已知抛物线y=x²-3x-四分之七的顶点为D,并与x轴交于A.B两点(A左B右)与y轴交于c,.关键是第3问,

已知抛物线y=x²-3x-四分之七的顶点为D,并与x轴交于A.B两点(A左B右)与y轴交于c,.关键是第3问,
已知抛物线y=x²-3x-四分之七的顶点为D,并与x轴交于A.B两点(A左B右)与y轴交于c,.

关键是第3问,

已知抛物线y=x²-3x-四分之七的顶点为D,并与x轴交于A.B两点(A左B右)与y轴交于c,.关键是第3问,
(1) y = (x - 3/2)² - 4 = (x + 1/2)(x - 7/2)
D(3/2,-4),A(-1/2,0),B(7/2,0),C(0,-7/4)
(2) AOC为直角三角形,两直角边的比为1/2 :7/4 = 2 :7
P(0,p),p > 0
(a) OP :OA = p :1/2 = 2 :7,p = 1/7,P(0,1/7)
(b) OA :OP = 1/2 :p = 2 :7,p = 7/4,P(0,7/4)
(3) G(5/2,-2)
l:x/(-3/2) + y/(-3/4) = 1
G的坐标满足上述方程,在l上.
设对称点为M'(m,0),l的斜率为(-3/4)/(0 + 3/2) = -1/2,MM'的斜率为2,方程为y = 2(x - m)
与l的交点为((8m - 3)/10,-(2m + 3)/5)
M(u,v),(u + m)/2 = (8m - 3)/10,u = 3(m - 1)/5
(v + 0)/2 = -(2m + 3)/5,v = -2(2m + 3)/5
带入抛物线得36m² - 172m + 161 = (18m - 23)(2m - 7) = 0
M(3/2,-4),(1/6,-20/9)
最后一部分请自己验算一下.