.【解析】(1)设椭圆的左焦点F(-c,0)(c>0),F1F2=2,则c=1当AB⊥F1F2时,tan∠AFF=y则M4F/见,AF|=、2+1=3M2,则由椭圆定义AF1+1AF21=2,:a=V,b=1,故椭圆的标准方程为。+y2=1(2)由题意直线AB的斜率必定不为零,于是可设直线AB:x=ty+1,,要三由【大空直个一县面,圆的等直景面,ty+1分联立方程{x212+y=1(2+2)y2+21y-1=0,用面为是x直线AB交椭圆于A(∴△=42+4(2+2)=8(2+1)>0,=8出设N为AB中点,则y=-2+2…xN=时水银,·由韦达定理y+y=z+2yy厘t2+22-+1=+t2x+2因为△MAB是以∠AMB为顶角的等腰三角形,MN⊥AB-一,Ax,一22=千,22又AN-号AB-2)32=√1+r+2∴tan∠MAN1AN=y2(x2+3)-(√r+1+_2≥2·2v2+1,·非当且仅当√+1元千即=士1时取等号,此时∠AMB最大+68AN|=√1+1√·√+1-2√2I MN31+23822√2二32所以此时△MAB的面积为S39

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