Ex 8.1, 10 - Find area bounded by x2 = 4y and line x = 4y - 2 - Ex 8.1

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Ex 8.1, 10 Find the area bounded by the curve 2=4 and the line =4 2 Let AB represent the line =4 2 AOB represent the curve 2 =4 First we find Points A and B Points A & B are the intersection of curve and line We know that, =4 2 Putting in equation of curve , we get 2 =4 4 2 2 =4 16 2 +4 16 =4 16 2 16 4 +4=0 16 2 20 +4=0 4 4 2 5 +1 =0 4 2 5 +1=0 4 2 4 +1=0 4 1 1 1 =0 4 1 1 =0 So, y = 1 4 , y = 1 As Point A is in 2nd Quadrant A = 1 , 1 4 & Point B is in 1st Quadrant B = 2 , 1 Finding required area Required Area = Area APBQ Area APOQBA = 1 2 1 1 2 2 Required Area = 1 2 + 2 4 1 2 2 4 = 1 4 1 2 +2 1 4 1 2 2 = 1 4 2 2 +2 1 2 1 4 3 3 1 2 = 1 4 2 2 1 2 2 +2 2 1 1 4 2 3 1 3 3 = 1 4 4 1 2 +2 3 1 4 8 + 1 3 = 1 4 3 2 +6 1 4 9 3 = 1 4 3 2 +6 3 = 1 4 3 2 +3 = 1 4 9 2 = 9 8 Required Area = Square units

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