jmc

Let the equation of a tangent line be $y=mx+b.$

Substituting into the equation $x^2+y^2=2,$ we get $$x^2+(mx+b{)}^2=2.$$Then $(m^2+1)x^2+2bmx+(b^2-2)=0.$ Since we have a tangent, this quadratic has a double root, meaning that its discriminant is 0. This gives us $$(2bm{)}^2-4(m^2+1)(b^2-2)=0,$$which simplifies to $b^2=2m^2+2.$

Solving for $x$ in $y=mx+b,$ we get $x=\frac{y-b}{m}.$ Substituting into $y^2=8x,$ we get $$y^2=\frac{8y-8b}{m},$$so $m{y}^2-8y+8b=0.$ Again, the discriminant of this quadratic will also be 0, so $$64-4(m)(8b)=0.$$Hence, $bm=2.$

Then $b=\frac{2}{m}.$ Substituting into $b^2=2m^2+2,$ we get $$\frac{4}{m^2}=2m^2+2.$$Then $4=2m^4+2m^2,$ so $m^4+m^2-2=0.$ This factors as $(m^2-1)(m^2+2)=0.$ Hence, $m^2=1,$ so $m=\pm 1.$

If $m=1,$ then $b=2.$ If $m=-1,$ then $b=-2.$ Thus, the two tangents are $y=x+2$ and $y=-x-2.$



We look at the tangent $y=x+2.$ Substituting into $x^2+y^2=2,$ we get $$x^2+(x+2{)}^2=2.$$This simplifies to $x^2+2x+1=(x+1{)}^2=0,$ so $x=-1.$ Hence, the tangent point on the circle is $(-1,1).$

We have that $x=y-2.$ Substituting into $y^2=8x,$ we get $$y^2=8(y-2).$$This simplifies to $(y-4{)}^2=0,$ so $y=4.$ Hence, the tangent point on the parabola is $(2,4).$

By symmetry, the other two tangent points are $(-1,-1)$ and $(2,-4).$

The quadrilateral in question is a trapezoid with bases 2 and 8, and height 3, so its area is $\frac{2+8}{2}\cdot 3=\boxed{15}.$