To find area given equations are
y2 = 6ax ...(i)
x2 + y2 = 16a2 ...(ii)
On solving Equation (i) and (ii)
Or x2 + (6ax)2 = 16a2
Or x2 + (6ax)2 – 16a2 = 0
Or (x + 8a) (x – 2a) = 0
Or x = 2a or x = – 8a is not possible solution.
Then y2 = 6a(2a) = 12a2 = \(\pm\)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 x2 + y2 = 16a2 and the parabola y2 = 6ax is \(\frac{4a^2}{3}(4\pi + \sqrt{3})\) sq. units