Question

如图,抛物线y=ax2+bx+c经过A(-1,0)B(3,0)C(0,3)三点,对称轴与抛物线交于点P，与直线BC相交于点M，连接PB。

Answer 1

1
A(-1,0)B(3,0)C(0,3)
x1=-1,x2=3
x=0,y=c=3
x1x2=c/a=-3
a=-1
x1+x2=2=-b/a
b=2
y=
-x^2+2x+3
对称轴x=-b/2a=(x1+x2)/圆手2=1
c-b^2/4a=3-4/(-4)=4
顶点P(1,4)
直线BC：y-3=[(3-0)/(0-3)]x

y=-x+3

x=1,y=2
M(1,2)
P(1,4)
2
过P平行BC直线
y-4=-(x-1)

y=5-x

y=-x^2+2x+3
5-x=-x^2+2x+3
x^2-3x+2=0
(x-2)(x-1)=0
x=1,y=4或x=2,y=3
Q(1,4)或Q(2,3)
3
|PM|=4-2=2
|BM|^2=(3-1)^2+2^2=8
|BM|=√橘旁嫌8
R(x0,y0)
过R垂直BC直线y-y0=x-x0

y=-x+3

-x+3-y0=x-x0
x=(x0+3-y0)/2

y-y0=3-y-x0
y=(3+y0-x0)/2

R到BM距离d^2=[(x0+3-y0)/2-x0]^2+[(3+y0-x0)/2-y0]^2

=2[(3-x0-y0)/2]^2

d=√2|3-x0-y0|/2
d*√8=2*(x0-1)
4|3-x0-y0|/启禅2=2(x0-1)
|3-x0-y0|=x0-1
a
|3-x0-y0|=3-x0-y0时
3-x0-y0=x0-1
y0=-x0^2+2x0+3
-x0^2+4x0-1=0
(x0-2)^2=3
x0=2-√3
y0=7-2√3-7+4√3=2√3，3-x0-y0<0
(舍去）
b
|3-x0-y0|=x0+y0-3
x0+y0-3=x0-1

y0=2

2=-x^2+2x+3

-x^2+2x+1=0
(x-1)^2=2
x=√2+1
R(√2+1,
2)