The equation of the latusrectum of the parabola ${\mathrm{x}}^2-12\mathrm{x}-8\mathrm{y}+52=0$

${\mathrm{x}}^2-12\mathrm{x}-8\mathrm{y}+52=0$

$$\begin{gathered} {\mathrm{x}}^2-12\mathrm{x}=8\mathrm{y}-52 \\ {\mathrm{x}}^2-2\left(6\right)\mathrm{x}=8\mathrm{y}-52 \\ {\left(\mathrm{x}-6\right)}^2=8\mathrm{y}-52+36 \\ {\left(\mathrm{x}-6\right)}^2=8\mathrm{y}-16 \\ {\left(\mathrm{x}-6\right)}^2=8\left(\mathrm{y}-2\right) \\ \mathrm{Vertex}=\left(6,2\right) \\ and 4\mathrm{a}=8 \\ \Rightarrow \mathrm{a}=2 \end{gathered}$$

 $$\begin{gathered} focus=\left(6,2+2\right) \\ =\left(6,4\right) \end{gathered}$$

Here Latusrectum parallel to X-axis  so,it's equation is $\mathrm{y}=4$