(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
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(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
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∵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