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Ex 11.4, 11 - Find hyperbola: foci (0, 13), conjugate axis 24

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

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Ex 11.4, 11 Find the equation of the hyperbola satisfying the given conditions: Foci (0, ±13), the conjugate axis is of length 24. We need to find equation of hyperbola Given foci (0, ±13) & conjugate axis is of length 24. Since foci is on the y−axis So required equation of hyperbola is 𝒚𝟐/𝒂𝟐 – 𝒙𝟐/𝒃𝟐 = 1 Now, Co-ordinates of foci = (0, ± c) & given foci = (0, ±13) So, (0, ± c) = (0, ±13) c = 13 Length of conjugate axis = 2b Given length of conjugate axis = 24 So, 2b = 24 b = 24/2 b = 12 We know that c2 = a2 + b2 Putting Values (13)2= a2 + (12)2 (13)2 – (12)2 = a2 169 − 144 = a2 25 = a2 a2 = 25 Thus, the required equation of ellipse 𝑦^2/𝑎^2 − 𝑥^2/𝑏^2 =1 Putting values 𝑦^2/25 − 𝑥^2/〖12〗^2 =1 𝒚^𝟐/𝟐𝟓 − 𝒙^𝟐/𝟏𝟒𝟒 = 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.