# Ex 8.1, 10

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 8.1, 10 Find the area bounded by the curve 𝑥2=4𝑦 and the line 𝑥=4𝑦 – 2 Let AB represent the line 𝑥=4𝑦−2 AOB represent the curve 𝑥2=4𝑦 First we find Points A and B Points A & B are the intersection of curve and line We know that, 𝑥=4𝑦−2 Putting in equation of curve , we get 𝑥2=4𝑦 4𝑦−22=4𝑦 16 𝑦2+4−16𝑦=4𝑦 16 𝑦2−16𝑦−4𝑦+4=0 16 𝑦2−20𝑦+4=0 4 4 𝑦2−5𝑦+1=0 4 𝑦2−5𝑦+1=0 4 𝑦2−4𝑦−𝑦+1=0 4𝑦 𝑦−1−1 𝑦−1=0 4𝑦−1 𝑦−1=0 So, y = 14 , y = 1 As Point A is in 2nd Quadrant ∴ A = −1 , 14 & Point B is in 1st Quadrant ∴ B = 2 , 1 Finding required area Required Area = Area APBQ – Area APOQBA = −12𝑦1𝑑𝑥− −12𝑦2𝑑𝑥 Required Area = −12 𝑥 + 24𝑑𝑥− −12 𝑥24𝑑𝑥 = 14 −12 𝑥+2𝑑𝑥 − 14 −12 𝑥2 𝑑𝑥 = 14 𝑥22+2𝑥−12− 14 𝑥33−12 = 14 22 − −122+2 2− −1− 14 23 − −133 = 14 4 − 12+2 3− 14 8 + 13 = 14 32+6− 14 93 = 14 32+6−3 = 14 32+3 = 14 92 = 98 ∴ Required Area = 𝟗𝟖 Square units

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