. Two common tangents to the circles $x^2 + y^2 = 2a^2$ and parabola $y^2 = 8ax$ are

Given circle: $x^2 + y^2 = 2a^2$
Given parabola: $y^2 = 8ax$
**Equation of tangent to parabola:**
For parabola $y^2 = 8ax$, tangent is:
$y = mx + \dfrac{2a}{m}$ ...(i)
**Condition for tangent to circle:**
For line $y = mx + \dfrac{2a}{m}$ to be tangent to circle $x^2 + y^2 = 2a^2$,
distance from centre $(0,0)$ = radius $= \sqrt{2}\ a$
$\Rightarrow \dfrac{\left|\dfrac{2a}{m}\right|}{\sqrt{1 + m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2a}{|m|\sqrt{1+m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2}{|m|\sqrt{1+m^2}} = \sqrt{2}$
$\Rightarrow |m|\sqrt{1+m^2} = \sqrt{2}$
Squaring both sides:
$\Rightarrow m^2(1+m^2) = 2$
$\Rightarrow m^4 + m^2 - 2 = 0$
$\Rightarrow (m^2 + 2)(m^2 - 1) = 0$
$\Rightarrow m^2 = 1 \quad (\because m^2 = -2 \text{ is not possible})$
$\Rightarrow m = \pm 1$
**Substituting in (i):**
When $m = 1$: $y = x + 2a$
When $m = -1$: $y = -x - 2a$
$\therefore$ The two common tangents are:
$\therefore \boxed{y = x + 2a \quad \text{and} \quad y = -(x + 2a)}$