发布网友 发布时间:2022-04-22 19:37
共2个回答
热心网友 时间:2023-09-14 16:46
(1) C(0, 3), 令B(t, 0), t > 2
tan∠CBA = 1/2 = OC/OB = 3/t = 1/2, t = 6
B(6, 0)
y = a(x - 2)(x - 6)
代入x = 0, y = 3, 得a = 1/4
y = (1/4)x² - 2x + 3
(2)
抛物线的对称轴为x = (2+6)/2 = 4, 顶点D(4, -1)
四边形ACBD的面积 = △ABC的面积+△ABD的面积 = (1/2)*AB*OC + (1/2)*AB*|D的纵坐标|
= (1/2)*(6-2)*3 + (1/2)*(6-2)*1 = 9
(3)
tan∠CBA = 1/2, ∠CBA与BC的倾斜角互补,BC的斜率 = -tan∠CBA = -1/2
如果B为直角顶点,则BE的斜率为-1/(-1/2) = 2, 方程为y = 2(x - 6)
与抛物线联立,得E(10, 8)
如果C为直角顶点,则CE的斜率为-1/(-1/2) = 2, 方程为y = 2x + 3, 与抛物线联立,得E(16, 35)
热心网友 时间:2023-09-14 16:46
B点坐标为(6,0)