Answer:
The given equation is

Which represents an elipse.
To find its elements, we need to divide the equation by 100

Where
and
. Remember that the greatest denominator is
, and the least is
. So, we extract the square root on each equation.
and
.
In a elipse, we have a major axis and a minor axis. In this case, the major axis is vertical and the minor axis is horizontal, that means this is a vertical elipse.
The length of the major axis is
.
The length of the minor axis is
.
The vertices are
and
.
Now, the main parameters of an elipse are related by
, which we are gonna use to find
, the parameter of the focus.

So, the coordinates of each focus are
and 
The eccentricity of a elipse is defined

The latus rectum is defined

Finally, the graph of the elipse is attached.