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

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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise
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Example 16 Find the equation of the hyperbola where foci are (0, 12) and the length of the latus rectum is 36. We need to find equation of hyperbola given foci (0, 12) & length of latus rectum 36. Since foci is on the y axis So required equation of hyperbola is 2 2 2 2 = 1 Now, Co-ordinates of foci = (0, c) & given foci = (0, 12) So, (0, c) = (0, 12) c = 12 We know that Length of latus rectum = 2 2 Given latus rectum = 36 36 = 2 2 36a = 2b2 2b2 = 36 a b2 = 36 2 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 144 = 0 a2 + 24a 6a 144 = 0 a + (a + 24) 6 (a + 24) = 0 (a 6) (a + 24) = 0 So , a 6 = 0 or a + 24 = 0 a = 6 or a = 24 Since a is distance, it cannot be negative , So a = 24 is not possible a = 6, From (1) b2 = 18a Putting a = 6 b2 = 18 (6) b2 = 108 Required Equation of hyperbola is 2 2 2 2 = 1 Putting values 2 62 2 108 = 1 = 1

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