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Ex 11.4, 12 - Find hyperbola: foci (35, 0), latus rectum 8

Ex 11.4,  12 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.4,  12 - Chapter 11 Class 11 Conic Sections - Part 3 Ex 11.4,  12 - Chapter 11 Class 11 Conic Sections - Part 4 Ex 11.4,  12 - Chapter 11 Class 11 Conic Sections - Part 5

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.