# Find the asymptotes of the hyperbola. $xy=4x+3y$

$\begin{array}{1 1}(A)\;x-3=0\;y+4=0 \\(B)\;x-3=0\;y-4=0 \\(C)\;x+3=0\;y+4=0\\(D)\;x+3=0\;y-4=0 \end{array}$

Given $xy=4x+3y$
=> $x(y-4)=3(y-4) +12$
=> $(x-4)(y-4)=12$
$\therefore$ joint equation of asymptotes is $(x-3)(y-4)=0$
Hence $x-3=0,y-4=0$
Hence B is the correct answer.