Given the foci points of a hyperbola are $\left(1\pm \sqrt{2},0\right)$

So, $S\left(1+\sqrt{2},0\right),S^{\text{'}}\left(1-\sqrt{2},0\right)$

$S{S}^{\text{'}}=2\mathrm{ae}=(1+\sqrt{2})-(1-\sqrt{2})$

$\Rightarrow 2ae=2\sqrt{2}$

$\Rightarrow 2\times a\times \sqrt{2}=2\sqrt{2}, \left(\text{given} e=\sqrt{2}\right)$

$\Rightarrow a=1$

Since $b^2=a^2(e^2-1)$

$\Rightarrow b^2=1(2-1)=1$

So, the latus rectum of a hyperbola is given by,

L.R. $=\frac{2b^2}{a}$

L.R. $=2$