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Last updated at Feb. 6, 2020 by Teachoo

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Ex 11.4, 15 Find the equation of the hyperbola satisfying the given conditions: Foci (0,Β±β10), passing through (2, 3) Since Foci is on the yβaxis So required equation of hyperbola is π¦2/π2 β π₯2/π2 = 1 Now, Co-ordinates of foci = (0, Β± c) & given foci = (0, Β±β10) So, (0, Β± c) = (0, Β±β10) c = βππ Also, c2 = a2 + b2 Putting value of c (β10)2 = a2 + b2 10 = a2 + b2 a2 + b2 = 10 b2 = 10 β a2 Since point (2,3) passes through the hyperbola, it satisfies the equation of hyperbola Putting values x = 2, y = 3 in equation π¦^2/π^2 β π₯^2/π^2 = 1 a 3^2/π^2 β 2^2/π^2 = 1 3^2/π^2 β 2^2/π^2 = 1 Also putting b2 = 10 β a2 3^2/π^2 β 2^2/(10 β π^2 ) = 1 9/π^2 β 4/(10 β π^2 ) = 1 ((10 β π^2 ) 9 β 4π^2)/(π^2 (10 β π^2)) = 1 (90 β 9π^2 β 4π^2)/(π^2 (10 β π^2)) = 1 90 β 13a2 = a2(10 β a2) 90 β 13a2 = 10a2 β a4 a4 β 23a2 + 90 = 0 Let a2 = t So, our equation becomes t2 β 23t + 90 = 0 t2 β 18t β 5t + 90 = 0 t(t β 18) β 5(t β 18) = 0 (t β 5) (t β 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 β a2 b2 = 10 β 5 b2 = 5 For a2 = 18 b2 = 10 β a2 b2 = 10 β 18 b2 = β8 Since square cannot be negative a2 = 18 is not possible Thus, a2 = 5, b2 = 5 Required equation of hyperbola π¦^2/π^2 β π₯^2/π^2 =1 Putting values π^π/π β π^π/π =π

Chapter 11 Class 11 Conic Sections

Serial order wise

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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.