一条渐近线方程为4y=3x,即b/a=3/4.
设方程是x^2/(4k)^2-y^2/(3k)^2=1.(k>0)
那么c=根号（a^2+b^2)=5k.
焦点坐标是（5k,0)
|3*5k|/根号（9+16)=6
|15k|=30
k=2.
所以,方程是x^2/64-y^2/36=1