Question 472808
<pre>
4x² + y² - 8x - 2y + 1 = 0

Get the constant off the left side:

    4x² + y² - 8x - 2y = -1

Swap the 2nd and 3rd terms

    4x² - 8x + y² - 2y = -1

Out of the first two terms factor out the coefficient of x²
which is 4

  4(x² - 2x) + y² - 2y = -1

We complete the square inside the parentheses:
Multiply the coefficient of x, which is -2, by {{{1/2}}},
getting -1, Then square that result, (-1)² = 1. So we add
1 inside the parentheses, getting 4(x² - 2x + 1), which,
because of the 4*1 in front amount to adding 4 to the left
side so we have to add 4 to the right side:

  4(x² - 2x + 1) + y² - 2y = -1 + 4

Next we complete the square for the y terms:
Multiply the coefficient of y, which is -2, by {{{1/2}}},
getting -1, Then square that result, (-1)² = 1. So we add
1 to both sides:

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

Now we factor "x²-2x+1" as (x-1)² and "y²-2y+1" as "(y-1)²"
and combine the numbers on the right as 4

  4(x - 1)² + (y - 1)² = 4

Then we divide through by 4 to get 1 on the right

  {{{4(x - 1)²/4 + (y - 1)²/4 = 4/4}}}

  {{{(x-1)²/1 + (y-1)/4 = 1}}}

Compare that to the standard form:

  {{{(x-h)²/b² + (y-k)/a² = 1}}}

since a > b, so a² > b² in an ellipse. a=1, b=2

So this is an ellipse with center (h,k) = (1,1),

That's one question answered:  center = (1,1)

Major axis vertical which is 2a or 2(2) or 4

Minor axis horizontal which is 2b or 2(1) = 2

So we draw the center (1,1) and the major and minor axes,
respective 4 and 2 units long with the center (1,1) as the
midoint of each.

{{{drawing(400,400,-3,5,-3,5,

graph(400,400,-3,5,-3,5), green(line(1,-1,1,3), line(0,1,2,1)) )}}}

Then we draw in the ellipse:

{{{drawing(400,400,-3,5,-3,5,

graph(400,400,-3,5,-3,5), green(line(1,-1,1,3), line(0,1,2,1)),
arc(1,1,2,4)

 )}}}

The vertices are at the top and bottom.  They are (1,3) and (1,-1)

{{{drawing(400,400,-3,5,-3,5,

graph(400,400,-3,5,-3,5), green(line(1,-1,1,3), line(0,1,2,1)),
arc(1,1,2,4), locate(1,3.3,"(1,3)"), locate(1,-1,"(1,-1)")

 )}}}

We have to calculate the focal points which are (h,k±c) inside 
the ellipse on the major axis:

First we calculate c from

c² = a² - b²

c² = 2² - 1²

c² = 4 - 1

c² = 3
      _
c = ±<font face = "symbol">Ö</font>3
                      _            _
So the foci are (1,1+<font face = "symbol">Ö</font>3) and (1,1-<font face = "symbol">Ö</font>3). They are the two points plotted
below inside the ellipse on the major axis.

{{{drawing(400,400,-3,5,-3,5,

graph(400,400,-3,5,-3,5), green(line(1,-1,1,3), line(0,1,2,1)),
arc(1,1,2,4), locate(1,3.3,"(1,3)"), locate(1,-1,"(1,-1)"),
circle(1,1-sqrt(3),.05), circle(1,1+sqrt(3),.05)


 )}}}

Edwin</pre>