# Ex 11.4, 10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 11.4, 10 Find the equation of the hyperbola satisfying the given conditions: Foci (±5, 0), the transverse axis is of length 8. Co-ordinates of foci is (±5, 0) Which is of form (±c, 0) Hence c = 5 Also , foci lies on the x-axis So, Equation of hyperbola is 𝑥2𝑎2 – 𝑦2𝑏2 = 1 We know that c2 = a2 + b2 Putting c = 5 25 = a2 + b2 a2 + b2 = 25 Now Transverse axis is of length 8 and we know that Length of traverse axis = 2a ∴ 2a = 8 a = 82 a = 4 So, a2 = (4)2= 16 Putting value of a2 in (1) a2 + b2 = 25 (16) + b2 = 25 b2 = 25 − 16 b2 = 9 Required equation of hyperbola 𝑥2𝑎2 − 𝑦2𝑏2 = 1 Putting value of a2 & b2 𝒙𝟐𝟏𝟔 − 𝒚𝟐𝟗 = 1

Chapter 11 Class 11 Conic Sections

Serial order wise

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