如图,抛物线y=ax²+bx+4的对称轴是直线x=3/2,与x轴交于C,并且点A的坐标为（-1,0）1）求抛物线的解析式（2）过点C作CD∥x轴交抛物线于点D,链接AD交Y轴于点E,连接AC,设△AEC的面积为S1,△DEC的面积

Question

如图,抛物线y=ax²+bx+4的对称轴是直线x=3/2,与x轴交于C,并且点A的坐标为（-1,0）
1）求抛物线的解析式
（2）过点C作CD∥x轴交抛物线于点D,链接AD交Y轴于点E,连接AC,设△AEC的面积为S1,△DEC的面积为S2,求S1,S2的值
（3）点F坐标为（6,0）,连接DF,在（2）的条件下,点P从点E出发,以每秒3个单位的速度沿E→C→D→F匀速运动,点Q从点F出发,以每秒2个单位的速度沿F→A匀速运动,当其中一点到达终点,另外一点也随之停止运动,若点P,Q同时出发,设运动时间为t秒,当t为何值时,以D,P,Q为顶点的三角形是直角三角形?请直接写出所有符合条件的t值

Answer 1

(1)
对称轴为x = -b/(2a) = 3/2, b = -3a
y = ax² - 3ax + 4
x = -1, y = a + 3a + 4 = 0, a = -1
y = -x² + 3x + 4
(2)
C(0, 4), C, D关于x = 3/2对称,D(3, 4)
AD的方程: (y - 0)/(4 - 0) = (x + 1)/(3 + 1)
x = 0, y = 1
E(0, 1)
EC = 4 - 1 = 3
EC上的高 = AO = 1
S1 = (1/2)*EC*AO = 3/2
S2 = (1/2)*CD *EC = (1/2)*3*3 = 9/2
(3)
AF = 7, Q从F到A需7/2秒
从D向x轴作垂线,垂足D'(3, 0)；D'F = 3, D'D = 4, DF = 5
E->C->D->F长为3 + 3 + 5 = 11, P从F到A需11/3秒 > 7/2, 即Q先到终点
t秒时,Q(6 - 2t, 0), 0 ≤ t ≤ 7/2
EC = 3: P(0, 1 + 3t), 0 ≤ t < 1
CD = 3: P(3(t - 1), 4), 1 ≤ t < 2
DF = 5: P在DF上时,从P做x轴的平行线,与D'D交于P’(x, y)
∆DP'P与∆DF'相似, DP = 3(t - 2)
P'P/D'F = DP/DF, P'P/3 = 3(t - 2)/5
P'P = 9(t - 2)/5, P的横坐标为3 + 9(t - 2)/5 = 3(3t - 1)/5
DP'/DD' = DP/DF, DP'/4 = 3(t - 2)/5, DP' = 12(t - 2)/5
P的纵坐标 = D'D - DP' = 4 - 12(t - 2)/5 = 4(11 - 3t)/5
P(3(3t - 1)/5, 4(11 - 3t)/5), 2 ≤ t ≤ 7/2
(i) 0 ≤ t < 1
PQ² = 13t² - 18t + 37
PD² = 9t² - 18t + 18
DQ² = 4t² - 12t + 25
分别令三者为斜边,用勾股定理,只有PQ为斜边时才有解: t = 1/2
(ii) 1 ≤ t < 2
与(i)类似,PQ为斜边时,t = 3/2 (舍去t = 2, 此时P, D重合)
DQ为斜边时,t = 9/5 (舍去t = 2)
(iii) 2 ≤ t ≤ 7/2
PQ为斜边时, t = 2, t = 25/6 > 4, 均舍去
DQ为斜边时, t = 2 (舍去), t = 55/21
三者结合, t = 1/2, 3/2, 或55/21
具体算挺繁琐,但做法不难.