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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


Transcript

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 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.