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Ex 11.3,  5 - x2/49 + y2/36 = 1 Find foci, eccentricity - Ex 11.3

Ex 11.3,  5 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.3,  5 - Chapter 11 Class 11 Conic Sections - Part 3

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Transcript

Ex 11.3, 5 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse x2/49 + y2/36 = 1 𝑥^2/49 + 𝑦^2/36 = 1 Since 49 > 36 Hence the above equation is of the form 𝑥^2/𝑎^2 + 𝑦^2/𝑏^2 = 1 Comparing (1) & (2) We know that c = √(a2−b2) c = √(49−36) c = √𝟏𝟑 Coordinate of foci = (± c, 0) = (± √𝟏𝟑, 0) So coordinate of foci are (√13, 0), (−√13, 0) Vertices = (± a, 0) = (±7, 0) So vertices are (7, 0) & (−7, 0) Length of major axis = 2a = 2 × 7 = 14 Length of minor axis = 2b = 2 × 6 = 12 Eccentricity e = 𝑐/𝑎 = √𝟏𝟑/𝟕 Latus rectum = (2𝑏^2)/𝑎 = (2 × 36)/7 = 𝟕𝟐/𝟕

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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.