阳光家教网 www.ygjj.com 中考（初三复习）数学资料 若四边形MEFN为正方形，则ME＝EF．
4x21 即 ?7－2x．解，得 x?． ……………………………………………11分
3102114∴ EF＝7?2x?7?2??＜4．
105∴ 四边形MEFN能为正方形，其面积为S正方形MEFN?14?196． ????525??214. 解：（1）由题意可知，m?m?1???m?3??m?1?．
解，得 m＝3． ………………………………3分
∴ A（3，4），B（6，2）；
y ∴ k＝4×3=12． ……………………………4分
A （2）存在两种情况，如图：
N1 ①当M点在x轴的正半轴上，N点在y轴的正半轴 B 上时，设M1点坐标为（x1，0），N1点坐标为（0，y1）．
M2 O x M1 ∵ 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， N2 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
2设直线M1N1的函数表达式为y?k1x?2，把x＝3，y＝0代入，解得k1??．
32∴ 直线M1N1的函数表达式为y??x?2． ……………………………………8分
3②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
2设直线M2N2的函数表达式为y?k2x?2，把x＝-3，y＝0代入，解得k2??，
32∴ 直线M2N2的函数表达式为y??x?2．
322所以，直线MN的函数表达式为y??x?2或y??x?2． ………………11分
33（3）选做题：（9，2），（4，5）． ………………………………………………2分 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y?a(x?1)(x?3)(a≠0)
又点D(0，-3)在抛物线上，∴a(0+1)(0-3)=-3，解之得：a=1
∴y=x-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y?ax2?bx?c(a≠0)
2
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 根据题意可知，A(-1,0)，B(3,0)，D(0，-3)三点都在抛物线上
?a?b?c?0?a?1?? ∴?9a?3b?c?0，解之得：?b??2
?c??3?c??3??∴y=x-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=3 在Rt△MCE中，∵OC=2，∠CMO=60°，∴ME=4
∴点C、E的坐标分别为(0，3)，(-3,0) ······················································ 6分
2
∴切线CE的解析式为y?3······························································· 8分 x?3·
3y
C
A B x M O E
D
解图12
(3)设过点D(0，-3)，“蛋圆”切线的解析式为：y=kx-3(k≠0) ···························· 9分
??y?kx?3 由题意可知方程组?只有一组解 2?y?x?2x?3? 即kx?3?x2?2x?3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
16.
解：（1）OP?6?t，OQ?t?2． 3
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 y C Q O D B C Q y B C Q E P 图3 F A x
y B D1 P 图1
A x O 图2 P A x O （2）当t?1时，过D点作DD1?OA，交OA于D1，如图1， 则DQ?QO?54，QC?， 33?CD?1，?D(1，3)．
（3）①PQ能与AC平行．
若PQ∥AC，如图2，则
OPOA?， OQOC1476?t6?，?t?，而0≤t≤， 2393t?314?t?．
9②PE不能与AC垂直．
即
若PE?AC，延长QE交OA于F，如图3，
则
QFOQQF???ACOC353t?23．
?2??QF?5?t??．
?3??EF?QF?QE?QF?OQ
?2??2??5?t????t??
?3??3?2?(5?1)t?(5?1)．
3又?Rt△EPF∽Rt△OCA，?PEOC?， EFOA
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 ?6?t3?，
?2?6(5?1)?t???3?7， 3?t?3.45，而0≤t≤?t不存在．
17. 解：（1）?直线y??3x?3与x轴交于点A，与y轴交于点C．
······································································································ 1分 ?A(?1，0)，C(0，?3) ·
?点A，C都在抛物线上，
??233?c?0?a??a??? ??3 3?c??3??3?c???抛物线的解析式为y?3223··································································· 3分 x?x?3 ·
33?43?1，?············································································································· 4分 ?顶点F???? ·3??（2）存在 ································································································································ 5分 ···························································································································· 7分 P，?3) ·1(0···························································································································· 9分 P，?3) ·2(2（3）存在 ······························································································································ 10分
理由： 解法一：
延长BC到点B?，使B?C?BC，连接B?F交直线AC于点M，则点M就是所求的点． ··································································································· 11分 过点B?作B?H?AB于点H．
y ?B点在抛物线y?32230) x?x?3上，?B(3，33H A C B O B x
3在Rt△BOC中，tan?OBC?，
3??OBC?30?，BC?23，
在Rt△BB?H中，B?H?M F 图9 1BB??23， 2
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 ························································· 12分 BH?3B?H?6，?OH?3，?B?(?3，?23) ·设直线B?F的解析式为y?kx?b
?3??23??3k?bk????6??43 解得?
?k?b???b??33?3??2?y?333 ··············································································································· 13分 x?62?3?? ?3??y??3x?3x???3107?????M，? 解得 ??333?77103y?x????y??，62??7??3103??． ·········· 14分 ?在直线AC上存在点M，使得△MBF的周长最小，此时M???7，?7??解法二：
过点F作AC的垂线交y轴于点H，则点H为点F关于直线AC的对称点．连接BH交
AC于点M，则点M即为所求． ········································· 11分
过点F作FG?y轴于点G，则OB∥FG，BC∥FH．
y ??BOC??FGH?90?，?BCO??FHG ??HFG??CBO
0)． 同方法一可求得B(3，在Rt△BOC中，tan?OBC?A O C M G F H 图10 B x
33?，??OBC?30，可求得GH?GC?， 33?GF为线段CH的垂直平分线，可证得△CFH为等边三角形，
?AC垂直平分FH．
?即点H为点F关于AC的对称点．?H?0，设直线BH的解析式为y?kx?b，由题意得
???53? ······················································· 12分 ??3?
2010中考数学专题复习 - 压轴题（含答案）
