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Example 5 - Find area bounded by ellipse x2/a2 + y2/b2 = 1

Example 5 - Chapter 8 Class 12 Application of Integrals - Part 2
Example 5 - Chapter 8 Class 12 Application of Integrals - Part 3
Example 5 - Chapter 8 Class 12 Application of Integrals - Part 4

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Transcript

Example 5 Find the area bounded by the ellipse π‘₯^2/π‘Ž^2 +𝑦^2/𝑏^2 =1 and the ordinates π‘₯=0 and π‘₯=π‘Žπ‘’, where, 𝑏2=π‘Ž2 (1 – 𝑒2) and e < 1 Required Area = Area of shaded region = Area BORQSP = 2 Γ— Area OBPS = 2 Γ— ∫_0^π‘Žπ‘’β–’γ€–π‘¦.𝑑π‘₯γ€— We know that , π‘₯^2/π‘Ž^2 +𝑦^2/𝑏^2 =1 (As ellipse is symmetric about its axis ) 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 ) ) 𝑦=±𝑏/π‘Ž √((π‘Ž^2βˆ’π‘₯^2 ) ) Since OBPS is in 1st quadrant, value of y is positive ∴ 𝑦=𝑏/π‘Ž √(π‘Ž^2βˆ’π‘₯^2 ) Required Area = 2 Γ— ∫_0^π‘Žπ‘’β–’γ€–π‘¦.𝑑π‘₯γ€— = 2∫_0^π‘Žπ‘’β–’γ€–π‘/π‘Ž √(π‘Ž^2βˆ’π‘₯^2 )γ€— 𝑑π‘₯ = 2𝑏/π‘Ž ∫_0^π‘Žπ‘’β–’βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯ = 2𝑏/π‘Ž [1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€— ]_0^π‘Žπ‘’ =2𝑏/π‘Ž [(π‘Žπ‘’/2 √(π‘Ž^2βˆ’(π‘Žπ‘’)^2 )+π‘Ž^2/2 sin^(βˆ’1)β‘γ€–π‘Žπ‘’/π‘Žγ€— )βˆ’(0/2 √(π‘Ž^2βˆ’0)+π‘Ž^2/2 sin^(βˆ’1)⁑(0/π‘Ž) )] =2𝑏/π‘Ž [π‘Žπ‘’/2 √(π‘Ž^2βˆ’π‘Ž^2 𝑒^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑〖(𝑒)βˆ’0βˆ’π‘Ž^2/2 sin^(βˆ’1)⁑(0) γ€— ] =2𝑏/π‘Ž [π‘Žπ‘’/2.π‘Žβˆš(1βˆ’π‘’^2 )+π‘Ž^2/2 sin^(βˆ’1)β‘γ€–π‘’βˆ’0γ€— ] 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〗 =2𝑏/π‘Ž [(π‘Ž^2 𝑒)/2 √(1βˆ’π‘’^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑𝑒 ] =2𝑏/π‘Ž (π‘Ž^2/2)[π‘’βˆš(1βˆ’π‘’^2 )+sin^(βˆ’1)⁑𝑒 ] =π‘Žπ‘[π‘’βˆš(1βˆ’π‘’^2 )+sin^(βˆ’1)⁑𝑒 ] ∴ Required Area =𝒂𝒃[π’†βˆš(πŸβˆ’π’†^𝟐 )+γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑𝒆 ] square units

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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.