数学选修4-4-极坐标与参数方程

10、|AB|=√[(4-3)²+(-π/4-π/3)²]=√(1+49π²/144)直线AB是(y-π/3)/(-π/4-π/3)=(x-3)/(4-3)即(-7π/12)x+y-8π/3=0所以O到直线距离=|0+0-8π/3|/√(49π²/144+1)=(8π/3)/√(49π²/144+1)所以面积=AB*O到AB距离÷2=4π/312、2t=-1-x所以y=3+2(-1-x)=-2x+1即AB是2x+y-1=0所以M到AB距离=|-2+2-1|/√(2²+1²)=√5/5把y=-2x+1代入曲线(-2x-1)²-x²=13x²+4x=0x1+x2=-4/3y=-2x+1y1+y2=-2(x1+x2)+2=14/3中点则x=(x1+x2)/2,y=(y1+y2)/2所以中点是(-2/3,7/3)