Slide58.JPG

Slide59.JPG
Slide60.JPG Slide61.JPG Slide62.JPG

Subscribe to our Youtube Channel - https://you.tube/teachoo

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

Transcript

Ex 11.3, 20 Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4, 3) and (6, 2). Since Major axis is on the 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 ellipse passes through point (4, 3) & (6, 2) Points (4, 3) & (6, 2) will satisfy the equation of ellipse Putting x = 4 & y = 3 in (1) ใ€–(4)ใ€—^2/๐‘Ž^2 + ใ€–(3)ใ€—^2/๐‘^2 = 1 16/๐‘Ž^2 + 9/๐‘^2 = 1 Putting x = 6 & y = 2 in (1) ใ€–(6)ใ€—^2/๐‘Ž^2 + ใ€–(2)ใ€—^2/๐‘^2 = 1 36/๐‘Ž^2 + 4/๐‘^2 = 1 From (3) 16/๐‘Ž^2 = 1 โˆ’ 9/๐‘^2 1/๐‘Ž^2 = 1/16 (1 โˆ’ 9/๐‘^2 ) Putting value of 1/๐‘Ž^2 in (2) 36/๐‘Ž^2 + 4/๐‘^2 = 1 36(1/๐‘Ž^2 ) + 4/๐‘^2 = 1 36(1/16 (1โˆ’9/๐‘^2 )) + 4/๐‘^2 = 1 36/16 (1โˆ’9/๐‘^2 ) + 4/๐‘^2 = 1 9/4 (1โˆ’9/๐‘^2 ) + 4/๐‘^2 = 1 9/4 โˆ’ 81/ใ€–4๐‘ใ€—^2 + 4/๐‘^2 = 1 (โˆ’81)/(4๐‘^2 ) + 4/๐‘^2 = 1 โˆ’ 9/4 (โˆ’81 + 16)/(4๐‘^2 ) = (4 โˆ’ 9)/4 (โˆ’65)/(4๐‘^2 ) = (โˆ’5)/4 (โˆ’5)/4 (13/๐‘^2 )= (โˆ’5)/4 13/๐‘^2 = 1 1/๐‘^2 = 1/13 b2 = 13 Putting value of b2 in 1/๐‘Ž^2 = 1/16 (1 โˆ’ 9/๐‘^2 ) 1/๐‘Ž^2 = 1/16 (1 โˆ’ 9/13) 1/๐‘Ž^2 = 1/16 ( (13 โˆ’ 9)/13) 1/๐‘Ž^2 = 1/16 ( 4/13) 1/๐‘Ž^2 = 1/52a a2 = 52 Equation of ellipse is ๐‘ฅ^2/๐‘Ž^2 + ๐‘ฆ^2/๐‘^2 = 1 Putting value of a2 & b2 ๐’™^๐Ÿ/๐Ÿ“๐Ÿ + ๐’š^๐Ÿ/๐Ÿ๐Ÿ‘ = 1

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.