12x,因为线段A′B′和CD的长是定值,所以要使四边形A′B′CD的2周长最短,只要使A′D+CB′最短;
y ……1分 A′ 8 第一种情况:如果将抛物线向右平移,显然有A′D+CB′>AD+CB,因此不存在
6 某个位置,使四边形A′B′CD的周长最短.……1分 4 B′′ B′ 第二种情况:设抛物线向左平移了b个单位,则点A′和点B′的坐标分别为2 D C A′(-4-b,8)和B′(2-b,2). -4 -2 O 2 4 x -2 因为CD=2,因此将点B′向左平移2个单位得B′′(-b,2),
-4 要使A′D+CB′最短,只要使A′D+DB′′最短. ……1分 点A′关于x轴对称点的坐标为A′′(-4-b,-8), ② 左右平移抛物线y?55直线A′′B′′的解析式为y?x?b?2.要使A′D+DB′′最短,点D应在直
22线A′′B′′上,将点D(-4,0)代入直线A′′B′′的解析式,解得b?A′′ (第24题(2)②) 16. 5故将抛物线向左平移时,存在某个位置,使四边形A′B′CD的周长最短,此时抛物线的116函数解析式为y?(x?)2. ……1分
25109.(2009年宁夏)26.(10分)
已知:等边三角形ABC的边长为4厘米,长为1厘米的线段MN在△ABC的边AB上沿AB方向以1厘米/秒的速度向B点运动(运动开始时,点M与点A重合,点N到达点B时运动终止),过点M、N分别作AB边的垂线,与△ABC的其它边交于P、Q两点,线段
MN运动的时间为t秒.
(1)线段MN在运动的过程中,t为何值时,四边形MNQP恰为矩形?并求出该矩形的面积;
(2)线段MN在运动的过程中,四边形MNQP的面积为S,运动的时间为t.求四边形
MNQP的面积S随运动时间t变化的函数关系式,并写出自变量t的取值范围.
C Q
P
A M N
(2009年宁夏26题解析)(1)过点C作CD?AB,垂足为D. 则AD?2,
当MN运动到被CD垂直平分时,四边形MNQP是矩形,
Q C B
P
A M N
B
即AM?3时,四边形MNQP是矩形, 23秒时,四边形MNQP是矩形. 23?PM?AMtan60°=3,
23?S四边形MNQP?3 ·············································································································· 4分
2(2)1°当0?t?1时,
1S四边形MNQP?(PM?QN·)MN C
2Q 1????3t?3(t?1)? P 2?t??3t?3 ··································································· 6分 2A
M
N
B 2°当1≤t≤2时
1S四边形MNQP?(PM?QN·)MN
21??3t?3(3?t)?·1
??23?3 ·········································································· 8分 23°当2?t?3时,
1S四边形MNQP?(PM?QN·)MN
21??3(3?t)?3(4?t)??? 27??3t?3 ···························································· 10分
2
C P
Q A
M
N B