Find the area common to the circle x2 plus y2 equals 16a2 and the parabola y2 equals 6ax Hence find the larger of the area into which the circle is divided by the parabola?

Answer 1

Circle : x^2 + y^2 = 16a^2 , therefore, y = +√(16a^2 - x^2) (above X-axis)

Parabola : y^2 = 6ax , therefore, y = +√(6ax) (above X-axis)

Intersection point of circle with parabola :

√(16a^2 - x^2) = √(6ax)

16a^2 - x^2 = 6ax

x^2 + 6ax - 16a^2 = 0

(x - 2a)(x + 8a) = 0

x = 2a (as the only positive zero)

Total common area = 2 * common area above X-axis

Common area above X-axis

= ∫ (6ax) dx (from 0 to 2a) + ∫ √(16a^2 - x^2) dx (from 2a to 4)

= {2x√(6ax)/3} (from 0 to 2a) +

{(x/2)√(16a^2 - x^2) + 8arctan[x/√(16a^2 - x^2)]} (from 2a to 4)

= 8√(3)*a^2/3 + 8√(a^2 - 1) + 8arctan[1/√(a^2 - 1)] - 2√(3)*a^2 - 4π/3

Therefore, total common area

= 2{8√(3)*a^2/3 + 8√(a^2 - 1) + 8arctan[1/√(a^2 - 1)] - 2√(3)*a^2 - 4π/3}

Area of circle = πr^2 = 16π

Therefore, larger area of the circle

= 16π - 2{8√(3)*a^2/3 + 8√(a^2 - 1) + 8arctan[1/√(a^2 - 1)] - 2√(3)*a^2 - 4π/3}

= (4/3){14π - a^2√(3) - 12√(a^2 - 1) - 12arctan[1/√(a^2 - 1)]}

(Note: if a = 1, then arctan[1/√(a^2 - 1)] = π/2)