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Last updated at Feb. 6, 2020 by Teachoo
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Ex 11.3, 19 Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6). Since major axis is along y-axis & centre is at (0,0) So required equation of ellipse is ๐^๐/๐^๐ + ๐^๐/๐^๐ = 1 Given that ellipse passes through point (3, 2) & (1, 6) Points (3, 2) & (1, 6) will satisfy equation of ellipse. Putting x = 3 & y = 2 in (1) (3)^2/๐^2 + (2)^2/๐^2 = 1 9/๐^2 + 4/๐^2 = 1 Putting x = 1 & y = 6 in (1) ใ(1)ใ^2/๐^2 + ใ(6)ใ^2/๐^2 = 1 1/๐^2 + 36/๐^2 = 1 From (3) 1/๐^2 + 36/๐^2 = 1 1/๐^2 = 1 โ 36/๐^2 Putting value of b2 in (2) 9/๐^2 + 4/๐^2 = 1 9(1/๐^2 ) + 4/๐^2 = 1 9(1โ36/๐^2 ) + 4/๐^2 = 1 9 โ 324/๐^2 + 4/๐^2 = 1 (โ320)/๐^2 = 1 โ 9 (โ320)/๐^2 = โ8 1/๐^2 = (โ8)/(โ320) 1/๐^2 = 8/320 1/๐^2 = 1/40 a2 = 40 Putting value of ๐^2 in (3) 1/๐^2 + 36/๐^2 = 1 1/๐^2 = 1 โ 36/๐^2 1/๐^2 = 1 โ 36 (1/40) 1/๐^2 = (40 โ 36)/40 1/๐^2 = 4/40 1/๐^2 = 1/10 b2 = 10 Now required equation of ellipse is ๐ฅ^2/๐^2 + ๐ฆ^2/๐^2 = 1 Putting value of b2 & a2 ๐^๐/๐๐ + ๐^๐/๐๐ = 1
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