Example 16 - Chapter 10 Class 11 Conic Sections
Last updated at May 6, 2024 by Teachoo
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Chapter 10 Class 11 Conic Sections
Serial order wise
- Ex 10.1
- Ex 10.2
- Ex 10.3
- Ex 10.4
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Transcript
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
