Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Last updated at Dec. 12, 2019 by Teachoo

Transcript

Ex 8.1, 10 Find the area bounded by the curve ๐ฅ2=4๐ฆ and the line ๐ฅ=4๐ฆ โ 2 Here, ๐ฅ2=4๐ฆ is a parabola And, x = 4y โ 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, ๐ฅ=4๐ฆโ2 Putting in equation of curve , we get ๐ฅ^2=4๐ฆ (4๐ฆโ2)^2=4๐ฆ 16๐ฆ^2+4โ16๐ฆ=4๐ฆ 16๐ฆ^2โ16๐ฆโ4๐ฆ+4=0 16๐ฆ^2โ20๐ฆ+4=0 4[4๐ฆ^2โ5๐ฆ+1]=0 4๐ฆ^2โ5๐ฆ+1=0 4๐ฆ^2โ4๐ฆโ๐ฆ+1=0 4๐ฆ(๐ฆโ1)โ1(๐ฆโ1)=0 (4๐ฆโ1)(๐ฆโ1)=0 So, y = 1/4 , y = 1 For ๐=๐/๐ ๐ฅ=4๐ฆโ2 ๐ฅ=4(1/4)โ2 ๐ฅ =โ1 So, point is (โ1, 1/4) For y = 1 ๐ฅ=4๐ฆโ2 ๐ฅ=4(1)โ2 ๐ฅ =2 So, point is (2, 1) As Point A is in 2nd Quadrant โด A = (โ1 , 1/4) & Point B is in 1st Quadrant โด B = (2 , 1) Finding required area Required Area = Area APBQ โ Area APOQBA Area APBQ Area APBQ = โซ_(โ1)^2โ๐ฆ๐๐ฅ Here, ๐ฆ โ Equation of Line ๐ฅ=4๐ฆโ2 ๐ฅ+2=4๐ฆ ๐ฆ=(๐ฅ + 2)/4 Area APBQ = โซ_(โ1)^2โ(๐ฅ + 2)/4 ๐๐ฅ = 1/4 โซ1_(โ1)^2โ(๐ฅ+2)๐๐ฅ = 1/4 [๐ฅ^2/2+2๐ฅ]_(โ1)^2 = 1/4 [(2^2/2+2(2))โ(ใ(โ1)ใ^2/2+2(โ1))] = 1/4 [(2+4)โ(1/2โ2))] = 1/4 [6โ1/2+2] = 1/4ร15/2 = 15/8 Area APOQBA Area APOQBA = โซ_(โ1)^2โใ๐ฆ ๐๐ฅใ Here, ๐ฆ โ Equation of Parabola ๐ฅ^2=4๐ฆ 4๐ฆ=๐ฅ^2 ๐ฆ=1/4 ๐ฅ^2 Area APOQBA = 1/4 โซ1_(โ1)^2โใ๐ฅ^2 ๐๐ฅใ = 1/4 [๐ฅ^3/3]_(โ1)^2 = 1/4 [((2)^3 โ (โ1)^3)/3] = 1/4 [(8 โ (โ1))/3] = 1/4 [(8 + 1)/3] = 3/4 Now, Required Area = Area APBQ โ Area APOQBA = 15/8 โ 3/4 " = " (15 โ 6)/8 = 9/8 โด Required Area = ๐/๐ Square units

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.