The larger of the two areas (in sq. units) into which the circle x 2 + y 2 = 16 a 2 is divided by the parabola y 2 = 6 a x , is

Question

The larger of the two areas (in sq. units) into which the circle x 2 + y 2 = 16 a 2 is divided by the parabola y 2 = 6 a x , is, 4 a 2 3 8 π - 3, 4 a 2 3 4 π - 3, 2 a 2 3 4 π + 3, 4 a 2 3 4 π + 3

MARKS App · Accepted Answer

For point of intersection, x 2 + 6 a x = 16 a 2 ⇒ x = 2 a Area = π ⋅ ( 4 a ) 2 − 2 ( ∫ 0 2 a 6 a x d x + ∫ 2 a 4 a 16 a 2 − x 2 d x ) = 16 π a 2 − 2 { ( 2 6 a x ⋅ 3 2 3 ) 0 2 a + ( x 2 16 a 2 − x 2 + 16 a 2 2 sin − 1 x 4 a ) 2 a 4 a } = 16 π a 2 − 2 { 2 6 a ⋅ 2 2 a a 3 + ( 8 a 2 ⋅ π 2 − ( a ⋅ 2 3 a + 8 a 2 ⋅ π 6 ) ) } = 16 π a 2 − 2 { 8 3 a 2 3 + 8 π a 2 3 − 2 3 a 2 } = 32 π a 2 3 − 4 3 a 2 3 = 4 a 2 3 ( 8 π − 3 ) sq . units

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