Answer:
Option B
Explanation:
we have,
Length of transverse axis 2a=6$\Rightarrow$ a=3
and length of latusrectum, $\frac{2b^{2}}{a}=\frac{8}{3}$
$\Rightarrow 2b^{2}=\frac{8}{3}\times3=8\Rightarrow b^{2}=4 $
$\therefore$ Required equation of hyperbola is
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\Rightarrow \frac{x^{2}}{9}-\frac{y^{2}}{4}=1\Rightarrow 4x^{2}-9y^{2}=36$