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Ex 11.3, 11 - Find equation Vertices (0, 13), foci (0, 5)

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


Transcript

Ex 11.3, 11 Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ±13), foci (0, ±5) Given Vertices (0, ±13) Hence The vertices are of the form (0, ±a) Hence, the major axis is along y-axis & Equation of ellipse is of the form 𝒙^𝟐/𝒃^𝟐 + 𝒚^𝟐/𝒂^𝟐 = 1 From (1) & (2) a = 13 Also given coordinate of foci = (0, ±5) We know that foci are = (0, ±c) So c = 5 We know that c2 = a2 − b2 (5) 2 = (13) 2 − b2 b2 = (13) 2 − (5) 2 b2 = 169 − 25 b2 = 144 Equation of ellipse is 𝑥^2/𝑏^2 + 𝑦^2/𝑎^2 = 1 Putting value 𝒙^𝟐/𝟏𝟒𝟒 + 𝒚^𝟐/𝟏𝟔𝟗 = 1 Equation of ellipse is 𝑥^2/𝑏^2 + 𝑦^2/𝑎^2 = 1 Putting value 𝒙^𝟐/𝟏𝟒𝟒 + 𝒚^𝟐/𝟏𝟔𝟗 = 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.