y C Q O D B C Q y B C Q y B E P 图3 F A x

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?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．

················································································ 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············································· 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 623??y??3x?3x???3107?????M，? 解得 ??333?77103y?x????y??，62??7??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 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?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??93 解得? ?M?， ?????77???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?

?在Rt△DOM中，DM??点D在第一象限，

13，OM? 22?31?·············································································· 5分 ?点D的坐标为???2，? ·2??由（1）知EO?AO?2，点E在y轴的正半轴上

2) ?点E的坐标为(0，················································································ 6分 ?点A的坐标为(?31)， ·

?抛物线y?ax2?bx?c经过点E，

?c?2

由题意，将A(?31)，，D??31?2，代入y?ax?bx?2中得 ??22???8??3a?3b?2?1a???9?? 解得 ?3?31b?2??a??b??53?422?9?853x?2 ·················································· 9分 ?所求抛物线表达式为：y??x2?99（3）存在符合条件的点P，点Q． ·································································· 10分 理由如下：?矩形ABOC的面积?AB?BO?3 ?以O，B，P，Q为顶点的平行四边形面积为23．

由题意可知OB为此平行四边形一边， 又?OB?3

?OB边上的高为2 ······················································································· 11分

2) 依题意设点P的坐标为(m，