Example 10 - 9x2 + 4y2 = 36, find foci, vertices, length - Examples

part 2 - Example 10 - Examples - Serial order wise - Chapter 10 Class 11 Conic Sections
part 3 - Example 10 - Examples - Serial order wise - Chapter 10 Class 11 Conic Sections part 4 - Example 10 - Examples - Serial order wise - Chapter 10 Class 11 Conic Sections

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Example 10 Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse 9x2 + 4y2 = 36. Given 9x2 + 4y2 = 36 Dividing whole equation by 36 (9๐‘ฅ^2 + 4๐‘ฆ^2)/36 = 36/36 9/36 x2 + (4๐‘ฆ^2)/36 = 1 ๐‘ฅ^2/4 + ๐‘ฆ^2/9 = 1 Since 4 < 9 Hence the above equation is of the form ๐‘ฅ^2/๐‘^2 + ๐‘ฆ^2/๐‘Ž^2 = 1 Comparing (1) & (2) We know that c = โˆš(a2โˆ’b2) c = โˆš(9โˆ’4) c = โˆš๐Ÿ“ Co-ordinate of foci = (0, ยฑ c) = (0, ยฑ โˆš5) So co-ordinates of foci (0, โˆš๐Ÿ“), & (0, โˆ’โˆš๐Ÿ“) Vertices = (0, ยฑ a) = (0, ยฑ 3) So, Vertices are (0, 3) & (0, โˆ’3) Length of major axis = 2a = 2 ร— 3 = 6 Length of minor axis = 2b = 2 ร— 2 = 4 Eccentricity e = c/a = โˆš๐Ÿ“/๐Ÿ‘ Length of latus rectum = 2b2/a = (2 ร— 4)/3 = ๐Ÿ–/๐Ÿ‘

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