1.设出A(x1,y1),B(x2,y2),F(c,0),因为AF=2FB,即(c-x1,-y1)=2(x2-c,y2),即y1=-2y2 x^2/a^2+y^2/b^2=1与y=√3(x-c)联立,得到(1/3b^2+a^2)y^2+(2b^2c/√3)y-b^4=0 y1+y2=-(2b^2c/√3)/ (1/3b^2+a^2)=-y2,y2=(2b^2c/√3)/ (1/3b^2+a^2) y1*y2=-b^4/(1/3b^2+a^2)=-2*y2^2,2*y2^2= b^4/(1/3b^2+a^2) 将y2代入上式,得到b^4/(1/3b^2+a^2)=2*(4b^4*c^2/3)/ (1/3b^2+a^2)^2 即8c^2=b^2+3a^2,即8c^2=a^2-c^2+3a^2,c^2=4/9a^2,e=2/3,
2.|AB|=√(1+k^2)|x1-x2|=√(1+1/k^2)|y1-y2|=√(1+1/k^2)|3y2|= 因为e=2/3,所以c=2/3a,b=√5/3a,代入y2=(2b^2c/√3)/ (1/3b^2+a^2) 所以y2=(2b^2c/√3)/ (1/3b^2+a^2)=5 √3/24a |AB|=√(1+1/k^2)|3y2|=(2√3/3)| 5 √3/8 a |=5/4a=15/4,所以a=3,b=√5 所以x^2/9+y^2/5=1