点A,B在抛物线y^2=4x上,已知线段AB的中点为(1,1)求直线AB的方程(2)线段AB的长

1、
直线y-1=k(x-1)
y=kx+(1-k)
代入
k²x²+2k(1-k)x+(1-k)²=4x
k²x²+2(-k²+k-2)x+(1-k)²=0
x1+x2=2(k²-k+2)/k²
中点横坐标=(x1+x2)/2=(k²-k+2)/k²=1
k=2
所以是2x-y-1=0
2、
2x-y-1=0
y=2x-1
代入
4x²-4x+1=4x
4x²-8x+1=0
x1+x2=2,x1x2=1/4
(x1-x2)²=(x1+x2)²-4x1x2=3
y=2x-1
(y1-y2)²=(2x1-1-2x2+1)²=4(x1-x2)²=12
所以AB=√(3+12)=√15

令x=3cosθ
则y²=4(1-cos²θ)=4sin²θ
y=2sinθ
T=6sinθ+6cosθ
=6√2(√2/2sinθ+√2/2cosθ)
=6√2(sinθcosπ/4+cosθsinπ/4)
=6√2sin(θ+π/4)
sin最大=1
所以T最大值=6√2 令x=3cosθ
则...