# Answer to Question #25466 in Analytic Geometry for scofeild ivan

Question #25466

find the equation of the ellipse with eccentricity of 1/3 and distance betwwen foci is 2 ?

Expert's answer

Let a,b be 2 semiaxis of ellipse. Let c be distance from origin to one of the focuses.

By definition of eccentricity we have that e=c/a = 1/3.

Since distance between focuses is 2, then 2c = 2.

Then c=1 => a= c/ e = 3

Since b^2=a^2-c^2=9-1=8 we have general equation of ellipse: x^2/9 +y^2/8 = 1

By definition of eccentricity we have that e=c/a = 1/3.

Since distance between focuses is 2, then 2c = 2.

Then c=1 => a= c/ e = 3

Since b^2=a^2-c^2=9-1=8 we have general equation of ellipse: x^2/9 +y^2/8 = 1

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