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Misc 9 - Find area of smaller region x2/a2 + y2/b2 = 1 - Miscellaneous

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

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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 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.