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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise

Transcript

Misc 9 Find the area of the smaller region bounded by the ellipse π‘₯^2/π‘Ž^2 +𝑦^2/𝑏^2 =1 & π‘₯/π‘Ž + 𝑦/𝑏 = 1 Let’s first draw the figure 𝒙^𝟐/𝒂^𝟐 +π’š^𝟐/𝒃^𝟐 =𝟏 is an which is a equation ellipse with x-axis as principal axis And, 𝒙/𝒂 + π’š/𝒃 = 1 is a line passing through A (a, 0) and B (0, b) Required Area Required Area = Area OACB – Area OAB Area OACB Area OACB = ∫_0^π‘Žβ–’γ€–π‘¦ 𝑑π‘₯γ€— 𝑦 β†’ Equation of ellipse π‘₯^2/π‘Ž^2 +𝑦^2/𝑏^2 =1 𝑦^2/𝑏^2 =1βˆ’π‘₯^2/π‘Ž^2 𝑦^2=𝑏^2 [1βˆ’π‘₯^2/π‘Ž^2 ] 𝑦=±√(𝑏^2 [1βˆ’π‘₯^2/π‘Ž^2 ] ) 𝑦=Β± π‘βˆš(1βˆ’π‘₯^2/π‘Ž^2 ) As OACB is in 1st quadrant, Value of 𝑦 will be positive ∴ 𝑦=π‘βˆš(1βˆ’π‘₯^2/π‘Ž^2 ) Now, Area OACB =∫_0^π‘Žβ–’γ€–π‘βˆš(1βˆ’π‘₯^2/π‘Ž^2 )γ€— 𝑑π‘₯ =b∫_0^π‘Žβ–’γ€–βˆš((π‘Ž^2 βˆ’ π‘₯^2)/π‘Ž^2 ) 𝑑π‘₯" " γ€— =𝑏/π‘Ž ∫_0^π‘Žβ–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯" " γ€— =𝑏/π‘Ž [1/2 π‘₯ √(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€— ]_0^π‘Ž =𝑏/π‘Ž [1/2.π‘Žβˆš(π‘Ž^2βˆ’π‘Ž^2 )+π‘Ž^2/2 sin^(βˆ’1)β‘γ€–π‘Ž/π‘Žγ€—βˆ’(1/2 0√(π‘Ž^2βˆ’0^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑〖0/π‘Žγ€— )] =𝑏/π‘Ž [0+π‘Ž^2/2.γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)β‘πŸβˆ’0βˆ’0] =𝑏/π‘Ž [0+π‘Ž^2/2.𝝅/𝟐 ] =𝑏/π‘Ž Γ— π‘Ž^2/2 " Γ— " πœ‹/2 =( πœ‹π‘Žπ‘ )/4 =𝑏/π‘Ž [0+π‘Ž^2/2.𝝅/𝟐 ] =𝑏/π‘Ž Γ— π‘Ž^2/2 " Γ— " πœ‹/2 =( πœ‹π‘Žπ‘ )/4 Area OAB Area OAB =∫_0^π‘Žβ–’γ€–π‘¦ 𝑑π‘₯γ€— 𝑦 β†’ Equation of line π‘₯/π‘Ž+𝑦/𝑏=1 𝑦/𝑏=1βˆ’π‘₯/π‘Ž 𝑦=𝑏[1βˆ’π‘₯/π‘Ž] Therefore, Area OAB =∫_0^π‘Žβ–’π‘[1βˆ’π‘₯/π‘Ž]𝑑π‘₯ = 〖𝑏[π‘₯βˆ’π‘₯^2/2π‘Ž]γ€—_0^π‘Ž = 𝑏[π‘Žβˆ’π‘Ž^2/2π‘Žβˆ’[0βˆ’0^2/2π‘Ž]] = 𝑏[π‘Žβˆ’π‘Ž/2βˆ’0] = π‘Γ—π‘Ž/2 =π‘Žπ‘/2 ∴ Area Required = Area OACB – Area OAB =( πœ‹ π‘Žπ‘ )/4βˆ’π‘Žπ‘/2 =π‘Žπ‘/2 [πœ‹/2βˆ’1] =π‘Žπ‘/2 [(πœ‹ βˆ’ 2)/2] =𝒂𝒃/πŸ’ [π…βˆ’πŸ] square units

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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.