Question 617296
STEP 1 - FIGURE OUT WHAT IT LOOKS LIKE:
The vertices are on a horizontal line, with y=-9, so the foci and center are on the same line (in between those vertices).
That line is the major axis, and we say that the ellipse has a horizontal major axis, meaning that is stretched out horizontally (short and fat).
 
STEP 2 - FIND THE CENTER:
The center of the ellipse is midway between the vertices, so it is the midpoint of the segment between (-9,3), and (-9,-23).
To find the coordinates of the center, you just average the coordinates of the vertices.
{{{(3+(-23))/2=-20/2=-10}}} so the center is at (-9,-10).
 
STEP 3 - FIND THE SEMI-MAJOR AXIS:
That is the distance from the vertices to the center, and it is 13 units:
{{{3-(-10)=3+10=13}}} or {{{-10-(-23)=-10+23=13}}}
We usually call it {{{a}}}, so {{{a=13}}} 
 
STEP 4 - FIND THE FOCAL DISTANCE:
That is the distance from the foci to the center. and it is 12 units:
{{{2-(-10)=2+10=12}}} or {{{-10-(-22)=-10+22=12}}}
We usually call it {{{c}}}, so {{{c=12}}}
 
STEP 5 - FIND THE SEMI-MINOR AXIS:
That is the distance from the co-vertices to the center.
We usually call it {{{b}}}, and we should know that in an ellipse
{{{a^2=b^2+c^2}}}
Substituting we get
{{{13^2=b^2+12^2}}} --> {{{169=b^2+144}}} --> {{{169-144=b^2}}} --> {{{b^2=25}}} --> {{{b=5}}}
 
STEP 6 - WRITE THE EQUATION:
We should know that the standard form of the equation of an ellipse with a horizontal major axis, centered at (h,k) is
{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}}
So we substitute the values for h,k,a, and b that we found before and write
{{{(x-(-10))^2/13^2+(y-(-9))^2/5^2=1}}} --> {{{highlight((x+10)^2/13^2+(y+9)^2/5^2=1)}}} or {{{highlight((x+10)^2/169+(y+9)^2/25=1)}}}
 
NOTE:
You do not need to exactly memorize the formulas.
What you do not remember, you can figure out because it makes sense.
It is easy to see that when you substitute the coordinates of the center,
Take the standard formula for the equation.
If you substitute the coordinates of the center (one at a time), you get vertices and covertices.
They are at distances a and b from the center, and we always choose the largest as a, so that we can calculate {{{c=sqrt(a^2-b^2)}}} all the time.
The {{{a^2}} is under {{{(x-h)^2}}} in the equation when the ellipse extends farther from the center horizontally. (Otherwise, it's under the {{{(y-k)^2}}} ).
The formula {{{a^2=b^2+c^2}}} takes a little more thinking.
The right triangle formed by the center, a focus, and a covertex gives you the formula
{{{a^2=b^2+c^2}}}
The legs of the triangle are b and c.
The hypotenuse is the distance from the focus to the covertex, and we can see that it is a (if someone illuminates the basis for that fact).
The sum of the distances to the foci is the same for all point on an ellipse.
(That's the definition of ellipse, and that should be remembered). 
The distance from one vertex to the foci are a-c (for the closest focus) and a+c (for the focus that is far away).
Their sum is (a-c)+(a+c)=2a.
So the sum of the distances form any point on the ellipse to the foci is 2a.
Back to that right triangle whose hypotenuse is the distance from the focus to the covertex.
Because the distance from the covertex to both foci is the same, the sum of the distances from the covertex to both foci (which is 2a) is twice that hypotenuse.