If the line $\mathrm{y}=\sqrt{2\mathrm{x}}-8\sqrt{2}$ is normal to the parabola ${\mathrm{y}}^2=8\mathrm{x}$ then its length is _____

$$\begin{gathered} \mathrm{Given} \mathrm{normal} \mathrm{y}=\sqrt{2}\mathrm{x}-8\sqrt{2} —\left(1\right), \\ and parabola {\mathrm{y}}^2=8\mathrm{x}—\left(2\right) \\ where \mathrm{m}=\sqrt{2}, \\ \mathrm{solving} \left(1\right) \& \left(2\right) \\ 2{\mathrm{x}}^2+128-32\mathrm{x}=8\mathrm{x} \\ 2{\mathrm{x}}^2-40\mathrm{x}+128=0 \\ {\mathrm{x}}^2-20\mathrm{x}+64=0 \\ (\mathrm{x}-16)\left(\mathrm{x}-4\right)=0 \\ \mathrm{x}=16 or x=4 \\ let two roots be x_1=16,x_2=4 \\ Now \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{normal} \mathrm{chord} = \left|{\mathrm{x}}_1-{\mathrm{x}}_2\right|\sqrt{1+{\mathrm{m}}^2} \\ =12\sqrt{3} \end{gathered}$$