Let $\mathrm{y}=\mathrm{mx}+\mathrm{c}, \mathrm{m}>0$ be the focal chord of $y^{2}=-64 x$, which is tangent to $(x+10)^{2}+y^{2}=4$ Then, the value of $4 \sqrt{2}(\mathrm{~m}+\mathrm{c})$ is equal to
$y^{2}=-64 x$
focus : $(-16,0)$
$\mathrm{y}=\mathrm{mx}+\mathrm{c}$ is focal chord
$\Rightarrow \mathrm{c}=16 \mathrm{~m}$..(1)
$\mathrm{y}=\mathrm{m} \mathrm{x}+\mathrm{c}$ is tangent to $(\mathrm{x}+10)^{2}+\mathrm{y}^{2}=4$
$\Rightarrow y=m(x+10) \pm 2 \sqrt{1+m^{2}}$
$\Rightarrow \mathrm{c}=10 \mathrm{~m} \pm 2 \sqrt{1+\mathrm{m}^{2}}$
$\Rightarrow 16 \mathrm{~m}=10 \mathrm{~m} \pm 2 \sqrt{1+\mathrm{m}^{2}}$
$\Rightarrow 6 \mathrm{~m}=2 \sqrt{1+\mathrm{m}^{2}} \quad(\mathrm{~m}>0)$
$\Rightarrow 9 \mathrm{~m}^{2}=1+\mathrm{m}^{2}$
$\Rightarrow \mathrm{m}=\frac{1}{2 \sqrt{2}} \& \mathrm{c}=\frac{8}{\sqrt{2}}$
$4 \sqrt{2}(\mathrm{~m}+\mathrm{c})=4 \sqrt{2}\left(\frac{17}{2 \sqrt{2}}\right)=34$