If the line 2x + 6 y = 2 touches the hyperbola x ^ 2 - 2 y ^ 2 = a...

The tangent to the hyperbola is $y = - \dfrac{2}{\sqrt{6}}x + \dfrac{2}{\sqrt{6}}$

Also, tangent to hyperbola in slope form is $y = mx \pm \sqrt{{a}^{2}{m}^{2} - {b}^{2}}$

i.e. $\dfrac{2}{\sqrt{6}} = \sqrt{{a}^{2}{\left(\dfrac{- 2}{\sqrt{6}}\right)}^{2} - \dfrac{{a}^{2}}{2}}$

On squaring, we get,
$\dfrac{4}{6} = {a}^{2} ⋅ \dfrac{4}{6} - \dfrac{{a}^{2}}{2} ⇒ \dfrac{2}{3} = \dfrac{{a}^{2}}{6} ⇒ {a}^{2} = 4$