Ex 11.4, 9 - Find hyperbola: Vertices (0, 3), foci (0, 5)

Ex 11.4,  9 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.4,  9 - Chapter 11 Class 11 Conic Sections - Part 3

  1. Chapter 11 Class 11 Conic Sections (Term 2)
  2. Serial order wise

Transcript

Ex 11.4, 9 Find the equation of the hyperbola satisfying the given conditions: Vertices (0, Β±3), foci (0, Β±5) We need to find equation of hyperbola Given Vertices (0, Β±3), foci (0, Β±5) Since Vertices are on the y-axis So required equation of hyperbola is π’šπŸ/π’‚πŸ – π’™πŸ/π’ƒπŸ = 1 ∴ Axis of hyperbola is yβˆ’axis We know that Vertices = (0, Β±a) & Given vertices = (0, Β±3) ∴ (0, Β±a) = (0, Β±3) So a = 3 Also, coordinates of foci are (0, Β± c) Given, foci are (0, Β±5) So, c = 5 We know that c2 = a2 + b2 (5)2 = (3)2 + b2 25 = 9 + b2 b2 = 25 βˆ’ 9 b2 = 16 Required equation of hyperbola is 𝑦2/π‘Ž2 βˆ’ π‘₯2/𝑏2 = 1 Putting values 𝑦2/32 βˆ’ π‘₯2/16 = 1 π’šπŸ/πŸ— βˆ’ π’™πŸ/πŸπŸ” = 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.