(1)x²+5y²-5=0
(2)右焦點F(2,0)
設直線AB：y=k(x-2)與x²+5y²-5=0聯(lián)立消去x得：
x²+ 5k²(x-2)²-5=0,即（5k²+1)x²-20k²x+20k²-5=0
設A(x1,y1),B(x2,y2),M(0,m)
X1+X2=20k²/(5K²+1),(1)X1X2=(20K²-5)/(5k²+1)(2)
∵向量MA=aAF, MB=bBF
∴(x1,y1-m)=a(2-x1,-y1),(x2,y2-m)=b(2-x2,-y2)
∴x1=a(2-x1),x2=b(2-x2)
∴a=x1/(2-x1),b=x2/(2-x2)
則 a+b=x1/(2-x1)+ x2/(2-x2)
=2[(x1+x2)-x1x2]/[4-2(x1+x2)+x1x2]
∵(x1+x2)-x1x2
=20k²/(5K²+1)-(20K²-5)/(5k²+1)
= 5/(5k²+1)
4-2(x1+x2)+x1x2
=4-40k²/(5K²+1)+(20K²-5)/(5k²+1)
=-1/(5k²+1)
∴a+b=[2*5/(5k²+1)]/[-1/(5k²+1)]=-10