Ex 11.3, 20 - 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, 20 Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4, 3) and (6, 2). Since Major axis is on the x-axis So required equation of ellipse is ๐^๐/๐^๐ + ๐^๐/๐^๐ = 1 Given that ellipse passes through point (4, 3) & (6, 2) Points (4, 3) & (6, 2) will satisfy the equation of ellipse Putting x = 4 & y = 3 in (1) ใ(4)ใ^2/๐^2 + ใ(3)ใ^2/๐^2 = 1 16/๐^2 + 9/๐^2 = 1 Putting x = 6 & y = 2 in (1) ใ(6)ใ^2/๐^2 + ใ(2)ใ^2/๐^2 = 1 36/๐^2 + 4/๐^2 = 1 From (3) 16/๐^2 = 1 โ 9/๐^2 1/๐^2 = 1/16 (1 โ 9/๐^2 ) Putting value of 1/๐^2 in (2) 36/๐^2 + 4/๐^2 = 1 36(1/๐^2 ) + 4/๐^2 = 1 36(1/16 (1โ9/๐^2 )) + 4/๐^2 = 1 36/16 (1โ9/๐^2 ) + 4/๐^2 = 1 9/4 (1โ9/๐^2 ) + 4/๐^2 = 1 9/4 โ 81/ใ4๐ใ^2 + 4/๐^2 = 1 (โ81)/(4๐^2 ) + 4/๐^2 = 1 โ 9/4 (โ81 + 16)/(4๐^2 ) = (4 โ 9)/4 (โ65)/(4๐^2 ) = (โ5)/4 (โ5)/4 (13/๐^2 )= (โ5)/4 13/๐^2 = 1 1/๐^2 = 1/13 b2 = 13 Putting value of b2 in 1/๐^2 = 1/16 (1 โ 9/๐^2 ) 1/๐^2 = 1/16 (1 โ 9/13) 1/๐^2 = 1/16 ( (13 โ 9)/13) 1/๐^2 = 1/16 ( 4/13) 1/๐^2 = 1/52a a2 = 52 Equation of ellipse is ๐ฅ^2/๐^2 + ๐ฆ^2/๐^2 = 1 Putting value of a2 & b2 ๐^๐/๐๐ + ๐^๐/๐๐ = 1
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