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??A(?1，0)，C(0，······················································································· 1分 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??43??顶点F?1，?3?33x?2233x?······················································· 3分 3 ·
? ······························································································ 4分 ???（2）存在 ··············································································································· 5分 P1(0，?P2(2，?··········································································································· 7分 3) ·
··········································································································· 9分 3) ·
（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?33x?2233x?3上，?B(3，0)
H 在Rt△BOC中，tan?OBC?33，
B A C O B x
???OBC?30，BC?23，
M F 图9 在Rt△BB?H中，B?H?BH?12BB??23，
3B?H?6，?OH?3，?B?(?3，?23)················································12分
设直线B?F的解析式为y?kx?b
?3??23??3k?bk????6??43 解得?
?k?b33???b??3???2
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?y?36x?332 ································································································13分
3??y??3x?3x??7???? 解得 ?M?333x??y??103，?y?62??7??310，???77??3 ?????在直线AC上存在点M，使得△MBF的周长最小，此时M?，??7?3103?． ···14分 ??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··················································· 11分 M，则点M即为所求． ·
过点F作FG?y轴于点G，则OB∥FG，BC∥FH．
???BOC??FGH?90，?BCO??FHG
y ??HFG??CBO
同方法一可求得B(3，0)．
33A O C M G F H 图10 33B x
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???3x?37??y?93?? 解得? ?M?y??103?y??3x?3??7?55?310，???77??3 ???
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??在直线AC上存在点M，使得△MBF的周长最小，此时M?，??7?3103?． ···14分 ??7?
85.（08内蒙古赤峰25题）（本题满分14分）
在平面直角坐标系中给定以下五个点A(?3，0)，B(?1，4)，C(0，3)，D?，?，E(1，0)．
?24??17?（1）请从五点中任选三点，求一条以平行于y轴的直线为对称轴的抛物线的解析式； （2）求该抛物线的顶点坐标和对称轴，并画出草图；
1715??（3）已知点F??1，?在抛物线的对称轴上，直线y?44??y G B(?1，4) F C(0，3) 过点G??1，?且垂直于对称轴．验证：以E(1，0)为圆心，
?4?EF为半径的圆与直线y??17??17?D?，??24?174相切．请你进一步验证，以抛
A(?3，0) E(1，0) H O x
物线上的点D?y?174?17?，?为圆心DF为半径的圆也与直线24??相切．由此你能猜想到怎样的结论．
（08内蒙古赤峰25题解析）25．解：（1）设抛物线的解析式为y?ax2?bx?c， 且过点A(?3，0)，C(0，3)，E(1，0)， 由(0，3)在y?ax?bx?cH ．
则c?3． ········································································································ （2分）
?9a?3b?3?0得方程组?，
a?b?c?0?17??G??1，? 4??2B(?1，4) y M N C(0，3)解得a??1，b??2．
··············· （4分） ?抛物线的解析式为y??x?2x?3 ·
2F Q （2）由y??x?2x?3??(x?1)?4············· （6分） 得顶点坐标为(?1，········· （8分） 4)，对称轴为x??1． ·（3）①连结EF，过点E作直线y?174A(?3，0) 22?17?D?，??24?E(1，0) H O x
的垂线，垂足为N，
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则EN?HG?174．
154在Rt△FHE中，HE?2，HF??EF?HE?HF22，
?174，
?EF?EN，
?以E点为圆心，EF为半径的?E与直线y?174相切． ·····························（10分）
②连结DF过点D作直线y?则DM?QG?174?74?10432?17452的垂线，垂足为M．过点D作DQ?GH垂足为Q， ．
154?74?84?2．
在Rt△FQD中，QD?FD?QF?QD22，QF??52．
174?以D点为圆心DF为半径的?D与直线y?相切． ································（12分）
174③以抛物线上任意一点P为圆心，以PF为半径的圆与直线y?
相切． ·····（14分）
86.（08青海西宁28题）如图14，已知半径为1的?O1与x轴交于A，B两点，OM为?O1的切线，切
2点为M，圆心O1的坐标为(2，0)，二次函数y??x?bx?c的图象经过A，B两点．
（1）求二次函数的解析式；
（2）求切线OM的函数解析式；
（3）线段OM上是否存在一点P，使得以P，O，A为顶点的三角形与△OO1M相似．若存在，请求出所有符合条件的点P的坐标；若不存在，请说明理由．
?O1（08青海西宁28题解析）解：（1）0)，?圆心O1的坐标为(2，y M O A O1 B x 图14
半径为1，?A(1，0)，B(3，0)……1分
?二次函数y??x?bx?c的图象经过点A，B，
??1?b?c?0 ··············································································· 2分 ?可得方程组??9?3b?c?0?2解得：?
?b?4?c??3··········································· 3分 ?二次函数解析式为y??x?4x?3 ·
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（2）过点M作MF?x轴，垂足为F． ······························································· 4分
?OM是?O1的切线，M为切点，?O1M?OM（圆的切线垂直于经过切点的半径）．
在Rt△OO1M中，sin?O1OM?O1MOO1??12
y P1 M P2 O H A F O1 B x ···························5分 ??O1OM为锐角，??O1OM?30 ·
32?OM?OO1?cos30?2???3，
在Rt△MOF中，OF?OM?cos30??3?32?32．
MF?OM?sin30??3?12?32．
?33?························································································ 6分 ?点M坐标为?，?·
?22???设切线OM的函数解析式为y?kx(k?0)，由题意可知32?32k，?k?33 ····· 7分
?切线OM的函数解析式为y?33··································································· 8分 x·
（3）存在． ··········································································································· 9分
①过点A作AP1?x轴，与OM交于点P1．可得Rt△AP1O∽Rt△MO1O（两角对应相等两三角形相似）
P1A?OA?tan?AOP1?tan30??33，?P1?1，???3? ············································10分 ??3?②过点A作AP2?OM，垂足为P2，过P2点作P2H?OA，垂足为H． 可得Rt△AP2O∽Rt△O1MO（两角对应相等两三角开相似）
32?在Rt△OP2A中，?OA?1，?OP2?OA?cos30?，
在Rt△OP2H中，OH?OP2?cos?AOP2?32?32?34，
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