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?在Rt△DOM中，DM?13，OM? 22?点D在第一象限，
?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
?31?2，由题意，将A(?31)代入y?ax?bx?2中得 ，，D???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，853x?2上 ?点P在抛物线y??x2?99853??m2?m?2?2
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解得，m1?0，m2??53 8?53?2??P2)，P2?1(0，??8，?
???以O，B，P，Q为顶点的四边形是平行四边形，
?PQ∥OB，PQ?OB?3， 2)时， ?当点P1的坐标为(0，点Q的坐标分别为Q1(?3，2)，Q2(3，2)； 当点P2的坐标为??A B F C D x O M y E ?53?2??8，?时，
???133??33?，2?2?点Q的坐标分别为Q3??······················································ 14分
??，Q4??8，?． ·8????（以上答案仅供参考，如有其它做法，可参照给分） 19. 解：（1）在y??32x?3中，令y?0 43??x2?3?0
4?x1?2，x2??2
?A(?2，0)，B(2，0) ·························································· 1分 又?点B在y??C E y 3x?b上 4A N 3?0???b
23b?
2M D O P B x 33?BC的解析式为y??x? ··························································································· 2分
4232?y??x?3?x1??1???4（2）由?，得?9
33y?1?y??x???4??42?x2?2 ································································ 4分 ?y?0?29??0) ?C??1，?，B(2，4??新课标第一网----免费课件、教案、试题下载
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9 ·············································································································· 5分 4199?S△ABC??4?? ······································································································· 6分
242（3）过点N作NP?MB于点P ?EO?MB ?NP∥EO
?△BNP∽△BEO ·············································································································· 7分 BNNP?? ·························································································································· 8分 BEEO?AB?4，CD?由直线y??33?3?x?可得：E?0，? 42?2?35，则BE? 22?在△BEO中，BO?2，EO??62tNP，?NP?t ····································································································· 9分 ?5352216?S??t?(4?t)
25312S??t2?t(0?t?4) ·································································································· 10分
55312S??(t?2)2? ············································································································ 11分
5512?此抛物线开口向下，?当t?2时，S最大?
512?当点M运动2秒时，△MNB的面积达到最大，最大为．
520. 解：（1）如图，过点B作BD⊥OA于点D. 在Rt△ABD中， ∵∣AB∣=35,sin∠OAB=
5, 5 ∴∣BD∣=∣AB∣·sin∠OAB =35×
又由勾股定理，得 AD? ?5=3. 52AB?BD (35)2?32?6
2∴∣OD∣=∣OA∣-∣AD∣=10-6=4.
∵点B在第一象限，∴点B的坐标为（4，3）. ??3分
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设经过O(0,0)、C（4，-3）、A(10,0)三点的抛物线的函数表达式为
2
y=ax+bx(a≠0).
1?a?,??16a?4b??3?8??由?
100a?10b?05??b??.??4∴经过O、C、A三点的抛物线的函数表达式为y?125x?x. ??2分 84（2）假设在（1）中的抛物线上存在点P，使以P、O、C、A为顶点的四边形为梯形 ①∵点C（4，-3）不是抛物线y?125x?x的顶点， 84∴过点C做直线OA的平行线与抛物线交于点P1 .
则直线CP1的函数表达式为y=-3. 对于y?125x?x，令y=-3?x=4或x=6. 84∴??x1?4,?x2?6, ??y1??3;?y2??3.而点C（4，-3），∴P1(6,-3).
在四边形P1AOC中，CP1∥OA,显然∣CP1∣≠∣OA∣.
∴点P1（6，-3）是符合要求的点. ??1分 ②若AP2∥CO.设直线CO的函数表达式为y?k1x. 将点C（4，-3）代入，得4k1??3.?k1??. ∴直线CO的函数表达式为y??343x. 43x?b1. 4 于是可设直线AP2的函数表达式为y??将点A（10，0）代入，得?315x?. 42315∴直线AP2的函数表达式为y??x?.
42315?y??x?.??42?x2?4x?60?0，即（x-10）（x+6）=0. 由??y?1x2?5x?84??x1?10,?x2??6∴? ?y?0;y?12;?1?2而点A（10，0），∴P2（-6，12）.
过点P2作P2E⊥x轴于点E，则∣P2E∣=12.
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在Rt△AP2E中，由勾股定理，得
AP2?P2E?AE?122?162?20.
22而∣CO∣=∣OB∣=5.
∴在四边形P2OCA中，AP2∥CO,但∣AP2∣≠∣CO∣.
∴点P2（-6，12）是符合要求的点. ??1分 ③若OP3∥CA,设直线CA的函数表达式为y=k2x+b2 将点A(10,0)、C(4,-3)代入，得
1?k?,?10k2?b2?0?2?2 ???4k2?b2??3?b??5.?21x?5. 21∴直线OP3的函数表达式为y?x
2∴直线CA的函数表达式为y?1?y?x??2?x2?14x?0,即x(x-14)=0. 由??y?1x2?5x?84?∴??x1?0,?x2?14, ??y1?0;?y2?7.而点O(0,0),∴P3（14，7）.
过点P3作P3E⊥x轴于点E，则∣P3E∣=7. 在Rt△OP3E中，由勾股定理，得
OP3?P?72?142?75. 3F?OF22而∣CA∣=∣AB∣=35.
∴在四边形P3OCA中，OP3∥CA,但∣OP3∣≠∣CA∣.
∴点P3（14，7）是符合要求的点. ??1分 综上可知，在（1）中的抛物线上存在点P1(6,-3)、P2(-6,12)、P3(14,7),
使以P、O、C、A为顶点的四边形为梯形. ??1分 （3）由题知，抛物线的开口可能向上，也可能向下.
①当抛物线开口向上时，则此抛物线与y轴的副半轴交与点N. 可设抛物线的函数表达式为y?a(x?2k)(x?5k)（a＞0）.
即y?ax?3akx?10ak
22349?a(x?k)2?ak2.
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