Check sibling questions

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


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

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.