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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise

Transcript

Ex 11.3, 19 Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6). Since major axis is along y-axis & centre is at (0,0) So required equation of ellipse is ๐’™^๐Ÿ/๐’ƒ^๐Ÿ + ๐’š^๐Ÿ/๐’‚^๐Ÿ = 1 Given that ellipse passes through point (3, 2) & (1, 6) Points (3, 2) & (1, 6) will satisfy equation of ellipse. Putting x = 3 & y = 2 in (1) (3)^2/๐‘^2 + (2)^2/๐‘Ž^2 = 1 9/๐‘^2 + 4/๐‘Ž^2 = 1 Putting x = 1 & y = 6 in (1) ใ€–(1)ใ€—^2/๐‘^2 + ใ€–(6)ใ€—^2/๐‘Ž^2 = 1 1/๐‘^2 + 36/๐‘Ž^2 = 1 From (3) 1/๐‘^2 + 36/๐‘Ž^2 = 1 1/๐‘^2 = 1 โˆ’ 36/๐‘Ž^2 Putting value of b2 in (2) 9/๐‘^2 + 4/๐‘Ž^2 = 1 9(1/๐‘^2 ) + 4/๐‘Ž^2 = 1 9(1โˆ’36/๐‘Ž^2 ) + 4/๐‘Ž^2 = 1 9 โˆ’ 324/๐‘Ž^2 + 4/๐‘Ž^2 = 1 (โˆ’320)/๐‘Ž^2 = 1 โˆ’ 9 (โˆ’320)/๐‘Ž^2 = โˆ’8 1/๐‘Ž^2 = (โˆ’8)/(โˆ’320) 1/๐‘Ž^2 = 8/320 1/๐‘Ž^2 = 1/40 a2 = 40 Putting value of ๐‘Ž^2 in (3) 1/๐‘^2 + 36/๐‘Ž^2 = 1 1/๐‘^2 = 1 โˆ’ 36/๐‘Ž^2 1/๐‘^2 = 1 โˆ’ 36 (1/40) 1/๐‘^2 = (40 โˆ’ 36)/40 1/๐‘^2 = 4/40 1/๐‘^2 = 1/10 b2 = 10 Now required equation of ellipse is ๐‘ฅ^2/๐‘^2 + ๐‘ฆ^2/๐‘Ž^2 = 1 Putting value of b2 & a2 ๐’™^๐Ÿ/๐Ÿ๐ŸŽ + ๐’š^๐Ÿ/๐Ÿ’๐ŸŽ = 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.