Check sibling questions

Example 2 - Find area enclosed by ellipse x2/a2 + y2/b2 = 1

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

Get Real time Doubt solving from 8pm to 12 am!


Transcript

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‘Žβ–’γ€–π‘¦ 𝑑π‘₯γ€— 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’π‘₯^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 OAB is in 1st quadrant, value of y is positive ∴ 𝑦=𝑏/π‘Ž √(π‘Ž^2βˆ’π‘₯^2 ) Area of ellipse = 4 Γ— ∫_0^π‘Žβ–’γ€–π‘¦.𝑑π‘₯γ€— = 4∫_0^π‘Žβ–’π‘/π‘Ž √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯ = 4𝑏/π‘Ž ∫_0^π‘Žβ–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯" " γ€— = 4𝑏/π‘Ž [π‘₯/2 √(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€— ]_0^π‘Ž = 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〗 = 4𝑏/π‘Ž Γ— π‘Ž^2/2 sin^(βˆ’1)⁑(1) = 2π‘Žπ‘ Γ—sin^(βˆ’1)⁑(1) = 2π‘Žπ‘ Γ— πœ‹/2 = πœ‹π‘Žπ‘ ∴ Required Area = 𝝅𝒂𝒃 square units

Davneet Singh's photo - Co-founder, Teachoo

Made by

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.