find the center, vertices, and foci of the ellipse
4x² + 25y² - 8x + 100y + 4 = 0. graph

We have to get it in one of these two forms:

   (x - h)²    (y + k)²     
  ---------- + ---------- = 1
      a²          b²

which is an ellipse that looks like an egg
sitting on a table, with the more pointed
ends horizontal.

or

   (x - h)²    (y + k)²     
  ---------- + ---------- = 1
      b²          a² 

which is an ellipse that looks like an egg
which is being held with the more pointed ends
vertical, that is, it looks like the character
zero, "0". 

We can tell this by looking to see whether the 
larger denominator is under the term with x and 
the smaller denominator is under the term in y, 
or vice-versa.  a is always larger than b, so
a² is always larger than b², except when they
are equal, (and then the ellipse is really a 
circle.  But they will not be equal here.)

4x² + 25y² - 8x + 100y + 4 = 0

Get the x terms and the y terms together in descending
order:

4x² - 8x + 25y² + 100y + 4 = 0

Subtract the constant term 4 from both sides:

    4x² - 8x + 25y² + 100y = -4

Factor 4 out of the first two terms on the left and 
factor out 25 out of the last two terms on the left:

   4(x² - 2x) + 25(y² + 4y) = -4

Now inside each of those parenthese we want to add
what is necessary to complete the square.

To complete the square inside (x² - 2x) we must
multiply the coefficient of x by 1/2, then square it.
So we multiply 2 by 1/2 and get 1, then when we
square 1 we get 1.  So 1 is what we want to add
inside the first parentheses.

To complete the square inside (y² + 4y) we must
multiply the coefficient of y by 1/2, then square it.
So we multiply 4 by 1/2 and get 2, then when we
square 2 we get 4.  So 4 is what we want to add
inside the second parentheses.

But when we add those numbers inside the parentheses
we have no choice but to add whatever is necessary to
to the right side also to offset it, so let's
temporarily put blanks on the right and then we will
see what we need to add to the right side to offset
what we have added inside the two sets of parentheses:

   4(x² - 2x + 1) + 25(y² + 4y + 4) = -4 +     +    

When we add 1 inside the first parentheses we really
are adding 4×1 or 4, because the 4 on the outside of the
first parentheses will multiply by the 1, showing that
we really have added 4, not just 1, to the left side. So
to offset we place 4 in the first blank on the right:

   4(x² - 2x + 1) + 25(y² + 4y + 4) = -4 +  4  +      

Similarly, when we add 4 inside the second parentheses 
we really are adding 25×4 or 100, because the 25 on the 
outside of the second parentheses will multiply by the 4,
showing that we really have added 100, not just 4, to the
left side, so to offset we place 100 in the second blank 
on the right:

   4(x² - 2x + 1) + 25(y² + 4y + 4) = -4 +  4  + 100 

Now we factor each of the parentheses as a perfect square,
and we combine the numbers on the right side:

   4(x - 1)² + 25(y + 2)² = 100

Now we divide every term by 100 to get 1 on the right

   4(x - 1)²    25(y + 2)²     100
  ---------- + ------------ = -----
      100          100         100

Divide numerator and denominator of the first fraction by 4
Divide numerator and denominator of the second fraction by 25
The fraction on the right is just 1

   (x - 1)²    (y + 2)²     
  ---------- + ---------- = 1
      25          4         

Now we can determine that a² = 25 and b² = 4 since a² must
be larger than b², so this ellipse is horizontal, like an egg 
lying on a table, and NOT vertical like the number zero. So
we compare it to

   (x - h)²    (y + k)²     
  ---------- + ---------- = 1
      a²          b²

and see that the center = (h,k) = (1, -2).
Since a² = 25, then a = 5 and that is the semi-major axis.
Since b² = 4, then b = 2 and that is the semi-minor axis.

So we plot the center (1, -2).  Then we find the two vertices
by finding the points which are a=5 units right and left of
the center.  The point 5 units right of the center (1,-2) is
the point (6,-2) and the point 5 units left of (1,-2) is the
point (-4,-2),

Now we find the top and bottom points of the ellipse, by
finding the points which are b=2 units directly above and 
directly below the center.  The point 2 units above the 
center (1,-2) is the point (1,0) and the point 2 units below
the center (1,-2) is the point (1,-4).  

So after plotting those five points, like this

  

we sketch in the ellipse like this:

 

Now to find the foci.  These are two point inside the ellipse which are
"c" units left and right of the center.  We have to calculate c from the
equation c² = a² - b²

c² = a² - b²
c² = 5² - 2²
c² = 25 - 4 
c² = 21
     __
c = Ö21

So we find two points this distance left and right of the center
(1,-2).  They are the points: 
    __              __
(1+Ö21, -2) and (1-Ö21, -2) or about (-3.6, -2) and (5.6, -2)

 

Edwin