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

Ex 8.1, 10 - Chapter 8 Class 12 Application of Integrals - Part 2
Ex 8.1, 10 - Chapter 8 Class 12 Application of Integrals - Part 3
Ex 8.1, 10 - Chapter 8 Class 12 Application of Integrals - Part 4
Ex 8.1, 10 - Chapter 8 Class 12 Application of Integrals - Part 5 Ex 8.1, 10 - Chapter 8 Class 12 Application of Integrals - Part 6 Ex 8.1, 10 - Chapter 8 Class 12 Application of Integrals - Part 7 Ex 8.1, 10 - Chapter 8 Class 12 Application of Integrals - Part 8

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Question 8 Find the area bounded by the curve 𝑥2=4𝑦 and the line 𝑥=4𝑦 – 2 Here, 𝑥2=4𝑦 is a parabola And, x = 4y – 2 is a line which intersects the parabola at points A and B We need to find Area of shaded region First we find Points A and B Finding 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 𝑥^2=4𝑦 (4𝑦−2)^2=4𝑦 16𝑦^2+4−16𝑦=4𝑦 16𝑦^2−16𝑦−4𝑦+4=0 16𝑦^2−20𝑦+4=0 4[4𝑦^2−5𝑦+1]=0 4𝑦^2−5𝑦+1=0 4𝑦^2−4𝑦−𝑦+1=0 4𝑦(𝑦−1)−1(𝑦−1)=0 (4𝑦−1)(𝑦−1)=0 So, y = 1/4 , y = 1 For 𝒚=𝟏/𝟒 𝑥=4𝑦−2 𝑥=4(1/4)−2 𝑥 =−1 So, point is (–1, 1/4) For y = 1 𝑥=4𝑦−2 𝑥=4(1)−2 𝑥 =2 So, point is (2, 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 Area APBQ Area APBQ = ∫_(−1)^2▒𝑦𝑑𝑥 Here, 𝑦 → Equation of Line 𝑥=4𝑦−2 𝑥+2=4𝑦 𝑦=(𝑥 + 2)/4 Area APBQ = ∫_(−1)^2▒(𝑥 + 2)/4 𝑑𝑥 = 1/4 ∫1_(−1)^2▒(𝑥+2)𝑑𝑥 = 1/4 [𝑥^2/2+2𝑥]_(−1)^2 = 1/4 [(2^2/2+2(2))−(〖(−1)〗^2/2+2(−1))] = 1/4 [(2+4)−(1/2−2))] = 1/4 [6−1/2+2] = 1/4×15/2 = 15/8 Area APOQBA Area APOQBA = ∫_(−1)^2▒〖𝑦 𝑑𝑥〗 Here, 𝑦 → Equation of Parabola 𝑥^2=4𝑦 4𝑦=𝑥^2 𝑦=1/4 𝑥^2 Area APOQBA = 1/4 ∫1_(−1)^2▒〖𝑥^2 𝑑𝑥〗 = 1/4 [𝑥^3/3]_(−1)^2 = 1/4 [((2)^3 − (−1)^3)/3] = 1/4 [(8 − (−1))/3] = 1/4 [(8 + 1)/3] = 3/4 Now, Required Area = Area APBQ – Area APOQBA = 15/8 – 3/4 " = " (15 − 6)/8 = 9/8 ∴ Required Area = 𝟗/𝟖 Square units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo