（2）设点D的坐标为(xD，yD)，由于D为抛物线的顶点
2332?(－3)3∴xD＝－＝1，yD＝－
32383＝33． ×1 2＋×1＋
333∴点D的坐标为(1，33)．
如图，过点D作DN⊥x轴于N，则DN＝33，AN＝3，∴AD＝32＋(33)2＝6．
∴∠DAO＝60° ···················································································· 4分 ∵OM∥AD
①当AD＝OP时，四边形DAOP为平行四边形． ∴OP＝6
∴t＝6（s） ······························································ 5分 ②当DP⊥OM时，四边形DAOP为直角梯形． 过点O作OE⊥AD轴于E．
在Rt△AOE中，∵AO＝2，∠EAO＝60°，∴AE＝1． （注：也可通过Rt△AOE∽Rt△AND求出AE＝1） ∵四边形DEOP为矩形，∴OP＝DE＝6－1＝5．
∴t＝5（s） ·························································································· 6分 ③当PD＝OA时，四边形DAOP为等腰梯形，此时OP＝AD－2AE＝6－2＝
4．
∴t＝4（s）
综上所述，当t＝6s、5s、4s时，四边形DAOP分别为平行四边形、直角
梯形、等腰梯形．
····················································································· 7分
6
（3）∵∠DAO＝60°，OM∥AD，∴∠COB＝60°．
又∵OC＝OB，∴△COB是等边三角形，∴OB＝OC＝AD＝6． ∵BQ＝2t，∴OQ＝6－2t（0＜t＜3） 过点P作PF⊥x轴于F，则PF＝∴S四边形BCPQ ＝S△COB －S△POQ
＝
311×6×33－×(6－2t)×t
22236333(t－)2＋ ······················································ 9分 2823t． ··············································· 8分 2＝
6333∴当t＝（s）时，S四边形BCPQ的最小值为． ······························ 10分
8233339此时OQ＝6－2t＝6－2×＝3，OP＝，OF＝，∴QF＝3－＝，
44422PF＝
33． 4∴PQ＝PF2＋QF2＝(333329 ······································ 11分 )＋()2＝244119.如图，已知直线y＝－x＋1交坐标轴于A、B两点，以线段AB为边向上作正方形
2ABCD，过点A，D，C的抛物线与直线另一个交点为E．
（1）请直接写出点C，D的坐标； （2）求抛物线的解析式；
（3）若正方形以每秒5个单位长度的速度沿射线AB下滑，直至顶点D落在x轴上时停止．设正方形落在x轴下方部分的面积为S，求S关于滑行时间t的函数关系式，并写出相应自变量t的取值范围；
（4）在（3）的条件下，抛物线与正方形一起平移，直至顶点D落在x轴上时停止，求抛物线上C、E两点间的抛物线弧所扫过的面积．
7
解：（1）C(3，2)，D(1，3)； ·············································································· 2分
（2）设抛物线的解析式为y＝ax 2＋bx＋c，把A(0，1)，D(1，3)，C(3，2)代
入
5?a＝－?6?c＝1?17??得?a＋b＋c＝3 解得?b＝ ················································· 4分
6?9a＋3b＋c＝2???c＝1??517∴抛物线的解析式为y＝－x 2＋x＋1； ······································· 5分
66（3）①当点A运动到点F（F为原B点的位置）时
8
∵AF＝12＋22＝5，∴t＝当0＜ t ≤1时，如图1． B′F＝AA′＝5t
∵Rt△AOF∽Rt△∠GB ′F，∴∴B ′G＝
55＝1（秒）．
OAB?G． ＝B?FOF5OA1·B ′F＝×5t＝t
22OF正方形落在x轴下方部分的面积为S即为△B ′FG的面积S△B′FG
51′15∴S＝S△B′B F·B ′G＝×5t×t＝t 2 ··························· 7分 FG＝
2224②当点C运动到x轴上时
CC?OB∵Rt△BCC ′∽Rt△∠AOB，∴． ＝
BCOA∴CC ′＝
25OB2·BC＝×5＝25，∴t＝＝2（秒）．
1OA5当1＜ t ≤2时，如图2．
∵A ′B ′＝AB＝5，∴A ′F＝5t－5． ∴A ′G＝∵B ′H＝
5t－5 25t 2∴S＝S梯形A′B′HG＝
1 G＋B ′H)·A ′B ′(A ′
29
5t－551t)·5 ＝(＋
22255··································································· 9分 ＝t－ ·
24③当点D运动到x轴上时
DD′＝35 t＝
355＝3（秒）
当2＜ t ≤3时，如图3． ∵A ′G＝
5t－5 25t－535－5t∴GD′ ＝5－＝
22∴D′H＝35－5t ∴S△D′GH ＝
35－5t2135－5t()(35－5t)＝()
222 －S△D′∴S＝S正方形A′B′C′D′GH
＝(5)2－(
35－5t2
)251525＝－t 2＋t－ ··································································· 11分
424（4）如图4，抛物线上C、E两点间的抛物线弧所扫过的面积为图中阴影部分
的面积．
∵t＝3，BB′＝AA′＝DD′＝35
∴S阴影＝S矩形BB′·········································································· 13分 C′C ·
10
2012中考数学压轴题及答案40例(5)
