To find area given equations are

y^{2} = 6ax ...(i)

x^{2} + y^{2} = 16a^{2} ...(ii)

On solving Equation (i) and (ii)

Or x^{2} + (6ax)^{2} = 16a^{2}

Or x^{2} + (6ax)^{2} – 16a^{2} = 0

Or (x + 8a) (x – 2a) = 0

Or x = 2a or x = – 8a is not possible solution.

Then y^{2} = 6a(2a) = 12a^{2} = \(\pm\)2√3a

Equation (i) represents a parabola with vertex (0, 0) and axis as x - axis.

Equation (ii) represents a with centre (0, 0) and meets axes (±4a, 0), (0, ±4a).

Point of intersection of circle and parabola are (2a, 2√3a), (2a, – 2√3a).

**These are shown in the graph below: -**

Required area = 2[Region ODCO + Region BCDB]

The area common to the circle x^{2} + y^{2} = 16a^{2} and the parabola y^{2} = 6ax is \(\frac{4a^2}{3}(4\pi + \sqrt{3})\) sq. units