Example 16 - Chapter 10 Class 11 Conic Sections
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
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 yβaxis So required equation of hyperbola is ππ/ππ β ππ/ππ = 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 = 2π2/π Given latus rectum = 36 36 = 2π2/π 36a = 2b2 2b2 = 36 a b2 = 36/2 π 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 β 144 = 0 a2 + 24a β 6a β 144 = 0 a (a + 24) β 6 (a + 24) = 0 (a β 6) (a + 24) = 0 So , a = 6 or a = β24 Since βaβ is distance, it cannot be negative , So a = β24 is not possible β΄ a = 6, From (1) b2 = 18a Putting a = 6 b2 = 18 (6) b2 = 108 Required Equation of hyperbola is π¦2/π2 β π₯2/π2 = 1 Putting values π¦2/62 β π₯2/108 = 1 ππ/ππ β ππ/πππ = 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo