Last updated at May 29, 2018 by Teachoo

Transcript

Ex 11.4, 14 Find the equation of the hyperbola satisfying the given conditions: Vertices (±7, 0), e = 43 Here, the vertices are on the x-axis. Therefore, the equation of the hyperbola is of the form 𝑥2𝑎2 – 𝑦2𝑏2 = 1 . Now, coordinates of vertices are (± a,0) & given vertices = (±7, 0), So, (± a,0) = (±7, 0), a = 7 We know that Eccentricity = e = 𝑐𝑎 Given that e = 43 43 = 𝑐𝑎 4a = 3c Putting a = 7 4 × 7=3 𝑐 28 = 3 c 3c = 28 c = 𝟐𝟖𝟑 Also, we know that c2 = a2 + b2 Putting values 2832 = 49 + b2 7849 = 49 + b2 b2 = 784 −4419 b2 = 𝟑𝟒𝟑𝟗 Required equation of hyperbola 𝑥2𝑎2 − 𝑦2𝑏2 =1 Putting values 𝑥249 − 𝑦23439 =1 𝒙𝟐𝟒𝟗 − 𝟗𝒚𝟐𝟑𝟒𝟑 = 1

Chapter 11 Class 11 Conic Sections

Serial order wise

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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.