如图,已知平面直角坐标系中,菱形ABCD的两个顶点C,D的坐标 分别为（4,0）,（0,3）.现有两动点P,Q分别从A,C同时出发,点P沿线段AD向终点D运动,点Q沿折线CBA向终点A运动,设运动时间为t秒. 1 若点P
如图,已知平面直角坐标系中,菱形ABCD的两个顶点C,D的坐标 分别为（4,0）,（0,3）.现有两动点P,Q分别从A,C同时出发,点P沿线段AD向终点D运动,点Q沿折线CBA向终点A运动,设运动时间为t秒. 1 若点P
如图,已知平面直角坐标系中,菱形ABCD的两个顶点C,D的坐标 分别为（4,0）,（0,3）.现有两动点P,Q分别从A,C同时出发,点P沿线段AD向终点D运动,点Q沿折线CBA向终点A运动,设运动时间为t秒.
1 若点P的速度为每秒1个单位,点Q的速度为每秒2个单位.当点Q在线段BA上时,求三角形APQ的面积S关于t的函数关系式,以及S的最大值.
2 若点P的速度为每秒1个单位,点Q的速度变为每秒k个单位,则当t=4秒,且PA=PQ时,求出k的值.
如图,已知平面直角坐标系中,菱形ABCD的两个顶点C,D的坐标 分别为（4,0）,（0,3）.现有两动点P,Q分别从A,C同时出发,点P沿线段AD向终点D运动,点Q沿折线CBA向终点A运动,设运动时间为t秒. 1 若点P
1.
OC = OA = 4,OB = OC = 3
AD = BC = AB = 5
点Q在线段BA上时,BC/2 = 5/2 ≤ t ≤ (5+5)/2 = 5
t秒时,AP = t
P的横坐标 = -(AO - APcos∠OAD) = -(4 - t*OA/AD) = -(4 - 4t/5) = 4t/5 - 4
P的纵坐标 = -APsin∠OAD = -3t/5
P(4t/5 - 4,-3t/5)
t秒时,AQ = 10 - 2t
Q的纵坐标 = AQsin∠BAO = 6(5-t)/5 = 6 - 6t/5
Q的横坐标 = -(AO - AQcos∠BAO) = 4 - 8t/5
Q(4 - 8t/5,6(5-t)/5)
设PQ与x轴交于R(r,0)
PQ的方程:(y + 3t/5)/(6 - 6t/5 + 3t/5) = (x - 4t/5 + 4)/(4 - 8t/5 - 4t/5 + 4)
令y = 0,r = (-16t² + 100t - 200)/(50 - 5t)
AR = 4 + r = 16t(5 - t)/(50 - 5t)
三角形APQ的面积S
= 三角形APR的面积 + 三角形AQR的面积
= (1/2)AR*|P的纵坐标| + (1/2)AR*|Q的纵坐标|
= (1/2)AR*(3t/5 + 6 - 6t/5)
= (1/2)[16t(5 - t)/(50 - 5t)](6 - 3t/5)
= 24t(5 - t)/25
t = 5-t即t = 5/2时,S最大,= 6
2.
用1中的结果,t = 4,P(-4/5,-12/5)
BC的方程:x/4 + y/3 = 1,3x + 4y - 12 = 0
P与BC的距离d = |-12/5 - 48/5 - 12|/5 = 24/5 > AP = 4,Q在AB上
以P为圆心,半径为4的圆的方程:(x + 4/5)² + (y + 12/5)² = 16
与AB的方程(-x/4 + y/3 = 1,3x - 4y + 12 = 0)联立,得25x²/16 + 97x/10 + 69/5 = 0
(x + 4)(25x/16 + 69/20) = 0
x = -4 (点P)
x = -276/125
Q( -276/125,168/125)
AQ² = (-4 + 276/125)² + (168/125 - 0)² = (280/125)²
AQ = 280/125 = 10 - 4k
k = 97/50
天哪,真辛苦.