Example 2 Find the area enclosed by the ellipse π‘₯^2/π‘Ž^2 +𝑦^2/𝑏^2 =1 We have to find Area Enclosed by ellipse Since Ellipse is symmetrical about both x-axis and y-axis ∴ Area of ellipse = 4 Γ— Area of OAB = 4 Γ— ∫_𝟎^π’‚β–’γ€–π’š 𝒅𝒙〗 We know that , π‘₯^2/π‘Ž^2 +𝑦^2/𝑏^2 =1 𝑦^2/𝑏^2 =1βˆ’π‘₯^2/π‘Ž^2 𝑦^2/𝑏^2 =(π‘Ž^2βˆ’γ€– π‘₯γ€—^2)/π‘Ž^2 𝑦^2=𝑏^2/π‘Ž^2 (π‘Ž^2βˆ’π‘₯^2 ) 𝑦=±√(𝑏^2/π‘Ž^2 (π‘Ž^2βˆ’π‘₯^2 ) ) π’š=±𝒃/𝒂 √((𝒂^πŸβˆ’π’™^𝟐 ) ) Since OAB is in 1st quadrant, value of y is positive ∴ π’š=𝒃/𝒂 √(𝒂^πŸβˆ’π’™^𝟐 ) Area of ellipse = 4 Γ— ∫_0^π‘Žβ–’γ€–π‘¦.𝑑π‘₯γ€— = 4∫_𝟎^𝒂▒𝒃/𝒂 √(𝒂^πŸβˆ’π’™^𝟐 ) 𝒅𝒙 = 4𝑏/π‘Ž ∫_0^π‘Žβ–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯" " γ€— = πŸ’π’ƒ/𝒂 [𝒙/𝟐 √(𝒂^πŸβˆ’π’™^𝟐 )+𝒂^𝟐/𝟐 γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑〖𝒙/𝒂〗 ]_𝟎^𝒂 = 4𝑏/π‘Ž [(π‘Ž/2 √(π‘Ž^2βˆ’π‘Ž^2 )+π‘Ž^2/2 sin^(βˆ’1)β‘γ€–π‘Ž/π‘Žγ€— )βˆ’(0/2 √(π‘Ž^2βˆ’0)βˆ’π‘Ž^2/2 sin^(βˆ’1)⁑(0) )] = 4𝑏/π‘Ž [0+π‘Ž^2/2 sin^(βˆ’1)⁑〖(1)γ€—βˆ’0βˆ’0] It is of form √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/π‘Ž+𝑐〗 = πŸ’π’ƒ/𝒂 Γ— 𝒂^𝟐/𝟐 〖𝐬𝐒𝐧〗^(βˆ’πŸ)⁑(𝟏) = 2π‘Žπ‘ Γ—sin^(βˆ’1)⁑(1) = 2π‘Žπ‘ Γ— πœ‹/2 = 𝝅𝒂𝒃 ∴ Required Area = 𝝅𝒂𝒃 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.