Check sibling questions

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


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.