# Example 16

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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 𝑦2𝑎2 – 𝑥2𝑏2 = 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 = 2𝑏2𝑎 Given latus rectum = 36 36 = 2𝑏2𝑎 36a = 2b2 2b2 = 36 a b2 = 362𝑎 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 𝑦2𝑎2 − 𝑥2𝑏2 = 1 Putting values 𝑦262 − 𝑥2108 = 1 𝒚𝟐𝟑𝟔 − 𝒙𝟐𝟏𝟎𝟖 = 1

Chapter 11 Class 11 Conic Sections

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .