The common tangents to the circle $x^2+y^2=2$ and the parabola $y^2=8x$ touch the circle at the points P, Q and the parabola at the points R, S. Then the area of the

quadrilateral PQRS is

The equation of any tangent to the parabola can be considered as  $y=mx+\frac{a}{m}=mx+\frac{2}{m}$

i.e., $m^{2}x-my=0$

As we know that, the length of the perpendicular from the center to the tangent to the circle is equal to the radius of a circle. 

Thus, $\frac{2}{\sqrt{m^4+m^2}}=\sqrt{2}$

$$\begin{gathered} m^4+m^2=2 \\ {m}^4+m^2-2=0 \\ (m^2+2) (m^2-1)=0 \end{gathered}$$

$m=1$

Hence, the equation of the tangents are

$y=x+2, y=-x-2$

Therefore, the points P, Q are (-1, 1), (-1, -1) and R, S are (2, 4) and (2, -4) respectively. 

Thus, the area of the quadrilateral PQRS 

$=\frac{1}{2}x(2+8)x3=15 sq. unit$