SOLUTION: Find the center, vertices, and foci for the ellipse: 4x^2+16y^2=64 Thanks:)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the center, vertices, and foci for the ellipse: 4x^2+16y^2=64 Thanks:)      Log On


   

Question 38780: Find the center, vertices, and foci for the ellipse: 4x^2+16y^2=64
Thanks:)
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
4x%5E2%2B16y%5E2=64
%284x%5E2%2F64%29%2B%2816y%5E2%2F64%29=64%2F64
%28x%5E2%2F16%29%2B%28y%5E2%2F4%29=1
Center: (0,0)
Transverse axis: horizontal
a=4
b=2
c^2=a^2-b^2
c^2=16-4
c^2=12
c+=+sqrt%2812%29
Vertex: (4,0),(-4,0),(0,2),(0,-2)
Foci: (4 - sqrt%2812%29,0),(-4 + sqrt%2812%29, 0)