Find an equation for the ellipse that satisfies the given conditions. Eccentricity: 0.25, foci: (±1.25, 0)

Question

Find an equation for the ellipse that satisfies the given conditions.
Eccentricity: 0.25,    foci: (±1.25, 0)

Answer 1

since foci are symetrically located on x-axis about origin, the equation of the ellipse must be of the following form:

 

   (x^2)/(a^2) - (y^2)/(b^2) = 1, where a = semi-major axis, and b = semi-minor axis,

 

and:  e = eccentricity = sqrt(a^2 - b^2)/a = 0.25;  foci located at (+/- sqrt(a^2 - b^2),0) = (+/- 1.25,0)

 

---> sqrt(a^2 - b^2) = 1.25 ---> 1.25/a = 0.25 ---> a = 1.25/0.25  ---> a = 5; and sqrt(a^2 - b^2) = 1.25 = 5/4

 

---> a^2 - b^2 = (5/4)^2 = 25/16;  or 5^2 - b^2 = 25/16 ---> 25 - b^2 = 25/16;  

 

---> b^2 = 25 - (25/16) = 25[1 - 1/16] = 25(15)/16 = 375/16

 

---> (x^2)/25 + (y^2)/(375/16) = 1 ---> (x^2)/25 + (16y^2)/375) = 1