Ex 10.3, 5 - Chapter 10 Class 11 Conic Sections

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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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Transcript

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 ๐‘ฅ^2/49 + ๐‘ฆ^2/36 = 1 Since 49 > 36 Hence the above equation is of the form ๐‘ฅ^2/๐‘Ž^2 + ๐‘ฆ^2/๐‘^2 = 1 Comparing (1) & (2) We know that c = โˆš(a2โˆ’b2) c = โˆš(49โˆ’36) c = โˆš๐Ÿ๐Ÿ‘ Coordinate of foci = (ยฑ c, 0) = (ยฑ โˆš๐Ÿ๐Ÿ‘, 0) So coordinate of foci are (โˆš13, 0), (โˆ’โˆš13, 0) Vertices = (ยฑ a, 0) = (ยฑ7, 0) So vertices are (7, 0) & (โˆ’7, 0) Length of major axis = 2a = 2 ร— 7 = 14 Length of minor axis = 2b = 2 ร— 6 = 12 Eccentricity e = ๐‘/๐‘Ž = โˆš๐Ÿ๐Ÿ‘/๐Ÿ• Latus rectum = (2๐‘^2)/๐‘Ž = (2 ร— 36)/7 = ๐Ÿ•๐Ÿ/๐Ÿ•

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