解:
x*sinα+y*cosα+m=0
y=-(sinα/cosα)*x-m/cosα 代入园方程
x^2+y^2=2
得
x^2+[-(sinα/cosα)*x-m/cosα]^2=2
(cos^2α+in^2α)*x^2+(2m*sinα)*x+m^2=2cos^2α
x^2+(2m*sinα)*x+m^2-2cos^2α=0
解上方程得直线x*sinα+y*cosβ+m=0与圆x^2+y^2=2的交点坐标:
x=-m*sinα±√(m^2*sin^2α-m^2+2cos^2α)
x1-x2=2√(m^2*sin^2α-m^2+2cos^2α)
y1-y2=-2(sinα/cosα)*√(m^2*sin^2α-m^2+2cos^2α)
已知直线x*sinα+y*cosβ+m=0被圆x^2+y^2=2截得的线段的长为三分之四根号三,则
(x1-x2)^2+(y1-y2)^2=(4*√3/3)^2=16/3
4(m^2*sin^2α-m^2+2cos^2α)*(1+sin^2α/cos^2α)=16/3
(m^2*sin^2α-m^2+2cos^2α)*(sin^2α+cos^2α)=4cos^2α/3
m^2*sin^2α-m^2+2cos^2α=4cos^2α/3
-m^2*(1-sin^2α)=-2cos^2α/3
m^2*cos^2α=2cos^2α/3
m^2=2/3
m=±√(2/3)=±√6/3