Question

直线4x-4y-k=0与抛物线y^2=x交于A、B两点,若|AB|=4则弦AB的中点到直线x+1/2=0的距离

Answer 1

由4kx-4y-k=0 与y^2=x消去y得k²x²-(1/2k²+1)x+1/16k²=0

设AB中点坐标为C(x0,y0),上式两根为x1,x2则

x0=(x1+x2)/2=(1/2k²+1)/(2k²)=1/(2k²)+1/4

y0=k(x0-1/4)=1/(2k)

又AB=(k²+1)^(1/2)|x1-x2|=(1+k²)/k²=4 得k²=1/3

AB中点C(1/(2k²)+1/4,1/(2k)）到直线x+1/2=0的距离为1/(2k²)+3/4=9/4