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Last updated at Feb. 6, 2020 by Teachoo

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Ex 11.4, 12 Find the equation of the hyperbola satisfying the given conditions: Foci (Β± 3β5, 0) , the latus rectum is of length 8. Co-ordinates of Foci is (Β±3β5, 0) Since foci is on the x-axis Hence equation of hyperbola is of the form π₯2/π2 β π¦2/π2 = 1 . Also, We know that co-ordinates of foci are (Β±c, 0) So, (Β±3β5, 0) = (Β±c, 0) 3β5 = c c = 3βπ Now, c2 = a2 + b2 Putting c = 3β5 a2 + b2 = (3β5)2 a2 + b2 = 3 Γ 3 Γ β5 Γ β5 a2 + b2 = 9 Γ 5 a2 + b2 = 45 Also, it is given that Latus Rectum = 8 (2π^2)/π = 8 2b2 = 8a b2 = 8π/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 β 45 = 0 a2 + 9a β 5a β 45 = 0 a(a + 9) β 5(a + 9) = 0 (a + 9) (a β 5) = 0 So, a = 5 or a = β9 Since a is distance, it canβt be negative So a = 5 From (2) b2 = 4a b2 = 4 Γ 5 b2 = 20 Equation of hyperbola is π₯^2/π^2 β π¦^2/π^2 = 1 Putting values π₯^2/5^2 β π¦^2/20 = 1 π^π/ππ β π^π/ππ = 1

Chapter 11 Class 11 Conic Sections

Serial order wise

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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.