Check sibling questions

Ex 11.4, 14 - Find hyperbola: vertices (7, 0), e = 4/3 - Ex 11.4

Ex 11.4,  14 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.4,  14 - Chapter 11 Class 11 Conic Sections - Part 3
Ex 11.4,  14 - Chapter 11 Class 11 Conic Sections - Part 4

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

Ex 11.4, 14 Find the equation of the hyperbola satisfying the given conditions: Vertices (Β±7, 0), e = 4/3 Here, the vertices are on the x-axis. Therefore, the equation of the hyperbola is of the form π’™πŸ/π’‚πŸ – π’šπŸ/π’ƒπŸ = 1 Now, coor#dinates of vertices are (Β± a,0) & Given vertices = (Β±7, 0), So, (Β± a,0) = (Β±7, 0), a = 7 We know that Eccentricity = e = 𝑐/π‘Ž Given that e = 4/3 4/3 = 𝑐/π‘Ž 4a = 3c Putting a = 7 4 Γ— 7=3 𝑐 28 = 3 c 3c = 28 c = πŸπŸ–/πŸ‘ Also, we know that c2 = a2 + b2 Putting values (28/3)^2 = 49 + b2 784/9 = 49 + b2 b2 = (784 βˆ’441)/9 b2 = πŸ‘πŸ’πŸ‘/πŸ— Required equation of hyperbola π‘₯2/π‘Ž2βˆ’ 𝑦2/𝑏2 =1 Putting values π‘₯2/7^2 βˆ’ 𝑦2/(343/9) =1 π’™πŸ/πŸ’πŸ— βˆ’ πŸ—π’šπŸ/πŸ‘πŸ’πŸ‘ = 1

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.