抛物线y=-x的平方+4x-3与坐标轴交于a,bc三点,将三角形aoc沿ac翻折得到三角形ace,直线ae交抛物线于p点,直线ap的解析式

Question

抛物线y=-x的平方+4x-3与坐标轴交于a,bc三点,将三角形aoc沿ac翻折得到三角形ace,直线ae交抛物线于p点,
直线ap的解析式

Answer 1

y=-x²+4x-3 = -(x - 1)(x - 3)
A(1,0),B(3,0),C(0,-3)
AC的解析式:x/1 + y/(-3) = 1,y = 3x - 3,斜率3
从O向AC做垂线,交于D
OD斜率 = -1/3
OD的解析式:y = -x/3
D(9/10,-3/10)
D为OE的中点,E(p,q)
9/10 = (0 + p)/2,p = 9/5
-3/10 = (0 + q)/2,q = -3/5
E(9/5,-3/5)
AE的解析式:(y - 0)/(-3/5 - 0) = (x - 1)/(9/5 - 1)
y = 3(1 - x)/4
与y= -x²+4x-3联立,x = 1(点A),x = 15/4,y = -33/16
P(15/4,-33/16)