抛物线y＝x²-4x＋3交x轴与M、N点（M在N左边）,交y轴于点D,E在第一象限抛物线上∠EMN＝2∠ODM,求E点

Question

抛物线y＝x²-4x＋3交x轴与M、N点（M在N左边）,交y轴于点D,E在第一象限抛物线上∠EMN＝2∠ODM,求E点

Answer 1

y = x²-4x＋3 = (x - 1)(x - 3)
M(1,0),N(3,0,D(0,3)
tan∠ODM = OM/OD = 1/3
ME的斜率k=tan∠EMN = tan(2∠ODM) = 2tan∠ODM/(1 - tan²∠ODM) = （2*1/3)/(1 _ 1/9) = 3/4
ME的方程:y - 0 = (3/4)(x - 1),y = (3/4)(x - 1)
(3/4)(x - 1) = (x - 1)(x - 3),x = 15/4 (舍去 x = 1）
与抛物线联立,的E(15/4,33/16)
如果没学过有关的倍角公式,则可先取直线y = (x - 1)/3,此为EMN的平分线,取线上一点,如A(4,1).设E为y = k(x - 1),则A与EM和x轴的距离均为1,这样可以得到k = 3/4