Diameters $y=m_1x$
$y=m_2x$ are conjugate diameters of the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$.
if $m_1m_2= \large\frac{b^2}{a^2}$
Here $a^2=9$
$b^2=16$
$m_1= \large\frac{1}{2}$
$\therefore m_1m_2=\large\frac{b^2}{a^2}$
$\qquad= \large\frac{1}{2}$$ (m_2)$
$\qquad= \large\frac{16}{9}$
$m_2=\large\frac{32}{9}$
Thus the required diameter is $y= \large\frac{32x }{9}$
Hence B is the correct answer.