新课标第一网（www.xkb1.com）--中小学教学资源共享平台
?在Rt△DOM中，DM??点D在第一象限， ?31，?点D的坐标为??22?12，OM?32
? ···························································································· 5分 ???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 1) ·?点A的坐标为(?3，?抛物线y?ax?bx?c经过点E， ?c?2
2由题意，将A(?3，1)，D??31，?22??代入y?ax2?bx?2中得 ???8??3a?3b?2?1a???9?? 解得 ??331b?2??b??53?a??422?9?892?所求抛物线表达式为：y??x?539························································· 9分 x?2 ·
（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
依题意设点P的坐标为(m，2)
89539?点P在抛物线y??x?2x?2上
??89m?2539m?2?2
新课标第一网----免费课件、教案、试题下载
新课标第一网（www.xkb1.com）--中小学教学资源共享平台
解得，m1?0，m2??538
?53??P1(0，2)，P2??，2?
??8???以O，B，P，Q为顶点的四边形是平行四边形， ?PQ∥OB，PQ?OB?2)时， ?当点P1的坐标为(0，3，
y E A B F C D O M x
点Q的坐标分别为Q1(?3，2)，Q2(3，2)； 当点P2的坐标为?????53?，2?时， ?8????1338?，2?，Q4??点Q的坐标分别为Q3???33?··················································14分 ，2?．·??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????34x?3?0
234x?3中，令y?0
2?x1?2，x2??2
y C E N ····················································1分 ?A(?2，0)，B(2，0) ·又?点B在y???0??b?3232?b
34x?b上
A M D O P B x
34x?32?BC的解析式为y?? ··················································································· 2分
32?y??x?3?x1??1??x2?2??4（2）由?，得? ·························································· 4分 9 ?y?033?2?y1??y??x??4??429??C??1，4??0) ?，B(2，?新课标第一网----免费课件、教案、试题下载
新课标第一网（www.xkb1.com）--中小学教学资源共享平台
?AB?4，CD?949 ···································································································· 5分
?9242（3）过点N作NP?MB于点P
?S△ABC?1?4? ······························································································ 6分
?EO?MB ?NP∥EO
?△BNP∽△BEO ····································································································· 7分
BNNP················································································································ 8分 ??BEEO由直线y??34x?32可得：E?0，?
?2?32?3??在△BEO中，BO?2，EO??2t52?NP3，则BE?52
，?NP?65t····························································································· 9分
216?S??t?(4?t)
253212S??t?t(0?t?4) ··························································································10分
553122S??(t?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=
55,
∴∣BD∣=∣AB∣·sin∠OAB =35×又由勾股定理，得 AD? ?AB255=3.
?BD222
(35)?3?6
∴∣OD∣=∣OA∣-∣AD∣=10-6=4.
∵点B在第一象限，∴点B的坐标为（4，3）. ??3分
新课标第一网----免费课件、教案、试题下载
新课标第一网（www.xkb1.com）--中小学教学资源共享平台
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 y=ax+bx(a≠0).
1?a?,??16a?4b??3?8由? ???100a?10b?0?b??5.??42
∴经过O、C、A三点的抛物线的函数表达式为y?18x?254x. ??2分
（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?18x?254x的顶点，
∴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?18x?254x，令y=-3?x=4或x=6.
?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??34x.
34x?b1. 34.
于是可设直线AP2的函数表达式为y??将点A（10，0）代入，得?34x?15234. 152∴直线AP2的函数表达式为y??x?.
315?y??x?.??422?x?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.
新课标第一网----免费课件、教案、试题下载
新课标第一网（www.xkb1.com）--中小学教学资源共享平台
在Rt△AP2E中，由勾股定理，得 AP2?P2E2?AE2?12?16?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.?2∴直线CA的函数表达式为y?∴直线OP3的函数表达式为y?1212x?5. x
1?y?x??22由??x?14x?0,即x(x-14)=0. ?y?1x2?5x?84??x1?0,?x2?14,∴? ?y?0;y?7.?1?2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?P3F2?OF2?7?14?75.
22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
?a(x?32k)?222494ak.
2新课标第一网----免费课件、教案、试题下载
相关推荐：
loading