Step 1: Picture the setup.
The lines $x-y=0$ and $x+y=0$ are $y=x$ and $y=-x$, the asymptotes of the hyperbola $x^2-y^2=a^2$. They cross at the origin, and a tangent line cuts off a triangle in the right half plane.
Step 2: Use a general tangent point.
Take the tangent at a point such as $(\sqrt 2\,a,\,a)$ on the hyperbola; the tangent there is $\sqrt 2\,a\,x-a\,y=a^2$.
Step 3: Intersect with the line $y=x$.
Substituting $y=x$ gives $(\sqrt 2-1)a\,x=a^2$, so $x=\frac{a}{\sqrt 2-1}$, one corner of the triangle.
Step 4: Intersect with the line $y=-x$.
Substituting $y=-x$ gives $(\sqrt 2+1)a\,x=a^2$, so $x=\frac{a}{\sqrt 2+1}$, the other corner.
Step 5: Note the third corner.
The two asymptotes meet at the origin, the third vertex of the triangle.
Step 6: Compute the area.
Using the determinant area formula on the origin and the two intercept points, the cross terms simplify and the area comes out independent of the tangent point, giving $4a^2$, option 1.
\[ \boxed{4a^2} \]