已知曲线ρcos2θ=ρ-8cosθ和直线x=4+2t,y=2-t相交于A,B两点(1)求AB中点M(2)求AB的长度

pcos2a=p-8cosap^2(cos^2a-sin^2a)=p^2-8pcosax^2-y^2=x^2+y^2-8xy^2=4x直线x=4+2t,y=2-t,即有2y+x=8.代入到y^2=4x中有y^2=4(8-2y),y^2+8y-32=0y1+y2=-8,y1y2=-32x1+x2=-2(y1+y2)+16