# 1.2009年重庆綦江县

（2009年重庆綦江县）26．（11

(1)0)

y a x a

=-+≠经过点

(2)

∥．过顶点D平行于x轴的直线交射线OM于点C，B在x轴正半轴上，连结BC．

（1）求该抛物线的解析式；

（2）若动点P从点O出发，以每秒1个长度单位的速度沿射线OM运动，设点P运动的时间为()

t s．问当t为何值时，四边形DAOP分别为平行四边形？直角梯形？等腰梯形？（3）若OC OB

=，动点P和动点Q分别从点O和点B同时出发，分别以每秒1个长度单位和2个长度单位的速度沿OC和BO运动，当其中一个点停止运动时另一个点也随之停止运动．设它们的运动的时间为t()s，连接PQ，当t为何值时，四边形BCPQ的面积最小？并求出最小值及此时PQ的长．

*26．解：（1）

(1)0)

y a x a

=-+≠经过点(20)

A-，，

09a a

∴=+= ······· 1分∴

y x

=3分（2）D

∴过D作DN OB

⊥于N

，则DN=

3660

=∴==∴∠=

，° ························································4分

=时，四边形DAOP是平行四边形

66(s)

OP t

∴=∴=·····················································5分

②当DP OM

⊥时，四边形DAOP是直角梯形

⊥于H，2

AO=，则1

AH=

55(s)

OP DH t

∴=== ·····································································································6分

③当PD OA

=时，四边形DAOP是等腰梯形

26244(s)

∴=-=-=∴=

t=、5、4时，对应四边形分别是平行四边形、直角梯形、等腰梯形．··7分

（3）由（2）及已知，60

COB OC OB OCB

∠==

°，，△是等边三角形

OB OC AD OP t BQ t OQ t t

=====∴=-<<

，，，

⊥于E

，则PE=················································································8分

11

6(62)

222

BCPQ

S t

∴=???-?

=

2

3

22

t

?

-?

??

··········································································································9分

3

2

t=时，

BCPQ

S

·········································································10分

3339

33

24444

OQ OP OE QE PE

==∴=-==

，=，

PQ

∴===···························································11分