若AB与y轴平行,则x=3,不符
于是可设AB：y=k(x-3)+1
xx/36+(kx-3k+1)^2/9=1
(4kk+1)xx+4(-6kk+2k)x+4(3k-1)^2-36=0
3=(x1+x2)/2
6=x1+x2=4(6kk-2k)/(4kk+1)
-8k=6
k=-4/3