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Ex 11.4, 5 - 5y2 - 9x2 = 36 Find vertices, latus rectum - Ex 11.4

Ex 11.4,  5 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.4,  5 - Chapter 11 Class 11 Conic Sections - Part 3
Ex 11.4,  5 - Chapter 11 Class 11 Conic Sections - Part 4
Ex 11.4,  5 - Chapter 11 Class 11 Conic Sections - Part 5

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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 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.