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23∴ 直线M1N1的函数表达式为y??x?2． ……………………………………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y?k2x?2，把x＝-3，y＝0代入，解得k2??∴ 直线M2N2的函数表达式为y??23x?22323，
． 或y??23x?2所以，直线MN的函数表达式为y??x?2． ………………11分
（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=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y?ax2?bx?c(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
?a?b?c?0?∴?9a?3b?c?0?c??3??a?1?，解之得：?b??2?c??3?
∴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=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
33∴切线CE的解析式为y?x?3 ·························································· 8分
y
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(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???y?kx?32??y?x?2x?3只有一组解
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?23． y y B C Q P A x O E P F A x
B y C Q O D1 D B C Q P 图1
A x O 图2 图3 （2）当t?1时，过D点作DD1?OA，交OA于D1，如图1， 则DQ?QO?53，QC?43，
?CD?1，?D(1，3)．
（3）①PQ能与AC平行．
OPOQOAOC若PQ∥AC，如图2，则
6?tt??t??，
73即
23149?63，?t?149，而0≤t≤，
．
②PE不能与AC垂直．
若PE?AC，延长QE交OA于F，如图3，
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则
QFAC?OQQF??OC35t?323．
?QF?2??5?t??．
3???EF?QF?QE?QF?OQ
2??2??5?t????t??
3??3??23(5?1)．
PEEF?OCOA??(5?1)t?又?Rt△EPF∽Rt△OCA，?6?t2??(5?1)?t??3??36，
??，
?t?3.45，而0≤t≤73，
?t不存在．
17. 解：（1）?直线y??3x???A(?1，0)，C(0，3与x轴交于点A，与y轴交于点C．
····························································································· 1分 3) ·
?点A，C都在抛物线上，
??2330?a??ca????? ??3 3???c??3??3?c?抛物线的解析式为y??43??顶点F?1，?3?33x?2233x?3 ······························································ 3分
? ···································································································· 4分 ???（2）存在 ····················································································································· 5分 P1(0，?P2(2，?3) ·················································································································· 7分 3) ·················································································································· 9分
（3）存在 ····················································································································10分
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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?33x?2233x?3上，?B(3，0)
在Rt△BOC中，tan?OBC?33H ，
B A C O M F B x
???OBC?30，BC?23，
图9
在Rt△BB?H中，B?H?BH?12BB??23，
3B?H?6，?OH?3，?B?(?3，·····················································12分 ?23)·
设直线B?F的解析式为y?kx?b
?3??23??3k?b?k???6??43 解得?
?k?b33???b??3???2?y?36x?332 ······································································································13分
3??y??3x?3x??7???? 解得 ?M?333x??y??103，?y?62??7??310，???77??3 ????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
y 过点F作FG?y轴于点G，则OB∥FG，BC∥FH．
???BOC??FGH?90，?BCO??FHG
??HFG??CBO
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A O C M G F H 图10 B x
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同方法一可求得B(3，0)．
3333在Rt△BOC中，tan?OBC?，??OBC?30?，可求得GH?GC?，
?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，????53? ···················································12分 ??3?设直线BH的解析式为y?kx?b，由题意得 5?k?3?0?3k?b???9 解得? 5?5b??3??b??33??3??y?593?533 ······································································································13分
3?x?55??3x?37??y?93 解得? ?M???y??103?y??3x?3??7??310，???77??3 ????3103?． 1 ??在直线AC上存在点M，使得△MBF的周长最小，此时M?，??7?7??18. 解：（1）点E在y轴上 ························································································· 1分 理由如下：
连接AO，如图所示，在Rt△ABO中，?AB?1，BO??sin?AOB?123，?AO?2
，??AOB?30
??由题意可知：?AOE?60
??BOE??AOB??AOE?30?60?90
???·········································································· 3分 ?点B在x轴上，?点E在y轴上． ·（2）过点D作DM?x轴于点M
?OD?1，?DOM?30?
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2011中考数学专题复习 - 压轴题（含答案）
