The area (in sq. units) of the region bounded by the curve x2=4y and the straight line x=4y−2 is:
A
34
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B
98
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C
54
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D
78
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Solution
The correct option is B98 x2=4y is a parabola with vertex (0,0) and focus (0,1) x=4y−2 is a straight line with intercepts (−2,0) and (0,12) Points of intersection of the given parabola and the line are : (−1,14) and (2,1)
Required area =2∫−1(x+24−x24)dx =((x+2)28−x312)2−1 =(2−23)−(18+112) ⇒ Area =98sq. units