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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 ﷐𝑦2﷮𝑎2﷯ – ﷐𝑥2﷮𝑏2﷯ = 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 = ﷐2𝑏2﷮𝑎﷯ Given latus rectum = 36 36 = ﷐2𝑏2﷮𝑎﷯ 36a = 2b2 2b2 = 36 a b2 = ﷐36﷮2﷯𝑎 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 ﷐𝑦2﷮𝑎2﷯ − ﷐𝑥2﷮𝑏2﷯ = 1 Putting values ﷐𝑦2﷮62﷯ − ﷐𝑥2﷮108﷯ = 1 ﷐𝒚𝟐﷮𝟑𝟔﷯ − ﷐𝒙𝟐﷮𝟏𝟎𝟖﷯ = 1

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