Equation of the tangent at $\left(3,-\frac{9}{2}\right)$ is $\frac{3x}{a^2}+\frac{-\frac{9}{2}y}{b^2}=1$

& given equation of the tangent is:

$x-2y=12 \Rightarrow \frac{x}{12}+ \frac{-y}{6}=1$

On comparing these equations:

$\frac{a^2}{3}=12\Rightarrow a^2=36\Rightarrow a=6$

$\frac{2b^2}{9}=6\Rightarrow b^2=27\Rightarrow b=3\sqrt{3}$

Therefore, the length of latus rectum 

$=\frac{2b^2}{a}=\frac{2\times 27}{6}=9$