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Misc 6 - Find area enclosed between parabola y2 = 4ax and y=mx

Misc 6 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 6 - Chapter 8 Class 12 Application of Integrals - Part 3
Misc 6 - Chapter 8 Class 12 Application of Integrals - Part 4
Misc 6 - Chapter 8 Class 12 Application of Integrals - Part 5
Misc 6 - Chapter 8 Class 12 Application of Integrals - Part 6
Misc 6 - Chapter 8 Class 12 Application of Integrals - Part 7
Misc 6 - Chapter 8 Class 12 Application of Integrals - Part 8

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Misc 6 Find the area enclosed between the parabola 𝑦2=4π‘Žπ‘₯ and the line 𝑦=π‘šπ‘₯ Let’s first draw the Figure Here, 𝑦2 =4ax is a Parabola And, 𝑦=π‘šπ‘₯ is a straight line Let A be point of intersection of line and parabola Finding point A Putting y = mx in equation of parabola 𝑦^2=4π‘Žπ‘₯ (π‘šπ‘₯)^2=4π‘Žπ‘₯ π‘š^2 π‘₯^2=4π‘Žπ‘₯ π‘š^2 π‘₯^2βˆ’4π‘Žπ‘₯=0 π‘₯(π‘š^2 π‘₯βˆ’4π‘Ž)=0 Therefore, π‘₯=0 π‘š^2 π‘₯βˆ’4π‘Ž=0 π‘š^2 π‘₯=4π‘Ž π‘₯=4π‘Ž/π‘š^2 Putting values of π‘₯ in 𝑦=π‘šπ‘₯ 𝑦=π‘š Γ—0=0 𝑦=π‘š Γ—4π‘Ž/π‘š^2 =4π‘Ž/π‘š So, the intersecting points are O(𝟎 , 𝟎) and A (πŸ’π’‚/π’Ž^𝟐 ,πŸ’π’‚/π’Ž) Finding Area Area Required = Area OBAD – Area OAD Area OBAD Area OBAD = ∫_0^(4π‘Ž/π‘š^2 )▒〖𝑦 𝑑π‘₯" " γ€— y β†’ Equation of parabola 𝑦^2 = 4ax 𝑦 = Β± √("4" π‘Žπ‘₯) Since OBAD is in 1st quadrant, value of y is positive ∴ y = √("4" π‘Žπ‘₯) Now, Area OBAD =∫_0^(4π‘Ž/π‘š^2 )β–’γ€–βˆš4π‘Žπ‘₯ 𝑑π‘₯" " γ€— =∫_0^(4π‘Ž/π‘š^2 )β–’γ€–βˆš4π‘Ž .√π‘₯ 𝑑π‘₯" " γ€— =√4π‘Ž ∫_0^(4π‘Ž/π‘š^2 )β–’γ€–βˆšπ‘₯ 𝑑π‘₯" " γ€— =2βˆšπ‘Ž [π‘₯^(1/2 + 1)/(1/2 + 1)]_0^(4π‘Ž/π‘š^2 ) =√4π‘Ž [π‘₯^(3/2)/(3/2)]_0^(4π‘Ž/π‘š^2 ) =√4π‘Ž Γ— 2/3 [π‘₯^(3/2) ]_0^(4π‘Ž/π‘š^2 ) =(2(2βˆšπ‘Ž))/3 [(4π‘Ž/π‘š^2 )^(3/2)βˆ’(0)^(3/2) ] =(4βˆšπ‘Ž)/3 [4π‘Ž/π‘š^2 √(4π‘Ž/π‘š^2 )βˆ’0] =(4βˆšπ‘Ž)/3 [4π‘Ž/π‘š^2 Γ—(2βˆšπ‘Ž)/π‘š] =(32 π‘Ž .βˆšπ‘Ž .βˆšπ‘Ž)/(3π‘š^3 ) =(32 π‘Ž^2)/(3π‘š^3 ) Area OAD Area OAD = ∫1_0^(4π‘Ž/π‘š^2 )▒〖𝑦 𝑑π‘₯γ€— y β†’ Equation of line y = mx Therefore, Area OAD = ∫1_0^(4π‘Ž/π‘š^2 )β–’γ€–"m" π‘₯ 𝑑π‘₯γ€— = m∫1_0^(4π‘Ž/π‘š^2 )β–’γ€–π‘₯ 𝑑π‘₯γ€— = π‘š[π‘₯^2/2]_0^(4π‘Ž/π‘š^2 ) = π‘š[π‘₯^2/2]_0^(4π‘Ž/π‘š^2 ) =π‘š/2 [(4π‘Ž/π‘š^2 )^2βˆ’0^2 ] =π‘š/2 (4π‘Ž)^2/π‘š^4 =(8π‘Ž^2)/π‘š^3 Thus, Area Required = Area OBAD – Area OAD = (32 π‘Ž^2)/(3π‘š^3 ) – (8π‘Ž^2)/π‘š^3 = ((32 βˆ’ 24) π‘Ž^2)/(3π‘š^3 ) = (πŸ–π’‚^𝟐)/γ€–πŸ‘π’Žγ€—^πŸ‘

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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.