Question 474815
<pre>
x²-2x+4y²+8y+1 = 0

The object is to make it look like this:

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

which is an ellipse that looks like this "&#4540;"
like an egg on a table, or:

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

which is an ellipse that looks like the number "0". 

We can tell because a² > b²

We start with this:

x²-2x+4y²+8y+1 = 0

We get the 1 on the other side as a -1:

x²-2x+4y²+8y = -1

It's good that the terms are in the order they are.
Sometimes we have to switch them so that the terms
are in the order "x², x, y², y" as they are here.

Write it this way: 

[x²-2x]+[4y²+8y] = -1 

The coefficient of x² is 1 so we don't do anything yet to
the first bracket. 
But the coefficient of y² is 4, so let's factor that out
in the 2nd bracket.

[x²-2x]+[4(y²+2y)] = -1

Now I'll dispense with the brackets and just have parentheses:

(x²-2x)+4(y²+2y) = -1

Next we want to make those two binomials into trinomials.

We skip some space after those binomials 

(x²-2x  )+4(y²+2y  ) = -1

so we can add a number in those two boxes to make those 
binomials into trinomials so they'll factor into squares 
of binomials.

(x²-2x+&#8591;)+4(y²+2y+&#8591;) = -1

Now let's figure out what number goes in the first box.

The coefficient of x is -2 so we take half of it, getting -1,
then we square -1, getting (-1)² or 1,
so we add 1 where the first box is, but we also have to add
1 to the other side of the equation, like this:

[x²-2x+4]+4(y²+2y+&#8591;) = -1+1

Now let's figure out what number goes in the second box.
 
The coefficient of y is 2 so we take half of it, getting 1,
then we square 1, getting 1² or 1, but wait!  See the 4 in 
front of the second parentheses?  If we put a 1 in that box,
It will get multiplied by the 4 in front of the parentheses.
In other words putting a 1 in that second box will in effect 
amount to the same as adding 4 times 1 or 4 to the left side,
not just 1.  So we have to add 4(1) to the right side to offset
adding 1 inside that parentheses since it will be multiplied
by the 4, so we have:

(x²-2x+4)+4(y²+2y+1) = -1+1+4(1)

Notice that what's in the first parentheses,
x²-2x+4 factors as (x-2)(x-2) or (x-2)²

Also notice that what's in the second parentheses
y²+2y+1 factors as (y+1)(y+1) or (y+1)².

So this

(x²-2x+4)+4(y²+2y+1) = -1+1+4(1)

after substituting their factorization for the parentheses
and combining the terms on the right, we have

(x-2)²+4(y+1)² = 4

Next we get a 1 on the right by dividing all three terms by 4:

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

And that simplifies to:

 (x-2)²  (y+1)²    
—————— + ——————— = 1
   4       1       
 
which is in the form:

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

because a² = 4 and b² = 1

The center is (h,k) = (2,-1)

Plot it:

{{{drawing(400,2000/7,-2,5,-3,2,

graph(400,2000/7,-2,5,-3,2), circle(2,-1,.05),
locate(2,-1,"(2,-1)") )}}}

Since a² = 4, a = 2
Since b² = 1, b = 1

a = 2 is the semi-major axis's length, so draw the
major axis 2a or 4 units long with the center as the midpoint.
We also draw the minor axis 2a or 4 units long with the center 
as the midpoint:

{{{drawing(400,2000/7,-2,5,-3,2,
green(line(0,-1,4,-1),line(2,-2,2,0)),
graph(400,2000/7,-2,5,-3,2), circle(2,-1,.05),
locate(2,-1,"(2,-1)") )}}}

Now we can sketch in the ellipse:

{{{drawing(400,2000/7,-2,5,-3,2,
green(line(0,-1,4,-1),line(2,-2,2,0)),
graph(400,2000/7,-2,5,-3,2), circle(2,-1,.05),
locate(2,-1,"(2,-1)"), arc(2,-1,4,-2) )}}}
 
The vertices are the endpoints of the major axes,
(0,-1) and (4,-1).

The covertices are the endpoints of the minor
axis, (2,0) and (2,-2).

Edwin</pre>