Ex 11.3, 19 - Chapter 11 Class 11 Conic Sections (Term 2)
Last updated at Feb. 6, 2020 by Teachoo
Last updated at Feb. 6, 2020 by Teachoo
Transcript
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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