发布时间 : 星期四 文章二次函数的应用与几何图形的建构更新完毕开始阅读
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23.(1) B(3,0),y?x2?2x?3
3 (2)P(,0)
739 (3)当E在第四象限,s??x2?x?6(0?x?3)
2211 当E在第三象限,s??x2?x?6(?1?x?0)
22 当E在第一象限或第二象限,s?2x2?4x(x??1或x?3)
六、综合运用(本题满分12分)
26.已知抛物线C1:y??x?2mx?n(m,n为常数,且m≠0,n?0)的顶点为A,与y轴交于点C;抛物线C2与抛物线C1关于y轴对称,其顶点为B,连接AC,BC,AB.
2?b4ac?b2?注:抛物线y?ax?bx?c?a≠0?的顶点坐标为??,?.
4a??2a2(1)请在横线上直接写出抛物线C2的解析式:________________________; (2)当m?1时,判定△ABC的形状,并说明理由;
(3)抛物线C1上是否存在点P,使得四边形ABCP为菱形?如果存在,请求出m的值;如果不存在,请说明理由.
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2O x
26.(1)y??x?2mx?n. ······················································· 2分 (2)当m?1时,△ABC为等腰直角三角形. ································· 3分 理由如下: 如图:
点A与点B关于y轴对称,点C又在y轴上,
··········································································· 4分 ?AC?BC. ·
过点A作抛物线C1的对称轴交x轴于D,过点C作CE?AD于E.
?当m?1时,顶点A的坐标为A?11,?n?,?CE?1.
又
点C的坐标为?0,n?,
?AE?1?n?n?1.?AE?CE.
从而∠ECA?45,?∠ACy?45.
由对称性知∠BCy?∠ACy?45,?∠ACB?90.
······················································ 7分 ?△ABC为等腰直角三角形. ·
(3)假设抛物线C1上存在点P,使得四边形ABCP为菱形,则PC?AB?BC. 由(2)知,AC?BC,?AB?BC?AC.
从而△ABC为等边三角形. ························································· 8分
?∠ACy?∠BCy?30.
四边形ABCP为菱形,且点P在C1上,?点P与点C关于AD对称.
?PC与AD的交点也为点E,因此∠ACE?90?30?60.
点A,C的坐标分别为Am,m?n,C?0,n?,
2???AE?m2?n?n?m2,CE?m. AEm2在Rt△ACE中,tan60???3.
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?m?3,?m??3.
故抛物线C1上存在点P,使得四边形ABCP为菱形,此时m??3.
····························································································· 12分 说明:只求出m的一个值扣2分.
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