Let H be the hyperbola, whose foci are (1 ± 2 , 0) and eccentricity...

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

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

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

$⇒ 2ae = 2\sqrt{2}$

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

$⇒ a = 1$

Since ${b}^{2} = {a}^{2} ( {e}^{2} - 1 )$

$⇒ {b}^{2} = 1 ( 2 - 1 ) = 1$

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

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

L.R. $= 2$