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Question

The area common to the circle x2+y2=16a2 and the parabola y2=6ax is

A
4a23(4π3)
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B
4a23(8π3)
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C
4a23(4π+3)
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D
None of these
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Solution

The correct option is D 4a23(4π+3)
Solving the two, we get x2+6a16a2=0
or (x+8a)(x2a)=0
x=2a
y=±23a
The required common area =2[APOA]
=2ydx+2ydx
=22a06axdx+24a2a(4a)2x2dx
=2.6a23[x3/2]2a0+2.[x2(4a)2x2+12(4a)2sin1x4a]4a2a
=2.6a23(2a)2a+2[(0a23a)+8a2(sin11sin112)]
=2.23.43a2+2[23a2(π2π6)]
=1633a243a2+16a2π3
=43a23+16πa23=4a23(4π+3)

470339_261309_ans_544198ed15314ee7b223f8b5f206a200.png

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