在平面直角坐标系中,O为原点,二次函数y=-1/4x²+bx+c的图像经过点A（4,0）,C（0,2）
3个回答
(1) y = -x²/4 + x/2 + 2 = -(x-1)²/4 + 9/4
(2)
对称轴: x = 1, D(1, 0)
tan∠A = OC/OA = 2/4 = 1/2
tan∠CDE = OD/OC = 1/2
∠A = ∠CDE
tan∠B = OC/BO = 2/2 = 1
∠B = 45°
所以要使两三角形相似,只需∠CED或∠DCE为45°即可
(a) ∠CED = 45°
因为∠B = 45°, ∠BDE = 90°, BC的延长线与对称轴的交点即为E,其横坐标为1
直线BC的方程为(截距式): x/(-2) + y/2 = 1, 取x = 1, y = 3
E(1, 3)
(b)∠DCE=45°
CD的斜率为k1 = (2-0)/(0-1) = -2
设CE的斜率为k2, tan∠DCE = 1 = |(k2 - k1)/(1+k1*k2)|
|k2 + 2| = |1-2k2|
k2 + 2 = 1 - 2k2, k2 = -1/3
或
k2 + 2 = 2k2 -1, k2 = 3 (画个草图可知,此时∠DCE=135°,舍去)
CE的方程为(斜截式): y = -x/3 + 2
取x = 1, y = 5/3
E(1, 5/3)
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