A(-2,0)设直线AM的斜率为k,显然k≠0,则AM:y=k(x+2) AN:y=-1/k(x+2)联立x^2/4+y^2=1解得M((2-8k^2)/(1+4k^2),4k/(1+4k^2))=(x1,y1)N((2k^2-8)/(k^2+4),-4k/(k^2+4))=(x2,y2)(1)令k=1,则M(-6/5,4/5)(2)直线MN:(y-y1)/(x-x1)=(y1-y2)/(x1-x2)化简得:y-y1=5kx/(4-4k^2)-5k(1-4k^2)/[2(1+4k^2)(1-k^2)]y-5kx/(4-4k^2)=3k/(2-2k^2)∴4(1-k^2)y-5kx=6k令y=0,∵k≠0 |