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

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.