寒假作业12答案(5)

∴△COE∽△EHP

COEH

·············································· 2′ OEHP

由题意知:CO 5 OE 3 EH EA AH 2 HP 52 HP∴ 得HP 3 3HP∴EH 5 ····················································································································· 3′ 在Rt△COE和Rt△EHP中

∴

∴CE

EP故CE EP ·················································································································· 5′ （2）CE EP仍成立.

COEH

····································································· 6′ OEHP

由题意知:CO 5 OE t EH 5 t HP 55 t HP∴ 整理得 5 t HP t 5 t tHP

∵点E不与点A重合 ∴5 t 0 ∴HP t EH 5 ∴在Rt△COE和Rt△EHP中

同理△COE∽△EHP. ∴

························································ 5′ CE

EP ∴CE EP ·

（3）y轴上存在点M，使得四边形BMEP是平行四边形. ··········································· 9′

过点B作BM∥EP交y轴于点M ∴ 5 CEP 90° ∴ 6 4

在△BCM和△COE中

6 4

∴△BCM≌△COE ∴BM CE BC OC

BCM COE

而CE EP ∴BM EP

由于BM∥EP ∴四边形BMEP是平行四边形. ·············································· 11′ 故△BCM≌△COE可得CM OE t ∴OM CO CM 5 t

故点M的坐标为 0，······································································································ 12′ 5 t ·18、解：（1）设该抛物线的解析式为y ax bx c，

由抛物线与y轴交于点C（0，－3）,可知c 3.

即抛物线的解析式为y ax bx 3． ………………………1分 把A（－1，0）、B（3，0）代入, 得

2

2

a b 3 0,

9a 3b 3 0.