# Ex 11.3, 14

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 11.3, 14 Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ± 5) , ends of minor axis (±1, 0) Given end of major axis (0, ± 5), & ends of minor axis (±1, 0) Major axis is along the y-axis So our required equation of ellipse is 𝑥2𝑏2 + 𝑦2𝑎2 = 1 We know that End of major axis is the vertices of the ellipse So vertices of ellipse = (0, ± 3) Also, Vertices of the ellipse is (0, ± a) Comparing (0, ± a) = (0, ± 3) a = 𝟑 We know that End of minor axis = (± b, 0) So, (±1, 0) = (± b, 0) So, b = 1 Required equation of ellipse is 𝑥2𝑏2 + 𝑦2𝑎2 = 1 Putting values 𝑥212 + 𝑦252 = 1 𝒙𝟐𝟏 + 𝒚𝟐𝟓 = 1

Chapter 11 Class 11 Conic Sections

Serial order wise

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CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .