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Ex 11.4, 13 - Find hyperbola: foci (4, 0), latus rectum 12

Ex 11.4,  13 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.4,  13 - Chapter 11 Class 11 Conic Sections - Part 3
Ex 11.4,  13 - Chapter 11 Class 11 Conic Sections - Part 4

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Ex 11.4, 13 Find the equation of the hyperbola satisfying the given conditions: Foci (Β±4, 0), the latus rectum is of length 12 Since the foci are on the x-axis. Hence, the required equation of the hyperbola is π’™πŸ/π’‚πŸ – π’šπŸ/π’ƒπŸ = 1 Now, coordinates of foci are (Β±c, 0) & given foci = (Β±4, 0) so, (Β±c,0) = (Β±4,0) c = 4 Now, Latus rectum =2𝑏2/π‘Ž Given latus rectum = 12 So, 2𝑏2/π‘Ž=12 2b2 = 12a b2 = 6a We know that c2 = a2 + b2 Putting value of c & b2 (4)2 = a2 + 6a 16 = a2 + 6a a2 + 6a βˆ’ 16 = 0 a2 + 8a βˆ’ 2a βˆ’16 = 0 a(a + 8) βˆ’ 2 (a + 8) = 0 (a βˆ’ 2) (a + 8) = 0 So, a = 2 or a = -8 Since β€˜a’ is distance, it cannot be negative , So a = βˆ’8 is not possible ∴ a = 2, From (1) b2 = 6a b2 = 6 Γ—2 b2 = 12 Thus, Required equation of hyperbola π‘₯2/π‘Ž2 βˆ’ 𝑦2/𝑏2=1 Putting values π‘₯2/22 βˆ’ 𝑦2/12=1 π’™πŸ/πŸ’ βˆ’ π’šπŸ/𝟏𝟐=𝟏

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.