欧美日韩黄网欧美日韩日B片|二区无码视频网站|欧美AAAA小视频|久久99爱视频播放|日本久久成人免费视频|性交黄色毛片特黄色性交毛片|91久久伊人日韩插穴|国产三级A片电影网站|亚州无码成人激情视频|国产又黄又粗又猛又爽的

22. 解：( 1)由已知得：解得

c=3,b=2

∴拋物線的線的解析式為

(2)由頂點(diǎn)坐標(biāo)公式得頂點(diǎn)坐標(biāo)為(1，4)

所以對(duì)稱軸為x=1,A,E關(guān)于x=1對(duì)稱，所以E(3,0)

設(shè)對(duì)稱軸與x軸的交點(diǎn)為F

所以四邊形ABDE的面積=

=

=

=9

(3)相似

如圖，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?yàn)辄c(diǎn)D為∠ACB的平分線，所以可設(shè)點(diǎn)D到邊AC,BC的距離均為h，

設(shè)△ABC AB邊上的高為H,

則利用面積法可得：

(CM+CN)h=MN﹒H

又 H=

化簡(jiǎn)可得　 (CM+CN)﹒

故 　　　　

20. 解：(1)如圖，過(guò)點(diǎn)B作BD⊥OA于點(diǎn)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.

∵點(diǎn)B在第一象限，∴點(diǎn)B的坐標(biāo)為(4，3).　　　　　　　　　　　　 ……3分

設(shè)經(jīng)過(guò)O(0,0)、C(4，-3)、A(10,0)三點(diǎn)的拋物線的函數(shù)表達(dá)式為

　　 y=ax²+bx(a≠0).

由

∴經(jīng)過(guò)O、C、A三點(diǎn)的拋物線的函數(shù)表達(dá)式為　　　　　　 ……2分

(2)假設(shè)在(1)中的拋物線上存在點(diǎn)P，使以P、O、C、A為頂點(diǎn)的四邊形為梯形

　 ①∵點(diǎn)C(4，-3)不是拋物線的頂點(diǎn)，

∴過(guò)點(diǎn)C做直線OA的平行線與拋物線交于點(diǎn)P₁ 　.

則直線CP₁的函數(shù)表達(dá)式為y=-3.

對(duì)于，令y=-3x=4或x=6.

∴

而點(diǎn)C(4，-3)，∴P₁(6,-3).

在四邊形P₁AOC中，CP₁∥OA,顯然∣CP₁∣≠∣OA∣.

∴點(diǎn)P₁(6，-3)是符合要求的點(diǎn).　　　　　　　　　　　　　　　　　 ……1分

②若AP₂∥CO.設(shè)直線CO的函數(shù)表達(dá)式為

　 將點(diǎn)C(4，-3)代入，得

∴直線CO的函數(shù)表達(dá)式為

　 于是可設(shè)直線AP₂的函數(shù)表達(dá)式為

將點(diǎn)A(10，0)代入，得

∴直線AP₂的函數(shù)表達(dá)式為

由，即(x-10)(x+6)=0.

∴

而點(diǎn)A(10，0)，∴P₂(-6，12).

過(guò)點(diǎn)P₂作P₂E⊥x軸于點(diǎn)E，則∣P₂E∣=12.

在Rt△AP₂E中，由勾股定理，得

而∣CO∣=∣OB∣=5.

∴在四邊形P₂OCA中，AP₂∥CO,但∣AP₂∣≠∣CO∣.

∴點(diǎn)P₂(-6，12)是符合要求的點(diǎn).　　　　　　　　　　　　　　　　　　　 ……1分

③若OP₃∥CA,設(shè)直線CA的函數(shù)表達(dá)式為y=k₂x+b₂

　 將點(diǎn)A(10,0)、C(4,-3)代入，得

∴直線CA的函數(shù)表達(dá)式為

∴直線OP₃的函數(shù)表達(dá)式為

由即x(x-14)=0.

∴

而點(diǎn)O(0,0),∴P₃(14，7).

過(guò)點(diǎn)P₃作P₃E⊥x軸于點(diǎn)E，則∣P₃E∣=7.

在Rt△OP₃E中，由勾股定理，得

而∣CA∣=∣AB∣=.

∴在四邊形P₃OCA中，OP₃∥CA,但∣OP₃∣≠∣CA∣.

∴點(diǎn)P₃(14，7)是符合要求的點(diǎn).　　　　　　　　　　　　　　　　　　　 ……1分

綜上可知，在(1)中的拋物線上存在點(diǎn)P₁(6,-3)、P₂(-6,12)、P₃(14,7),

使以P、O、C、A為頂點(diǎn)的四邊形為梯形.　　　　　　　　　　　　　　　　 ……1分

(3)由題知，拋物線的開(kāi)口可能向上，也可能向下.

　①當(dāng)拋物線開(kāi)口向上時(shí)，則此拋物線與y軸的副半軸交與點(diǎn)N.

可設(shè)拋物線的函數(shù)表達(dá)式為(a＞0).

即

如圖，過(guò)點(diǎn)M作MG⊥x軸于點(diǎn)G.

∵Q(-2k，0)、R(5k，0)、G(、N(0，-10ak²)、M

∴

∴　　　　　　　　　　　　　　　 ……2分

②當(dāng)拋物線開(kāi)口向下時(shí)，則此拋物線與y軸的正半軸交于點(diǎn)N，

　 同理，可得　　　　　　　　　　　　　　　　　 　　　……1分

綜上所知，的值為3：20.　　　　　　　　　　　　　　　　　　 ……1分

19. 解：(1)在中，令

，

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

又點(diǎn)在上

的解析式為···················································································· 2分

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

，

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

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

(3)過(guò)點(diǎn)作于點(diǎn)

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

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

由直線可得：

在中，，，則

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

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

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

此拋物線開(kāi)口向下，當(dāng)時(shí)，

當(dāng)點(diǎn)運(yùn)動(dòng)2秒時(shí)，的面積達(dá)到最大，最大為．　　　

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(guò)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(guò)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(zhǎng)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(guò)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(zhǎng)最小，此時(shí)．········· 14分

解法二：

過(guò)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(guò)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(zhǎng)最小，此時(shí)．　　 1

16.

解：(1)，．

(2)當(dāng)時(shí)，過(guò)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(zhǎng)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(guò)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(guò)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(guò)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(guò)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．