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Ex 8.2, 1 Find the area of circle 4π‘₯^2+4𝑦^2=9 which is interior to the parabola x2 = 4𝑦 Given Circle and a Parabola Circle 4π‘₯^2+4𝑦^2=9 π‘₯^2+𝑦^2=9/4 𝒙^𝟐+π’š^𝟐=(πŸ‘/𝟐)^𝟐 Hence, Center = (0, 0) Radius = 3/2 Parabola π‘₯^2=4𝑦 This is a parabola with vertical axis Our figure looks like Required Area = Area ABCO Finding point of intersection A and C Solving (1) and (2) 4π‘₯^2+4𝑦^2=9 …(1) π‘₯^2=4𝑦 …(2) Putting value of π‘₯^2 from (2) in (1) 4π‘₯^2+4𝑦^2=9 4(4𝑦)+4𝑦^2=9 16𝑦+4𝑦^2βˆ’9=0 γ€–πŸ’π’šγ€—^𝟐+πŸπŸ”π’šβˆ’πŸ—=𝟎 γ€–4𝑦〗^2+18π‘¦βˆ’2π‘¦βˆ’9=0 2y(2y+9)βˆ’1(2𝑦+9)=0 (πŸπ’š βˆ’πŸ)(πŸπ’š+πŸ—)=𝟎 Hence, π’š=𝟏/𝟐 & π’š=(βˆ’πŸ—)/𝟐 Putting values of y in (2) For π’š=𝟏/𝟐 π‘₯^2=4𝑦 π‘₯^2=4 Γ— 1/2 π‘₯^2=2 𝒙=±√𝟐 Hence, π‘₯ "= " √2 & π‘₯" ="βˆ’βˆš2 For π’š=(βˆ’πŸ—)/𝟐 π‘₯^2=4𝑦 π‘₯^2=4 Γ—((βˆ’9)/( 2)) π‘₯^2=βˆ’2 Γ—9 π‘₯^2=βˆ’18 As square cannot be negative, x has no real value Hence the points are A=(βˆ’βˆšπŸ , 𝟏/𝟐) & C=(√𝟐 , 𝟏/𝟐) Finding Area Area required = Area ABCO Since ABCO is symmetric in y – axis, Area ABCO = 2 Γ— Area BOC Area BOC = Area BCDO βˆ’ Area OCD Area BCDO Area BCDO = ∫_0^(√2)▒〖𝑦 𝑑π‘₯γ€— y β†’ Equation of circle 4x2 + 4y2 = 9 4y2 = 9 βˆ’ 4x2 y2 = 9/4 βˆ’ x2 y = ±√(9/4 " βˆ’ x2" ) Since BCDO is above x βˆ’ axis, we take positive value of y ∴ y = √(πŸ—/πŸ’ " βˆ’ x2" ) Area BCDO = ∫_𝟎^(√𝟐)β–’βˆš(πŸ—/πŸ’ " βˆ’ x2" ) dx = ∫_0^(√2)β–’βˆš((3/2)^2 " βˆ’ " "x" ^2 ) dx = [𝒙/𝟐 √((πŸ‘/𝟐)^𝟐 " βˆ’ " "x" ^𝟐 )+(πŸ‘/𝟐)^𝟐/𝟐 γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑〖𝒙/((πŸ‘/𝟐) )γ€— ]_𝟎^√𝟐 = [π‘₯/2 √((3/2)^2 " βˆ’ " "x" ^2 )+9/8 sin^(βˆ’1)⁑〖2π‘₯/3γ€— ]_0^√2 = [√2/2 √(9/4 " βˆ’ " γ€–"(" √2 ")" γ€—^2 ) + 9/8 sin^(βˆ’1)⁑((2√2)/3) ] βˆ’ [0/2 √(9/4 " βˆ’ " "0" ^2 )+9/8 sin^(βˆ’1)⁑((2Γ—0)/3) ] We know that ∫1β–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— =π‘₯/2 √(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖π‘₯/π‘Ž+𝑐〗 Putting a = 3/2 = √2/2 √(9/4 " βˆ’ " 2)+9/8 sin^(βˆ’1)⁑〖((2√2)/3)βˆ’9/8 sin^(βˆ’1)⁑〖(0)γ€— γ€— = √2/2Γ—βˆš(1/4)+9/8 sin^(βˆ’1)⁑〖((2√2)/3)βˆ’9/8Γ—0γ€— = √𝟐/πŸ’+ πŸ—/πŸ– sinβˆ’1((𝟐√𝟐)/πŸ‘) Area OCD Area OCD = ∫_0^(√2)▒𝑦 dx y β†’ Equation of parabola x2 = 4y y = 𝒙^𝟐/πŸ’ So, Area OCD = ∫_0^(√2)▒𝑦 dx = ∫_𝟎^(√𝟐)▒𝒙^𝟐/πŸ’ dx = 1/4 ∫_0^(√2)β–’π‘₯^2 dx = 𝟏/πŸ’ [𝒙^πŸ‘/πŸ‘]_𝟎^√𝟐 = 1/4 [(√2)^3/3βˆ’0^3/3] = 1/4 [(√2Γ—βˆš2Γ—βˆš2)/3βˆ’0] = 𝟏/πŸ’ [(𝟐√𝟐)/πŸ‘] = √𝟐/πŸ” Now, Area BOC = Area BCDO βˆ’ Area OCD = √𝟐/πŸ’ + πŸ—/πŸ– γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑((𝟐√𝟐)/πŸ‘) – √𝟐/πŸ” = √2/4 βˆ’ √2/6+9/8 sin^(βˆ’1)⁑((2√2)/3) = √𝟐/𝟏𝟐 +πŸ—/πŸ– γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑((𝟐√𝟐)/πŸ‘) Required Area = Area ABCO = 2 Γ— Area BOC = 2 Γ— [√𝟐/𝟏𝟐+πŸ—/πŸ– γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑((𝟐√𝟐)/πŸ‘) ] = √𝟐/πŸ”+πŸ—/πŸ’ γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑((𝟐√𝟐)/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.