Ex 11.3, 10 - Chapter 11 Class 11 Conic Sections
Last updated at Dec. 8, 2016 by Teachoo
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 11.3, 10 Find the equation for the ellipse that satisfies the given conditions: Vertices (±5, 0), foci (±4, 0) Given Vertices (±5, 0) Since the vertices are of form (± a, 0) Hence, Major axis is along x-axis and equation of ellipse is 𝑥2𝑎2 + 𝑦2𝑏2 = 1 From (1) & (2) a = 5 Also given coordinate of foci = (±4, 0) We know that foci = (± c, 0) So c = 4 We know that c2 = a2 − b2 (4) 2 = (5) 2 − b2 b2 = (5) 2 − (4) 2 b2 = 25 − 16 b2 = 9 b = 9 b = 3 Hence b = 3 Equation of ellipse is 𝑥2𝑎2 + 𝑦2𝑏2 = 1 Putting value 𝑥2(5)2 + 𝑦2(3)2 = 1 𝑥225 + 𝑦29 = 1 Which is required equation
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