Question 38780
{{{4x^2+16y^2=64}}}
{{{(4x^2/64)+(16y^2/64)=64/64}}}
{{{(x^2/16)+(y^2/4)=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(12)}}}
Vertex: (4,0),(-4,0),(0,2),(0,-2)
Foci: (4 - {{{sqrt(12)}}},0),(-4 + {{{sqrt(12)}}}, 0)