Question 1007364
{{{16x^2+y^2+2y=0}}}
{{{16x^2+(y^2+2y+1)-1=0}}}
{{{16x^2+(y+1)^2=1}}}
{{{x^2/(1/16)+(y+1)^2=1}}}


The long axis is vertical and the short axis is horizontal.  This is like {{{x^2/b^2+(y-k)^2/a^2=1}}}.



Your ellipse has center at (0,-1).
Minor vertices are ( -1/4, -1 ) and ( 1/4, -1).
Major vertices are  (0,0) and (0,-2).
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{{{system(a=1,b=1/4)}}}


(HINT: Each focus will be on the y-axis).
{{{c^2=a^2-b^2}}}, and c is the distance from the center to either focus.