Question 857855
The foci are on the y-axis, so the vertices, and the major axis will be on the y-axis too.
With the foci at (0,5) and (0,-5), the center is obviously at (0,0),
the midpoint of the segment connecting the foci.
(When not so obvious, the midpoint coordinates are calculated from the coordinates of the foci, by averaging them,
{{{x=(0+0)/2=0/2}}} and {{{y=(5+(-5))/2=0/2=0}}} ). 
The focal distance, the distance from either focus to the center is obviously {{{5}}} .
(When not so obvious, we calculate it as half the difference between the coordinates of the foci:
{{{c=5-(-5))/2=10/2=5}}} ).
We calculate the semi-major axis {{{a}}} as half the length of the major axis:
{{{a=14/2=7}}} ,
The equation of an ellipse centered on the origin, (0,0),
and with the major axis along the y-axis is
{{{x^2/b^2+y^2/a^2=1}}} .
We need to find {{{b^2}}} , the square of the semi-minor axis {{{b}}} .
It can be found from {{{a^2=b^2+c^2}}} :
{{{7^2=b^2+5^2}}}
{{{49=b^2+25}}}
{{{49-25=b^2}}}
{{{b^2=24}}}
The equation of the ellipse is
{{{highlight(x^2/24+y^2/49=1)}}}
{{{drawing(400,400,-8,8,-8,8,
grid(1),red(circle(0,0,0.2)),
red(arc(0,0,9.8,14,0,360)),
blue(line(0,0,0,5)),blue(line(0,0,4.9,0)),
blue(line(0,5,4.9,0)),blue(rectangle(0,0,0.4,0.4)),
red(circle(0,7,0.2)),red(circle(0,-7,0.2)),
blue(circle(0,5,0.2)),blue(circle(0,-5,0.2)),
blue(circle(4.9,0,0.2)),blue(circle(-4.9,0,0.2)),
locate(0.1,2.8,blue(c=5)),
locate(1.5,0.8,blue(b=sqrt(24))),
locate(2.5,2.9,blue(a=7))
)}}}