The latusrectum of the parabola 
$13\left[{\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}-4\right)}^2\right]={\left(2\mathrm{x}-3\mathrm{y}+5\right)}^2$

$$\begin{gathered} {\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}-4\right)}^2={\left(\frac{\left(2\mathrm{x}-3\mathrm{y}+5\right)}{\sqrt{9+4}}\right)}^2 \\ Focus=\left(3, 4\right) \end{gathered}$$

$$\begin{gathered} \mathrm{directrix} is 2\mathrm{x}-3\mathrm{y}+5=0 \\ \mathrm{length} \mathrm{of} \mathrm{latus} \mathrm{rectum} = 4\mathrm{a} \\ =2(\mathrm{perpendicular} \mathrm{distance} \mathrm{from} \mathrm{focus} \mathrm{to} \mathrm{directrix}) \\ =2\times \frac{\left|6-12+5\right|}{\sqrt{4+9}} \\ =\frac{2}{\sqrt{13}} \end{gathered}$$