Check sibling questions

Example 14 - Prove that y2 = 4x, x2 = 4y divide area of square

Example 14 - Chapter 8 Class 12 Application of Integrals - Part 2
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 3
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 4
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 5
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 6
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 7
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 8
Example 14 - Chapter 8 Class 12 Application of Integrals - Part 9

Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month


Transcript

Example 14 Prove that curves 𝑦2=4𝑥 and 𝑥2=4𝑦 divide the area of the square bounded by 𝑥=0, 𝑥=4, 𝑦=4 and 𝑦=0 into three equal parts Drawing figure Here, we have parabolas 𝑦^2=4𝑥 𝑥^2=4𝑦 And, Square made by the lines x = 4, y = 4, x = 0, y = 0 We need to prove that area of square is divided into 3 parts by the curve So, we need to prove Area OPQA = Area OAQB = Area OBQR Area OPQA Area OPQA = ∫_0^4▒〖𝑦 𝑑𝑥〗 Here, 4𝑦=𝑥^2 𝑦=𝑥^2/4 So, Area OPQA =∫_0^4▒〖𝑥^2/4 𝑑𝑥〗 = 1/4 [𝑥^3/3]_0^4 =1/12×[4^3−0^3 ] =1/12 × [64−0] =64/12=16/3 Area OPQA is the area by curve x2 = 4y in the x-axis from x = 0 to x = 4 Area OBQR Since Area is on 𝑦−𝑎𝑥𝑖𝑠 , we use formula ∫1▒〖𝑥 𝑑𝑦〗 Area OBQR = ∫_0^4▒〖𝑥 𝑑𝑦〗 Here, 𝑦^2=4𝑥 𝑥=𝑦^2/4 So, Area OBQR =∫_0^4▒〖𝑦^2/4 𝑑𝑦〗 = 1/4 [𝑦^3/3]_0^4=1/12×[4^3−0^3 ]=1/12 ×[64−0]=16/3 Area OBQR is the area by curve y2 = 4x in the y-axis from y = 0 to y = 4 Area OAQB Area OAQB = Area OBQP – Area OAQP Finding Area OBQP Area OBQP =∫_0^4▒〖𝑦 𝑑𝑥〗 Here, 𝑦^2=4𝑥 𝑦=±√4𝑥 As OBQP is in 1st quadrant, value of y is positive ∴ 𝑦=√4𝑥 Area OBQP =∫_0^4▒〖√4𝑥 𝑑𝑥〗 =2∫_0^4▒〖√𝑥 𝑑𝑥〗 =2∫_0^4▒〖𝑥^(1/2) 𝑑𝑥〗 =2 × [𝑥^(1/2+1)/(1/2+1)]_0^4 =2 × [𝑥^(3/2)/(3/2)]_0^4 =2 × 2/3 [𝑥^(3/2) ]_0^4 =4/3 [(4)^(3/2)−(0)^(3/2) ] =4/3 [8−0] =32/3 Area OAQP Area OAQP =∫_0^4▒〖𝑦 𝑑𝑥〗 Here, 𝑥^2=4𝑦 𝑦=𝑥^2/4 Area OAQP =∫_0^4▒〖𝑥^2/4 𝑑𝑥〗 =1/4 [𝑥^(2+1)/(2+1)]_0^4 =1/(4 ×3) [𝑥^3 ]_0^4 =1/12 [4^3−0^3 ] Area OAQP Area OAQP =∫_0^4▒〖𝑦 𝑑𝑥〗 Here, 𝑥^2=4𝑦 𝑦=𝑥^2/4 Area OAQP =∫_0^4▒〖𝑥^2/4 𝑑𝑥〗 =1/4 [𝑥^(2+1)/(2+1)]_0^4 =1/(4 ×3) [𝑥^3 ]_0^4 =1/12 [4^3−0^3 ] =1/12 × [64−0] =16/3 ∴ Area OAQB = Area OBQD – Area OAQP = 32/3−16/3 = 16/3 So, Area OAQB = Area OAQP = Area OBRQ = 𝟏𝟔/𝟑 square units Hence Proved.

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.