If the line $2x+\sqrt{6}y=2$ touches the hyperbola $x^2-2y^2=a^2,$ then $a^2$ is equal to 

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

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

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

On squaring, we get,
$$\begin{gathered} \frac{4}{6}=a^2\cdot \frac{4}{6}-\frac{a^2}{2} \\ \Rightarrow \frac{2}{3}=\frac{a^2}{6}\Rightarrow a^2=4 \end{gathered}$$