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Last updated at Feb. 6, 2020 by Teachoo

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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

Ex 11.3

Ex 11.3, 1

Ex 11.3, 2

Ex 11.3, 3

Ex 11.3, 4

Ex 11.3, 5 Important

Ex 11.3, 6

Ex 11.3, 7

Ex 11.3, 8

Ex 11.3, 9

Ex 11.3, 10

Ex 11.3, 11 Important

Ex 11.3, 12 Important

Ex 11.3, 13

Ex 11.3, 14 Important

Ex 11.3, 15

Ex 11.3, 16 Important

Ex 11.3, 17

Ex 11.3, 18 Important

Ex 11.3, 19 Important

Ex 11.3, 20 You are here

Chapter 11 Class 11 Conic Sections

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.