Ex 11.3, 19 - Find ellipse: Centre (0, 0), major axis y-axis

Ex 11.3,  19 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.3,  19 - Chapter 11 Class 11 Conic Sections - Part 3 Ex 11.3,  19 - Chapter 11 Class 11 Conic Sections - Part 4 Ex 11.3,  19 - Chapter 11 Class 11 Conic Sections - Part 5

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Transcript

Ex 10.3, 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 𝒙^𝟐/𝒃^𝟐 + 𝒚^𝟐/𝒂^𝟐 = 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/𝑏^2 + (2)^2/𝑎^2 = 1 9/𝑏^2 + 4/𝑎^2 = 1 Putting x = 1 & y = 6 in (1) 〖(1)〗^2/𝑏^2 + 〖(6)〗^2/𝑎^2 = 1 1/𝑏^2 + 36/𝑎^2 = 1 From (3) 1/𝑏^2 + 36/𝑎^2 = 1 1/𝑏^2 = 1 − 36/𝑎^2 Putting value of b2 in (2) 9/𝑏^2 + 4/𝑎^2 = 1 9(1/𝑏^2 ) + 4/𝑎^2 = 1 9(1−36/𝑎^2 ) + 4/𝑎^2 = 1 9 − 324/𝑎^2 + 4/𝑎^2 = 1 (−320)/𝑎^2 = 1 − 9 (−320)/𝑎^2 = −8 1/𝑎^2 = (−8)/(−320) 1/𝑎^2 = 8/320 1/𝑎^2 = 1/40 a2 = 40 Putting value of 𝑎^2 in (3) 1/𝑏^2 + 36/𝑎^2 = 1 1/𝑏^2 = 1 − 36/𝑎^2 1/𝑏^2 = 1 − 36 (1/40) 1/𝑏^2 = (40 − 36)/40 1/𝑏^2 = 4/40 1/𝑏^2 = 1/10 b2 = 10 Now required equation of ellipse is 𝑥^2/𝑏^2 + 𝑦^2/𝑎^2 = 1 Putting value of b2 & a2 𝒙^𝟐/𝟏𝟎 + 𝒚^𝟐/𝟒𝟎 = 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.