

在直線上存在點(diǎn)，使得的周長最小，此時(shí)．　　 1

16.

解：(1)，．

(2)當(dāng)時(shí)，過點(diǎn)作，交于，如圖1，

則，，

，．

(3)①能與平行．

若，如圖2，則，

即，，而，

．

②不能與垂直．

若，延長交于，如圖3，

則．

．

．

又，，

，

，而，

不存在．

15. 解：(1)解法1：根據(jù)題意可得：A(-1,0)，B(3,0)；

則設(shè)拋物線的解析式為(a≠0)

又點(diǎn)D(0，-3)在拋物線上，∴a(0+1)(0-3)=-3，解之得：a=1

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

自變量范圍：-1≤x≤3···················································································· 4分

　　　　　 解法2：設(shè)拋物線的解析式為(a≠0)

　　　　　　　　　 根據(jù)題意可知，A(-1,0)，B(3,0)，D(0，-3)三點(diǎn)都在拋物線上

　　　　　　　　　 ∴，解之得：

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

自變量范圍：-1≤x≤3······························································ 4分

　　　　　 (2)設(shè)經(jīng)過點(diǎn)C“蛋圓”的切線CE交x軸于點(diǎn)E，連結(jié)CM，

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

　　　　　　 在Rt△MCE中，∵OC=2，∠CMO=60°，∴ME=4

　∴點(diǎn)C、E的坐標(biāo)分別為(0，)，(-3,0) ·················································· 6分

∴切線CE的解析式為··························································· 8分

(3)設(shè)過點(diǎn)D(0，-3)，“蛋圓”切線的解析式為：y=kx-3(k≠0) ·························· 9分

　　　　　　　 由題意可知方程組只有一組解

　 即有兩個(gè)相等實(shí)根，∴k=-2············································· 11分

　 ∴過點(diǎn)D“蛋圓”切線的解析式y=-2x-3····················································· 12分

14. 解：(1)由題意可知，．

解，得 m＝3．　　　　 ………………………………3分

∴ A(3，4)，B(6，2)；

∴ k＝4×3=12．　 　　……………………………4分

(2)存在兩種情況，如圖：　

①當(dāng)M點(diǎn)在x軸的正半軸上，N點(diǎn)在y軸的正半軸

上時(shí)，設(shè)M₁點(diǎn)坐標(biāo)為(x₁，0)，N₁點(diǎn)坐標(biāo)為(0，y₁)．

∵ 四邊形AN₁M₁B為平行四邊形，

∴ 線段N₁M₁可看作由線段AB向左平移3個(gè)單位，

再向下平移2個(gè)單位得到的(也可看作向下平移2個(gè)單位，再向左平移3個(gè)單位得到的)．

由(1)知A點(diǎn)坐標(biāo)為(3，4)，B點(diǎn)坐標(biāo)為(6，2)，

∴ N₁點(diǎn)坐標(biāo)為(0，4－2)，即N₁(0，2)；　　　 ………………………………5分

M₁點(diǎn)坐標(biāo)為(6－3，0)，即M₁(3，0)．　　　 ………………………………6分

設(shè)直線M₁N₁的函數(shù)表達(dá)式為，把x＝3，y＝0代入，解得．

∴ 直線M₁N₁的函數(shù)表達(dá)式為． ……………………………………8分

②當(dāng)M點(diǎn)在x軸的負(fù)半軸上，N點(diǎn)在y軸的負(fù)半軸上時(shí)，設(shè)M₂點(diǎn)坐標(biāo)為(x₂，0)，N₂點(diǎn)坐標(biāo)為(0，y₂)．　

∵ AB∥N₁M₁，AB∥M₂N₂，AB＝N₁M₁，AB＝M₂N₂，

∴ N₁M₁∥M₂N₂，N₁M₁＝M₂N₂．　　

∴ 線段M₂N₂與線段N₁M₁關(guān)于原點(diǎn)O成中心對(duì)稱．　　　

∴ M₂點(diǎn)坐標(biāo)為(-3，0)，N₂點(diǎn)坐標(biāo)為(0，-2)．　　 ………………………9分

設(shè)直線M₂N₂的函數(shù)表達(dá)式為，把x＝-3，y＝0代入，解得，

∴ 直線M₂N₂的函數(shù)表達(dá)式為． 　　

所以，直線MN的函數(shù)表達(dá)式為或．　 ………………11分

(3)選做題：(9，2)，(4，5)．　 ………………………………………………2分

13. 解：(1)分別過D，C兩點(diǎn)作DG⊥AB于點(diǎn)G，CH⊥AB于點(diǎn)H． ……………1分

∵ AB∥CD，　

∴ DG＝CH，DG∥CH．　

∴ 四邊形DGHC為矩形，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．　

∴ 四邊形MEFN為矩形．　

∵ AB∥CD，AD＝BC，　　

∴ ∠A＝∠B．　

∵ ME＝NF，∠MEA＝∠NFB＝90°，　　

∴ △MEA≌△NFB(AAS)．

∴ AE＝BF． 　　　　……………………4分　

設(shè)AE＝x，則EF＝7－2x．　 ……………5分　

∵ ∠A＝∠A，∠MEA＝∠DGA＝90°，　　

∴ △MEA∽△DGA．

∴ ．

∴ ME＝．　　　　 …………………………………………………………6分

∴ ．　 ……………………8分

當(dāng)x＝時(shí)，ME＝＜4，∴四邊形MEFN面積的最大值為．……………9分

(3)能．　　 ……………………………………………………………………10分

由(2)可知，設(shè)AE＝x，則EF＝7－2x，ME＝．　

若四邊形MEFN為正方形，則ME＝EF．　

　　 即 7－2x．解，得 ．　 ……………………………………………11分

∴ EF＝＜4．　

∴ 四邊形MEFN能為正方形，其面積為．

12. 解：(1)．····················································································· 3分

(2)相等，比值為．················· 5分(無“相等”不扣分有“相等”，比值錯(cuò)給1分)

(3)設(shè)，

在矩形中，，

，

，

，

，

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

同理．

，

，

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

，

，······························································································ 8分

解得．

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

(4)，·············································································································· 10分

．　 12分

11. 解：(1)設(shè)地經(jīng)杭州灣跨海大橋到寧波港的路程為千米，

由題意得，································································································ 2分

解得．

地經(jīng)杭州灣跨海大橋到寧波港的路程為180千米．················································· 4分

(2)(元)，

該車貨物從地經(jīng)杭州灣跨海大橋到寧波港的運(yùn)輸費(fèi)用為380元．···························· 6分

(3)設(shè)這批貨物有車，

由題意得，···························································· 8分

整理得，

解得，(不合題意，舍去)，································································ 9分

這批貨物有8車．···································································································· 10分

10.

9.