Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Slide36.JPG

Slide37.JPG
Slide38.JPG Slide39.JPG Slide40.JPG

  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise

Transcript

Ex 11.4, 12 Find the equation of the hyperbola satisfying the given conditions: Foci (Β± 3√5, 0) , the latus rectum is of length 8. Co-ordinates of Foci is (Β±3√5, 0) Since foci is on the x-axis Hence equation of hyperbola is of the form π‘₯2/π‘Ž2 – 𝑦2/𝑏2 = 1 . Also, We know that co-ordinates of foci are (Β±c, 0) So, (Β±3√5, 0) = (Β±c, 0) 3√5 = c c = 3βˆšπŸ“ Now, c2 = a2 + b2 Putting c = 3√5 a2 + b2 = (3√5)2 a2 + b2 = 3 Γ— 3 Γ— √5 Γ— √5 a2 + b2 = 9 Γ— 5 a2 + b2 = 45 Also, it is given that Latus Rectum = 8 (2𝑏^2)/π‘Ž = 8 2b2 = 8a b2 = 8π‘Ž/2 b2 = 4a Now, our equations are a2 + b2 = 45 b2 = 4a Putting the value of b2 in (1) a2 + b2 = 45 a2 + 4a = 45 a2 + 4a βˆ’ 45 = 0 a2 + 9a βˆ’ 5a βˆ’ 45 = 0 a(a + 9) βˆ’ 5(a + 9) = 0 (a + 9) (a βˆ’ 5) = 0 So, a = 5 or a = βˆ’9 Since a is distance, it can’t be negative So a = 5 From (2) b2 = 4a b2 = 4 Γ— 5 b2 = 20 Equation of hyperbola is π‘₯^2/π‘Ž^2 βˆ’ 𝑦^2/𝑏^2 = 1 Putting values π‘₯^2/5^2 βˆ’ 𝑦^2/20 = 1 𝒙^𝟐/πŸπŸ“ βˆ’ π’š^𝟐/𝟐𝟎 = 1

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.