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

Transcript

Ex 11.4, 14 Find the equation of the hyperbola satisfying the given conditions: Vertices (ยฑ7, 0), e = 4/3 Here, the vertices are on the x-axis. Therefore, the equation of the hyperbola is of the form ๐’™๐Ÿ/๐’‚๐Ÿ โ€“ ๐’š๐Ÿ/๐’ƒ๐Ÿ = 1 Now, coor#dinates of vertices are (ยฑ a,0) & Given vertices = (ยฑ7, 0), So, (ยฑ a,0) = (ยฑ7, 0), a = 7 We know that Eccentricity = e = ๐‘/๐‘Ž Given that e = 4/3 4/3 = ๐‘/๐‘Ž 4a = 3c Putting a = 7 4 ร— 7=3 ๐‘ 28 = 3 c 3c = 28 c = ๐Ÿ๐Ÿ–/๐Ÿ‘ Also, we know that c2 = a2 + b2 Putting values (28/3)^2 = 49 + b2 784/9 = 49 + b2 b2 = (784 โˆ’441)/9 b2 = ๐Ÿ‘๐Ÿ’๐Ÿ‘/๐Ÿ— Required equation of hyperbola ๐‘ฅ2/๐‘Ž2โˆ’ ๐‘ฆ2/๐‘2 =1 Putting values ๐‘ฅ2/7^2 โˆ’ ๐‘ฆ2/(343/9) =1 ๐’™๐Ÿ/๐Ÿ’๐Ÿ— โˆ’ ๐Ÿ—๐’š๐Ÿ/๐Ÿ‘๐Ÿ’๐Ÿ‘ = 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.