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Last updated at Feb. 6, 2020 by Teachoo
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Example 13 Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (โ 1,4). Given that Major axis is along x-axis So required equation of ellipse is ๐^๐/๐^๐ + ๐^๐/๐^๐ = 1 Given that point (4, 3) & (โ1, 4) lie of the ellipse So, point (4, 3) & (โ1, 4) will satisfy equation of ellipse Putting x = 4 & y = 3 in (1) ๐ฅ^2/๐^2 + ๐ฆ^2/๐^2 = 1 ใ(4)ใ^2/๐^2 + ใ(3)ใ^2/๐^2 = 1 ๐๐/๐^๐ + ๐/๐^๐ = 1 Putting x = โ1, y = 4 is in (1) ๐ฅ^2/๐^2 + ๐ฆ^2/๐^2 = 1 ใ(โ1)ใ^2/๐^2 + ใ(4)ใ^2/๐^2 = 1 ๐/๐^๐ + ๐๐/๐^๐ = 1 Now, our equations are 16/๐^2 + 9/๐^2 = 1 1/๐^2 + 16/๐^2 = 1 From (3) 1/๐^2 +16/๐^2 = 1 1/๐^2 " "= 1โ16/๐^2 Putting value of 1/๐^2 in (2) 16/๐^2 + 9/๐^2 = 1 16(1/๐^2 ) + 9/๐^2 = 1 16(1โ16/๐^2 ) + 9/๐^2 = 1 16 โ 256/๐^2 + 9/๐^2 = 1 (โ256 + 9)/๐^2 = 1 โ16 (โ247)/๐^2 = โ15 b2 = (โ247)/(โ15) b2 = ๐๐๐/๐๐ Putting value of b2 = 247/15 in (3) 1/๐^2 " "= 1โ16/๐^2 1/๐^2 " "= 1โ16/(247/15) 1/๐^2 " "= 1โ(16 ร 15)/247 1/๐^2 " "= (247 โ 240)/247 1/๐^2 " "= 7/247 ๐๐ = ๐๐๐/๐ Thus, a2 = 247/7 & b2 = 247/15 Hence required of ellipse is ๐ฅ^2/๐^2 + ๐ฆ^2/๐^2 = 1 Putting values of a2 & b2 ๐ฅ^2/((247/7) ) + ๐ฆ^2/((247/15) ) = 1 1/๐^2 " "= 1โ(16 ร 15)/247 1/๐^2 " "= (247 โ 240)/247 1/๐^2 " "= 7/247 ๐๐ = ๐๐๐/๐ Thus, a2 = 247/7 & b2 = 247/15 Hence required of ellipse is ๐ฅ^2/๐^2 + ๐ฆ^2/๐^2 = 1 Putting values of a2 & b2 ๐ฅ^2/((247/7) ) + ๐ฆ^2/((247/15) ) = 1 (7๐ฅ^2)/247 + (15๐ฆ^2)/247 = 1 7x2 + 15y2 = 247
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