Question

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

Answer 1

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