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/3
12、
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=1
3x²+4x=0
x1+x2=-4/3
y=-2x+1
y1+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)