Ex 10.3, 11 - Chapter 10 Class 11 Conic Sections
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
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 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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