If the foci of hyperbola lies on the line $y=x,$ one asymptote is $y =2x$ and it is passing through the point  $(3,4),$then

Other asymptote is the image of y = 2x in line x= y. i e , x= 2y
=> Hyperbola is (x – 2y) (2x - y) = k
$\therefore$ It passes through (3, 4) => K = - 10
angle between asymptotes = 2sec-1 e
$=>{\tan }^{-1}\left(\frac{3}{4}\right)=2\mathrm{se}{\mathrm{e}}^{-1}\mathrm{e}=>\mathrm{e}=\frac{\sqrt{10}}{3}$