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Ex 11.4, 15 - Find hyperbola: foci (0, 10), passing (2, 3)

Ex 11.4,  15 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.4,  15 - Chapter 11 Class 11 Conic Sections - Part 3 Ex 11.4,  15 - Chapter 11 Class 11 Conic Sections - Part 4 Ex 11.4,  15 - Chapter 11 Class 11 Conic Sections - Part 5

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Transcript

Ex 10.4, 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.