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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 (Β± 3√5, 0) , the latus rectum is of length 8. Co-ordinates of Foci is (Β±3√5, 0) Since foci is on the x-axis Hence equation of hyperbola is of the form π‘₯2/π‘Ž2 – 𝑦2/𝑏2 = 1 . Also, We know that co-ordinates of foci are (Β±c, 0) So, (Β±3√5, 0) = (Β±c, 0) 3√5 = c c = 3βˆšπŸ“ Now, c2 = a2 + b2 Putting c = 3√5 a2 + b2 = (3√5)2 a2 + b2 = 3 Γ— 3 Γ— √5 Γ— √5 a2 + b2 = 9 Γ— 5 a2 + b2 = 45 Also, it is given that Latus Rectum = 8 (2𝑏^2)/π‘Ž = 8 2b2 = 8a b2 = 8π‘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 βˆ’ 45 = 0 a2 + 9a βˆ’ 5a βˆ’ 45 = 0 a(a + 9) βˆ’ 5(a + 9) = 0 (a + 9) (a βˆ’ 5) = 0 So, a = 5 or a = βˆ’9 Since a is distance, it can’t be negative So a = 5 From (2) b2 = 4a b2 = 4 Γ— 5 b2 = 20 Equation of hyperbola is π‘₯^2/π‘Ž^2 βˆ’ 𝑦^2/𝑏^2 = 1 Putting values π‘₯^2/5^2 βˆ’ 𝑦^2/20 = 1 𝒙^𝟐/πŸπŸ“ βˆ’ π’š^𝟐/𝟐𝟎 = 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.