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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 8.1, 2 Find the area of the region bounded by the ellipse ๐‘ฅ^2/4+๐‘ฆ^2/9=1Given Equation of Ellipse ๐‘ฅ^2/4+๐‘ฆ^2/9=1 ๐’™^๐Ÿ/(๐Ÿ)^๐Ÿ +๐’š^๐Ÿ/(๐Ÿ‘)^๐Ÿ =๐Ÿ Area of Ellipse = Area of ABCD = 2 ร— [Area of BCD] = 2 ร— โˆซ_(โˆ’๐Ÿ)^๐Ÿโ–’ใ€–๐’š ๐’…๐’™ใ€— We know that ๐‘ฅ^2/4+๐‘ฆ^2/9=1 ๐‘ฆ^2/9=1โˆ’๐‘ฅ^2/4 ๐‘ฆ^2/9=(4 โˆ’ ๐‘ฅ^2)/4 ๐‘ฆ^2=9/4 (4โˆ’๐‘ฅ^2 ) Taking Square Root on Both Sides ๐‘ฆ="ยฑ" โˆš(9/4 (4โˆ’๐‘ฅ^2 ) ) ๐’š ="ยฑ" ๐Ÿ‘/๐Ÿ โˆš(๐Ÿ’โˆ’๐’™^๐Ÿ ) Since Area BCD is above the x-axis, The value of y will be positive โˆด ๐’š=๐Ÿ‘/๐Ÿ โˆš(๐Ÿ’โˆ’๐’™^๐Ÿ ) Now, Area of Ellipse = 2 ร— โˆซ_(โˆ’2)^2โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— = 2 ร— โˆซ_(โˆ’2)^2โ–’ใ€–3/2 โˆš(4โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— = 3 โˆซ_(โˆ’๐Ÿ)^๐Ÿโ–’ใ€–โˆš((๐Ÿ)^๐Ÿโˆ’๐’™^๐Ÿ ) ๐’…๐’™ใ€— It is of form โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ=๐‘ฅ/2 โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+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 [๐‘ฅ/2 โˆš((2)^2โˆ’๐‘ฅ^2 )+(2)^2/2 sin^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— ]_(โˆ’2)^2 = 3 [๐‘ฅ/2 โˆš(4โˆ’๐‘ฅ^2 )+2 sin^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— ]_(โˆ’2)^2 = 3[(2/2 โˆš(4โˆ’2^2 )+2 sin^(โˆ’1)โกใ€–2/2ใ€— )โˆ’((โˆ’2)/2 โˆš(4โˆ’ใ€–(โˆ’2)ใ€—^2 )+2 sin^(โˆ’1)โกใ€–(โˆ’2)/2ใ€— )] = 3[(0+2 sin^(โˆ’1)โก1 )โˆ’(0+2 sin^(โˆ’1)โกใ€–(โˆ’1)ใ€— )] = 3[2 sin^(โˆ’1)โก1โˆ’2 ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โกใ€–(โˆ’๐Ÿ)ใ€— ] = 3[2 sin^(โˆ’1)โก1โˆ’2ร—โˆ’ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โก๐Ÿ ] = 3[2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โก1+2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โก1 ] = 3 [4 ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โก(๐Ÿ) ] = 3 ร— 4 ร— ๐…/๐Ÿ = 6ฯ€ โˆด Area of Ellipse = 6ฯ€ 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.