Chapter 11 Class 11 Conic Sections

Class 11
Important Questions for exams Class 11

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### Transcript

Ex 10.4, 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 − 9𝑥2/36 = 36/36 𝑦2/((36/5) ) − 𝑥2/4 = 1 The above equation is of the form 𝑦2/𝑎2 − 𝑥2/𝑏2 = 1 ∴ Axis of hyperbola is y-axis Comparing (1) & (2) a2 = 36/5 a = 𝟔/√𝟓 & b2 = 4 b = 2 Also, c2 = a2 + b2 c2 = 36/5 + 4 c2 = (36 + 20)/5 c2 = 56/5 c2 = √(56/5) c = (𝟐√𝟏𝟒)/√𝟓 Co−ordinate of foci = (0, ± c) = ("0, ± " (2√14)/√5) So, co-ordinates of foci are ("0, " (𝟐√𝟏𝟒)/√𝟓) & ("0," (−𝟐√𝟏𝟒)/√𝟓) Coordinates of vertices = (0, ±a) = ("0, " ±6/√5) So, co-ordinates of vertices are ("0, " 𝟔/√𝟓) & ("0," (−𝟔)/√𝟓) Eccentricity is e = 𝑐/𝑎 e = ((2√14)/√5)/(6/√5) e = (2√14)/√5 × √5/6 = √𝟏𝟒/𝟑 Latus rectum = 2𝑏2/𝑎 = (2 × 2^2)/(6/√5) = 2 × 4 × √5/6 = (𝟒√𝟓)/𝟑

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.