Example 16 - Find hyperbola: foci (0, 12), latus rectum 36

Example 16 - Chapter 11 Class 11 Conic Sections - Part 2
Example 16 - Chapter 11 Class 11 Conic Sections - Part 3 Example 16 - Chapter 11 Class 11 Conic Sections - Part 4 Example 16 - Chapter 11 Class 11 Conic Sections - Part 5

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Example 16 Find the equation of the hyperbola where foci are (0, Β±12) and the length of the latus rectum is 36. We need to find equation of hyperbola Given foci (0, Β±12) & length of latus rectum 36. Since foci is on the yβˆ’axis So required equation of hyperbola is π’šπŸ/π’‚πŸ – π’™πŸ/π’ƒπŸ = 1 Now, Co-ordinates of foci = (0, Β± c) & given foci = (0, Β±12) So, (0, Β± c) = (0, Β±12) c = 12 We know that Length of latus rectum = 2𝑏2/π‘Ž Given latus rectum = 36 36 = 2𝑏2/π‘Ž 36a = 2b2 2b2 = 36 a b2 = 36/2 π‘Ž b2 = 18a We know that c2 = b2 + a2 Putting value of c & b2 (12)2 = 18a + a2 144 = 18a + a2 a2 + 18a = 144 a + 18a βˆ’ 144 = 0 a2 + 24a βˆ’ 6a βˆ’ 144 = 0 a (a + 24) βˆ’ 6 (a + 24) = 0 (a βˆ’ 6) (a + 24) = 0 So , a = 6 or a = –24 Since β€˜a’ is distance, it cannot be negative , So a = βˆ’24 is not possible ∴ a = 6, From (1) b2 = 18a Putting a = 6 b2 = 18 (6) b2 = 108 Required Equation of hyperbola is 𝑦2/π‘Ž2 βˆ’ π‘₯2/𝑏2 = 1 Putting values 𝑦2/62 βˆ’ π‘₯2/108 = 1 π’šπŸ/πŸ‘πŸ” βˆ’ π’™πŸ/πŸπŸŽπŸ– = 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.