Ex 10.3,  5 - x2/49 + y2/36 = 1 Find foci, eccentricity - Ex 10.3 - Ex 10.3

part 2 - Ex 10.3,  5 - Ex 10.3 - Serial order wise - Chapter 10 Class 11 Conic Sections
part 3 - Ex 10.3,  5 - Ex 10.3 - Serial order wise - Chapter 10 Class 11 Conic Sections

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Ex 10.3, 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 š‘„^2/49 + š‘¦^2/36 = 1 Since 49 > 36 Hence the above equation is of the form š‘„^2/š‘Ž^2 + š‘¦^2/š‘^2 = 1 Comparing (1) & (2) We know that c = √(a2āˆ’b2) c = √(49āˆ’36) c = āˆššŸšŸ‘ Coordinate of foci = (± c, 0) = (± āˆššŸšŸ‘, 0) So coordinate of foci are (√13, 0), (āˆ’āˆš13, 0) Vertices = (± a, 0) = (±7, 0) So vertices are (7, 0) & (āˆ’7, 0) Length of major axis = 2a = 2 Ɨ 7 = 14 Length of minor axis = 2b = 2 Ɨ 6 = 12 Eccentricity e = š‘/š‘Ž = āˆššŸšŸ‘/šŸ• Latus rectum = (2š‘^2)/š‘Ž = (2 Ɨ 36)/7 = šŸ•šŸ/šŸ•

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