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2∴ 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)
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??2
∴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?3x?3································································ 8分 3y
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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
??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?y C Q O D B C Q 2． 3y 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?45，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，
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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?6?t3?，
?2?6(5?1)?t???3?7?t?3.45，而0≤t≤，
3?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．
?A(?1，0)，C(0，······································································································ 1分 ?3) ·
?点A，C都在抛物线上，
??233?c?0?a??a??? ??33 ??3?c?c??3???抛物线的解析式为y?3223x?x?3 ···································································· 3分 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
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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?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························································· 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?333x? ··············································································································· 13分 62?3?? ?3??y??3x?3x???3107?????M，? 解得 ??333?77103x???y??y??，62??7??3103?MAC△MBF上存在点，使得的周长最小，此时M?，········· 14分 ?在直线
?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
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A O C M G F H 图10 B x
新课标第一网（www.xkb1.com）--中小学教学资源共享平台
0)． 同方法一可求得B(3，在Rt△BOC中，tan?OBC?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?5?k?3?0?3k?b???9 解得? 5?5b??3??b??33??3??y?553?3 ··············································································································· 13分 933?55x???3x?3?3?1037??y??M，?93 解得 ???????7??7?y??103?y??3x?3??7??3103???在直线AC上存在点M，使得△MBF的周长最小，此时M???7，?． 1 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?3，?AO?2
?sin?AOB?1?，??AOB?30 2?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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