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

$$\begin{gathered} Given parabola13\left[{\left(\mathrm{x}-3\right)}^2+{(\mathrm{y}-4)}^2\right]={(2\mathrm{x}-3\mathrm{y}+5)}^2 \Rightarrow {\left(\mathrm{x}-3\right)}^2+{(\mathrm{y}-4)}^2=\left(\frac{2\mathrm{x}-3\mathrm{y}+5}{\sqrt{13}}\right) \\ It in the form of {\mathrm{SP}}^2={\mathrm{PM}}^2 \\ Now \mathrm{LLR}=4a \\ =2\left(perpendicular \mathrm{distance} \mathrm{from} \mathrm{focus} (3,4) \mathrm{to} \mathrm{directrix} 2\mathrm{x}-3\mathrm{y}+5=0\right) \\ =2\left(\frac{\left|6-12+5\right|}{\sqrt{13}}\right) \\ =\frac{2}{\sqrt{13}} \end{gathered}$$