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

Transcript

Ex 11.4, 15 Find the equation of the hyperbola satisfying the given conditions: Foci (0,±√10), passing through (2, 3) Since Foci is on the yβˆ’axis So required equation of hyperbola is 𝑦2/π‘Ž2 – π‘₯2/𝑏2 = 1 Now, Co-ordinates of foci = (0, Β± c) & given foci = (0, ±√10) So, (0, Β± c) = (0, ±√10) c = √𝟏𝟎 Also, c2 = a2 + b2 Putting value of c (√10)2 = a2 + b2 10 = a2 + b2 a2 + b2 = 10 b2 = 10 βˆ’ a2 Since point (2,3) passes through the hyperbola, it satisfies the equation of hyperbola Putting values x = 2, y = 3 in equation 𝑦^2/π‘Ž^2 βˆ’ π‘₯^2/𝑏^2 = 1 a 3^2/π‘Ž^2 βˆ’ 2^2/𝑏^2 = 1 3^2/π‘Ž^2 βˆ’ 2^2/𝑏^2 = 1 Also putting b2 = 10 βˆ’ a2 3^2/π‘Ž^2 βˆ’ 2^2/(10 βˆ’ π‘Ž^2 ) = 1 9/π‘Ž^2 βˆ’ 4/(10 βˆ’ π‘Ž^2 ) = 1 ((10 βˆ’ π‘Ž^2 ) 9 βˆ’ 4π‘Ž^2)/(π‘Ž^2 (10 βˆ’ π‘Ž^2)) = 1 (90 βˆ’ 9π‘Ž^2 βˆ’ 4π‘Ž^2)/(π‘Ž^2 (10 βˆ’ π‘Ž^2)) = 1 90 βˆ’ 13a2 = a2(10 βˆ’ a2) 90 βˆ’ 13a2 = 10a2 βˆ’ a4 a4 βˆ’ 23a2 + 90 = 0 Let a2 = t So, our equation becomes t2 βˆ’ 23t + 90 = 0 t2 βˆ’ 18t βˆ’ 5t + 90 = 0 t(t βˆ’ 18) βˆ’ 5(t βˆ’ 18) = 0 (t βˆ’ 5) (t βˆ’ 18) = 0 So, t = 5 or t = 18 Now, a2 = t Thus, a2 = 5, 18 Putting t = 5 a2 = 5 Putting t = 18 a2 = 18 For a2 = 5 b2 = 10 – a2 b2 = 10 – 5 b2 = 5 For a2 = 18 b2 = 10 – a2 b2 = 10 – 18 b2 = –8 Since square cannot be negative a2 = 18 is not possible Thus, a2 = 5, b2 = 5 Required equation of hyperbola 𝑦^2/π‘Ž^2 βˆ’ π‘₯^2/𝑏^2 =1 Putting values π’š^𝟐/πŸ“ βˆ’ 𝒙^𝟐/πŸ“ =𝟏

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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.