Check sibling questions

Example 8 - AOBA is part of ellipse 9x2 + y2 = 36, OA = 2

Example 8 - Chapter 8 Class 12 Application of Integrals - Part 2
Example 8 - Chapter 8 Class 12 Application of Integrals - Part 3
Example 8 - Chapter 8 Class 12 Application of Integrals - Part 4
Example 8 - Chapter 8 Class 12 Application of Integrals - Part 5
Example 8 - Chapter 8 Class 12 Application of Integrals - Part 6

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Transcript

Example 8 In Fig AOBA is the part of the ellipse 9๐‘ฅ2+๐‘ฆ2=36 in the first quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the chord AB Here, Area Area Required = Area ABC Area ABC Area ABC = Area OACB โ€“ Area โˆ† OAB Area OACB Area OACB = โˆซ_0^2โ–’๐‘ฆ๐‘‘๐‘ฅ Here, ๐‘ฆ โ†’ Equation of Ellpise 9๐‘ฅ2+๐‘ฆ2=36 ๐‘ฆ2=36โˆ’9๐‘ฅ2 ๐‘ฆ="ยฑ" โˆš(36โˆ’9๐‘ฅ^2 ) As OACB is in 1st quadrant, Value of ๐‘ฆ will be positive โˆด ๐‘ฆ=โˆš(36โˆ’9๐‘ฅ^2 ) Area OACB = โˆซ_0^2โ–’๐‘ฆ๐‘‘๐‘ฅ = โˆซ_0^2โ–’โˆš(36โˆ’9๐‘ฅ^2 ) = โˆซ_0^2โ–’โˆš(9(36/9โˆ’๐‘ฅ^2)) = โˆซ_0^2โ–’โˆš(9(4โˆ’๐‘ฅ^2)) = โˆซ_0^2โ–’โˆš(3^2 (2^2โˆ’๐‘ฅ^2)) = 3โˆซ_0^2โ–’ใ€–โˆš(2^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ ใ€— = 3[๐‘ฅ/2 โˆš(2^2โˆ’๐‘ฅ^2 )+2^2/2 sin^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— ]_0^2 = 3 [2/2 โˆš(2^2โˆ’2^2 )+2 sin^(โˆ’1)โกใ€–2/2โˆ’0/2ใ€— โˆš(2^2โˆ’0^2 )โˆ’2 sin^(โˆ’1)โก0 ] It is of form โˆซ1โ–’ใ€–โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ=1/2 ๐‘ฅโˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )ใ€—+๐‘Ž^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€–๐‘ฅ/๐‘Ž+๐‘ใ€— Replacing a by 2 , we get = 3 [0+2 sin^(โˆ’1)โกใ€–1โˆ’0โˆ’0ใ€— ] = 3 [2 sin^(โˆ’1)โก1 ] = 3 ร— 2 ร— ๐œ‹/2 = 3๐œ‹ Area OAB Area OAB =โˆซ_0^2โ–’๐‘ฆ๐‘‘๐‘ฅ Here, ๐‘ฆ โ†’ equation of line AB Equation of line between A(2, 0) & B(0, 6) is (๐‘ฆ โˆ’ 0)/(๐‘ฅ โˆ’ 2)=(6 โˆ’ 0)/(0 โˆ’ 2) ๐‘ฆ/(๐‘ฅโˆ’2)=6/(โˆ’2) Eq. of line b/w (x1, y1) & (x2, y2) is (๐‘ฆ โˆ’ ๐‘ฆ1)/(๐‘ฅ โˆ’ ๐‘ฅ1)=(๐‘ฆ2 โˆ’ ๐‘ฆ1)/(๐‘ฅ2 โˆ’ ๐‘ฅ1) ๐‘ฆ/(๐‘ฅ โˆ’ 2)=โˆ’3 ๐‘ฆ=โˆ’3(๐‘ฅโˆ’2) Area OBC =โˆซ_0^2โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— =โˆซ_0^2โ–’ใ€–โˆ’3(๐‘ฅโˆ’2) ๐‘‘๐‘ฅใ€— = โ€“3โˆซ_0^2โ–’ใ€–(๐‘ฅโˆ’2) ๐‘‘๐‘ฅใ€— = โ€“3 [๐‘ฅ^2/2 โˆ’2๐‘ฅ ]_0^2 = โ€“ 3 [2^2/2โˆ’2 ร—2โˆ’0^2/2โˆ’2 ร—0] = โ€“3 [โˆ’2] = 6 Area Required = Area OACB โ€“ Area โˆ† OAB =๐Ÿ‘๐…โˆ’๐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.