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Ex 11.3, 20 - Find ellipse: Major axis on x-axis, passes

Ex 11.3,  20 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.3,  20 - Chapter 11 Class 11 Conic Sections - Part 3
Ex 11.3,  20 - Chapter 11 Class 11 Conic Sections - Part 4
Ex 11.3,  20 - Chapter 11 Class 11 Conic Sections - Part 5

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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 𝒙^𝟐/𝒂^𝟐 + 𝒚^𝟐/𝒃^𝟐 = 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) 〖(4)〗^2/𝑎^2 + 〖(3)〗^2/𝑏^2 = 1 16/𝑎^2 + 9/𝑏^2 = 1 Putting x = 6 & y = 2 in (1) 〖(6)〗^2/𝑎^2 + 〖(2)〗^2/𝑏^2 = 1 36/𝑎^2 + 4/𝑏^2 = 1 From (3) 16/𝑎^2 = 1 − 9/𝑏^2 1/𝑎^2 = 1/16 (1 − 9/𝑏^2 ) Putting value of 1/𝑎^2 in (2) 36/𝑎^2 + 4/𝑏^2 = 1 36(1/𝑎^2 ) + 4/𝑏^2 = 1 36(1/16 (1−9/𝑏^2 )) + 4/𝑏^2 = 1 36/16 (1−9/𝑏^2 ) + 4/𝑏^2 = 1 9/4 (1−9/𝑏^2 ) + 4/𝑏^2 = 1 9/4 − 81/〖4𝑏〗^2 + 4/𝑏^2 = 1 (−81)/(4𝑏^2 ) + 4/𝑏^2 = 1 − 9/4 (−81 + 16)/(4𝑏^2 ) = (4 − 9)/4 (−65)/(4𝑏^2 ) = (−5)/4 (−5)/4 (13/𝑏^2 )= (−5)/4 13/𝑏^2 = 1 1/𝑏^2 = 1/13 b2 = 13 Putting value of b2 in 1/𝑎^2 = 1/16 (1 − 9/𝑏^2 ) 1/𝑎^2 = 1/16 (1 − 9/13) 1/𝑎^2 = 1/16 ( (13 − 9)/13) 1/𝑎^2 = 1/16 ( 4/13) 1/𝑎^2 = 1/52a a2 = 52 Equation of ellipse is 𝑥^2/𝑎^2 + 𝑦^2/𝑏^2 = 1 Putting value of a2 & b2 𝒙^𝟐/𝟓𝟐 + 𝒚^𝟐/𝟏𝟑 = 1

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.