Simone如图抛物线y＝ax²＋bx＋c与y轴交于点C（0,3）与x轴交于A（－3,0）点B（1,0）两点
的有关信息介绍如下：
(1) y = -x² - 2x + 3 = -(x + 3)(x - 1)(2) AC的斜率为k = (3 - 0)/[0 - (-3)] = 1, CD的斜率k' = -1/k = -1, CD的方程为y = -x + 3y = -x + 3 = -x² - 2x + 3x² + x = x(x + 1)＝0x = -1 (舍去x = 0, 此为点C)D(-1, 4)(3) AC = 3√2, CD = √2, 即△ACD两条直角边之比为3:1令F(e, 0), E(e, -(e + 3)(e - 1))EF = |(e+3)(e - 1)|, AF = |e - (-3)| = |e + 3|EF/AF = |e - 1| = 3或|e - 1| = 1/3共有四个解：e = -2: E(-2, 3)e = 2/3, E(2/3, 11/9)e = 4, E(4, -21)e = 4/3, E(4/3, -13/9)
