已知抛物线y＝ax 2＋4ax－c（a≠0）与y轴正半轴的交点为A,顶点D在x轴上且OD=2OA．

(1)

x = 0,y = -c > 0,c <0,A(0,-c)
y = ax² + 4ax - c = a(x + 2)² - 4a - c
D(-2,-4a - c)
2 = 2(-c),c = -1
-4a - c = 0,a = 1/4
y = x²/4 + x + 1


(2)
对称轴x = -2
旋转后, A和O分别变为A'和O'. PA = PA',PA⊥PA', 即PAA'为等腰直角三角形,P为直角顶点.
于是A'和A关于对称轴对称, A'(-4,1)
令P(-2,p)
A'A² = 4² = 16 = 2AP² = 2[(-2 - 0)² + (p - 1)²]
p = 3,P(-2,3)
p = -1, 此时为逆时针旋转,舍去
OP² = 13

OP的斜率为k = -3/2
O'P的斜率为k' = -1/k = 2/3
O'P的方程y - 3 = (2/3)(x + 2)
令O'(u,(2/3)(u + 2) + 3)
O'P² = (-2 - u)² + [3 - (2/3)(u + 2) - 3]² = (13/9)(u + 2)² = 13
u = -5,O'(-5,1)
u = 1, 显然不合题意,舍去



(3)
D(-2,0)

AD的斜率为k = 1/2
BD⊥AD,BD的斜率为-2, 方程为y - 0 = -2(x +2),y = -2x - 4

与抛物线联立:-2x - 4 = x²/4 + x + 1
(x + 2)(x + 10) = 0
x = -10,B(-10,16)

x = 2为D,不考虑