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

Transcript

Example 13 Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (โ€“ 1,4). Given that Major axis is along x-axis So required equation of ellipse is ๐’™^๐Ÿ/๐’‚^๐Ÿ + ๐’š^๐Ÿ/๐’ƒ^๐Ÿ = 1 Given that point (4, 3) & (โˆ’1, 4) lie of the ellipse So, point (4, 3) & (โˆ’1, 4) will satisfy equation of ellipse Putting x = 4 & y = 3 in (1) ๐‘ฅ^2/๐‘Ž^2 + ๐‘ฆ^2/๐‘^2 = 1 ใ€–(4)ใ€—^2/๐‘Ž^2 + ใ€–(3)ใ€—^2/๐‘^2 = 1 ๐Ÿ๐Ÿ”/๐’‚^๐Ÿ + ๐Ÿ—/๐’ƒ^๐Ÿ = 1 Putting x = โˆ’1, y = 4 is in (1) ๐‘ฅ^2/๐‘Ž^2 + ๐‘ฆ^2/๐‘^2 = 1 ใ€–(โˆ’1)ใ€—^2/๐‘Ž^2 + ใ€–(4)ใ€—^2/๐‘^2 = 1 ๐Ÿ/๐’‚^๐Ÿ + ๐Ÿ๐Ÿ”/๐’ƒ^๐Ÿ = 1 Now, our equations are 16/๐‘Ž^2 + 9/๐‘^2 = 1 1/๐‘Ž^2 + 16/๐‘^2 = 1 From (3) 1/๐‘Ž^2 +16/๐‘^2 = 1 1/๐‘Ž^2 " "= 1โˆ’16/๐‘^2 Putting value of 1/๐‘Ž^2 in (2) 16/๐‘Ž^2 + 9/๐‘^2 = 1 16(1/๐‘Ž^2 ) + 9/๐‘^2 = 1 16(1โˆ’16/๐‘^2 ) + 9/๐‘^2 = 1 16 โˆ’ 256/๐‘^2 + 9/๐‘^2 = 1 (โˆ’256 + 9)/๐‘^2 = 1 โˆ’16 (โˆ’247)/๐‘^2 = โˆ’15 b2 = (โˆ’247)/(โˆ’15) b2 = ๐Ÿ๐Ÿ’๐Ÿ•/๐Ÿ๐Ÿ“ Putting value of b2 = 247/15 in (3) 1/๐‘Ž^2 " "= 1โˆ’16/๐‘^2 1/๐‘Ž^2 " "= 1โˆ’16/(247/15) 1/๐‘Ž^2 " "= 1โˆ’(16 ร— 15)/247 1/๐‘Ž^2 " "= (247 โˆ’ 240)/247 1/๐‘Ž^2 " "= 7/247 ๐š๐Ÿ = ๐Ÿ๐Ÿ’๐Ÿ•/๐Ÿ• Thus, a2 = 247/7 & b2 = 247/15 Hence required of ellipse is ๐‘ฅ^2/๐‘Ž^2 + ๐‘ฆ^2/๐‘^2 = 1 Putting values of a2 & b2 ๐‘ฅ^2/((247/7) ) + ๐‘ฆ^2/((247/15) ) = 1 1/๐‘Ž^2 " "= 1โˆ’(16 ร— 15)/247 1/๐‘Ž^2 " "= (247 โˆ’ 240)/247 1/๐‘Ž^2 " "= 7/247 ๐š๐Ÿ = ๐Ÿ๐Ÿ’๐Ÿ•/๐Ÿ• Thus, a2 = 247/7 & b2 = 247/15 Hence required of ellipse is ๐‘ฅ^2/๐‘Ž^2 + ๐‘ฆ^2/๐‘^2 = 1 Putting values of a2 & b2 ๐‘ฅ^2/((247/7) ) + ๐‘ฆ^2/((247/15) ) = 1 (7๐‘ฅ^2)/247 + (15๐‘ฆ^2)/247 = 1 7x2 + 15y2 = 247

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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.