Check sibling questions

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Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.