If the equation $8[{(\mathrm{x}+1)}^2+{( \mathrm{y}-1)}^2 ] = {(\mathrm{x}-\mathrm{y}+3)}^2$ represents a conic. The equation of its latusrectum is

$$\begin{gathered} {(\mathrm{x}+1)}^2+{(\mathrm{y}-1)}^2=\frac{1}{8}{\left(\frac{\mathrm{x}-\mathrm{y}+3}{\sqrt{2}}\right)}^2\times 2 \\ {(\mathrm{x}+1)}^2+{(\mathrm{y}-1)}^2=\frac{1}{4}{\left(\frac{\mathrm{x}-\mathrm{y}+3}{\sqrt{2}}\right)}^2 \\ Focus \mathrm{S}=(-1, 1) ,\mathrm{e}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{directrix}=\mathrm{x}-\mathrm{y}+3=0 \\ \mathrm{eq}uation of \mathrm{L}.\mathrm{R} \mathrm{is} x-y+k=0 \\ \sin ce it passes through (-1,1) then \\ \Rightarrow -1-1+k=0 \\ \Rightarrow k=2 \\ \Rightarrow \mathrm{x}-\mathrm{y}+2=0 \end{gathered}$$