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

part 2 - Ex 10.3,  11 - Circle - Concept wise - Chapter 10 Class 11 Conic Sections
part 3 - Ex 10.3,  11 - Circle - Concept wise - Chapter 10 Class 11 Conic Sections

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Ex 10.3, 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 š’™^šŸ/š’ƒ^šŸ + š’š^šŸ/š’‚^šŸ = 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 āˆ’ b2 (5) 2 = (13) 2 āˆ’ b2 b2 = (13) 2 āˆ’ (5) 2 b2 = 169 āˆ’ 25 b2 = 144 Equation of ellipse is š‘„^2/š‘^2 + š‘¦^2/š‘Ž^2 = 1 Putting value š’™^šŸ/šŸšŸ’šŸ’ + š’š^šŸ/šŸšŸ”šŸ— = 1 Equation of ellipse is š‘„^2/š‘^2 + š‘¦^2/š‘Ž^2 = 1 Putting value š’™^šŸ/šŸšŸ’šŸ’ + š’š^šŸ/šŸšŸ”šŸ— = 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo