# Ex 11.4, 15

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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 y−axis So required equation of hyperbola is 𝑦2𝑎2 – 𝑥2𝑏2 = 1 Now, Co-ordinates of foci = (0, ± c) & given foci = (0, ±10) So, (0, ± c) = (0, ±10) ⇒ c = 𝟏𝟎 Also, c2 = a2 + b2 Putting value of c 102 = a2 + b2 10 = a2 + b2 a2 + b2 = 10 b2 = 10 − a2 Since point (2,3) passes through the hyperbola, it satisfies the equation of hyperbola Putting values x = 2, y = 3 in equation 𝑦2𝑎2 − 𝑥2𝑏2 = 1 32𝑎2 − 22𝑏2 = 1 32𝑎2 − 22𝑏2 = 1 Also putting b2 = 10 − a2 32𝑎2 − 2210 − 𝑎2 = 1 9𝑎2 − 410 − 𝑎2 = 1 10 − 𝑎29 − 4𝑎2(10 − 𝑎2) = 1 90 − 9𝑎2 − 4𝑎2𝑎2(10 − 𝑎2) 90 − 13a2 = a2(10 − a2) 90 − 13a2 = 10a2 − a4 a4 − 10a2 − 13a2 + 90 = 0 a4 − 23a2 + 90 = 0 a4 − 23a2 + 90 = 0 Let a2 = t So, our equation becomes t2 − 23t + 90 = 0 t2 − 18t − 5t + 90 = 0 t(t − 18) − 5(t − 18) = 0 (t − 5) (t − 18) = 0 So, t = 5 or t = 18 Now, a2 = t Thus, a2 = 5, 18 Thus, a2 = 5, b2 = 5 Required equation of hyperbola 𝑦2𝑎2 − 𝑥2𝑏2 =1 Putting values 𝒚𝟐𝟓 − 𝒙𝟐𝟓 =𝟏

Chapter 11 Class 11 Conic Sections

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .