解:(1)椭圆C: 直线AB:y=k(x-m), ,(10k2+6)x2-20k2mx+10k2m2-15m2=0. 设A(x1, y1)、B(x2,y2),则x1+x2=,x1x2= 则xm= 若存在,使为ON的中点,∴. ∴, 即N点坐标为. 由N点在椭圆上,则 即5k4-2k2-3=0.∴或(舍). 故存在,使.··········5分 (2)=x1x2+k2(x1-m)(x2-m) =(1+k2)x1x2-k2m(x1+x2)+k2m2 =(1+k2)· 由得 即k2-15≤-20k2-12,且k≠0.··········10分 |