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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise

Transcript

Ex 11.4, 5 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y2 – 9x2 = 36 Given equation is 5y2 – 9x2 = 36. Dividing whole equation by 36 5𝑦2/36 βˆ’ 9π‘₯2/36 = 36/36 𝑦2/((36/5) ) βˆ’ π‘₯2/4 = 1 The above equation is of the form 𝑦2/π‘Ž2 βˆ’ π‘₯2/𝑏2 = 1 ∴ Axis of hyperbola is y-axis Comparing (1) & (2) a2 = 36/5 a = πŸ”/βˆšπŸ“ & b2 = 4 b = 2 Also, c2 = a2 + b2 c2 = 36/5 + 4 c2 = (36 + 20)/5 c2 = 56/5 c2 = √(56/5) c = (πŸβˆšπŸπŸ’)/βˆšπŸ“ Coβˆ’ordinate of foci = (0, Β± c) = ("0, Β± " (2√14)/√5) So, co-ordinates of foci are ("0, " (πŸβˆšπŸπŸ’)/βˆšπŸ“) & ("0," (βˆ’πŸβˆšπŸπŸ’)/βˆšπŸ“) Coordinates of vertices = (0, Β±a) = ("0, " Β±6/√5) So, co-ordinates of vertices are ("0, " πŸ”/βˆšπŸ“) & ("0," (βˆ’πŸ”)/βˆšπŸ“) Eccentricity is e = 𝑐/π‘Ž e = ((2√14)/√5)/(6/√5) e = (2√14)/√5 Γ— √5/6 = βˆšπŸπŸ’/πŸ‘ Latus rectum = 2𝑏2/π‘Ž = (2 Γ— 2^2)/(6/√5) = 2 Γ— 4 Γ— √5/6 = (πŸ’βˆšπŸ“)/πŸ‘

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.