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

Transcript

Ex 11.4, 12 Find the equation of the hyperbola satisfying the given conditions: Foci (ยฑ 3โˆš5, 0) , the latus rectum is of length 8. Co-ordinates of Foci is (ยฑ3โˆš5, 0) Since foci is on the x-axis Hence equation of hyperbola is of the form ๐‘ฅ2/๐‘Ž2 โ€“ ๐‘ฆ2/๐‘2 = 1 . Also, We know that co-ordinates of foci are (ยฑc, 0) So, (ยฑ3โˆš5, 0) = (ยฑc, 0) 3โˆš5 = c c = 3โˆš๐Ÿ“ Now, c2 = a2 + b2 Putting c = 3โˆš5 a2 + b2 = (3โˆš5)2 a2 + b2 = 3 ร— 3 ร— โˆš5 ร— โˆš5 a2 + b2 = 9 ร— 5 a2 + b2 = 45 Also, it is given that Latus Rectum = 8 (2๐‘^2)/๐‘Ž = 8 2b2 = 8a b2 = 8๐‘Ž/2 b2 = 4a Now, our equations are a2 + b2 = 45 b2 = 4a Putting the value of b2 in (1) a2 + b2 = 45 a2 + 4a = 45 a2 + 4a โˆ’ 45 = 0 a2 + 9a โˆ’ 5a โˆ’ 45 = 0 a(a + 9) โˆ’ 5(a + 9) = 0 (a + 9) (a โˆ’ 5) = 0 So, a = 5 or a = โˆ’9 Since a is distance, it canโ€™t be negative So a = 5 From (2) b2 = 4a b2 = 4 ร— 5 b2 = 20 Equation of hyperbola is ๐‘ฅ^2/๐‘Ž^2 โˆ’ ๐‘ฆ^2/๐‘^2 = 1 Putting values ๐‘ฅ^2/5^2 โˆ’ ๐‘ฆ^2/20 = 1 ๐’™^๐Ÿ/๐Ÿ๐Ÿ“ โˆ’ 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 10 years. He provides courses for Maths and Science at Teachoo.