If the line $x-2 y=12 is tangent to the ellipse

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

$\frac{3 x}{a^{2}}-\frac{9 y}{2 b^{2}}=1$

But given equation of tangent is, $x-2 y=12$

$\therefore \frac{3}{a^{2}}=\frac{-9}{2 b^{2} \cdot(-2)}=\frac{1}{12}$ (On comparing)

$\Rightarrow a^{2}=3 \times 12$ and $b^{2}=\frac{9 \times 12}{4}$

$\Rightarrow a=6$ and $b=3 \sqrt{3}$

Therefore, latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 27}{6}=9$