Question 738632
<pre>

The idea is to get the equation like one of these

{{{(x-h)^2/a^2}}}{{{""+""}}}{{{(y-k)^2/b^2}}}{{{""=""}}}{{{1}}} or {{{(x-h)^2/b^2}}}{{{""+""}}}{{{(y-k)^2/a^2}}}{{{""=""}}}{{{1}}}

a is one-half the major axis and b is one-half the minor axis.
So a² is always larger than b².  If a² is under the term in x,
the ellipse is like this: {{{drawing(20,10,-2,2,-1,1,arc(0,0,-3.9,1.9) )}}} and if b² is under the term in x, 
the ellipse looks like this {{{drawing(10,20,-1,1,-2,2,arc(0,0,1.9,-3.9) )}}}.
 
 9x² - 18x + 4y² - 27 = 0
 
Add 28 to both sides to get the constant 
term off the left side:

      9x² - 18x + 4y² = 27

Factor 9 out of of the first two terms:

     9(x² - 2x) + 4y² = 27

Complete the square in the parentheses:
1. Multiply the coefficient of x, which is -2 
   by {{{1/2}}}, getting -1
2. Square -1, getting (-1)² or +1
3. Add +1 to the expression in the parentheses.
4. Since the parentheses is multiplied by 9,
   this amouts to adding 9 to the left side,
   so we must add 9·1 or 9 to the right side.

 9(x² - 2x + 1) + 4y² = 27 + 9

We factor x² - 2x + 1 as (x - 1)(x - 1) and as (x - 1)²

      9(x - 1)² + 4y² = 36

There is no y term so we do not need to complete
the square.  We just write y as (y - 0)

9(x - 1)² + 4(y - 0)² = 36  

Next we get 1 on the right side by dividing every 
term by 36

     {{{(9(x-1)^2)/36}}}{{{""+""}}}{{{(4(y-0)^2)/36}}}{{{""=""}}}{{{36/36}}}

Simplify:

     {{{(x-1)^2/4}}}{{{""+""}}}{{{(y-0)^2/9}}}{{{""=""}}}{{{1}}}

Since 4 is less than 9 we know that b²=4 is under the 
term in x, so the ellipse looks like this {{{drawing(10,20,-1,1,-2,2,arc(0,0,1.9,-3.9) )}}}.  We compare that to

     {{{(x-h)^2/b^2}}}{{{""+""}}}{{{(y-k)^2/a^2}}}{{{""=""}}}{{{1}}}

and get h=1, b²=4, k=0, a²=9, and of course b=2 and a=3

So the center = (h,k) = (1,0), the major axis is vertical, a=3 units
above the center and a=3 units below the center, like this green line,
which is the whole major axis:

{{{drawing(300,400,-2,4,-4,4,graph(300,400,-2,4,-4,4),
circle(1,0,.08),green(line(1,-3,1,3)) )}}}

The minor axis is horizontal, b=2 units left of the center and b=2
units right of the center, like this green line, which is the whole
minor axis:

{{{drawing(300,400,-2,4,-4,4,graph(300,400,-2,4,-4,4),
circle(1,0,.08),green(line(1,-3,1,3),line(-1,0,3,0)) )}}}

So we can sketch in the ellipse:

{{{drawing(300,400,-2,4,-4,4,graph(300,400,-2,4,-4,4),
circle(1,0,.08),green(line(1,-3,1,3),line(-1,0,3,0)),
arc(1,0,4,6),locate(1,3.4,"(1,3)"),locate(1,-3,"(1,-3)")

 )}}}

 To calculate the foci, we need the quantity c, the distance
from the center to the foci, given by the equation:

c² = a² - b²
c² = 9 - 4
c² = 5
 c = &#8730;<span style="text-decoration: overline">5</span>

So the foci are the points (1,±&#8730;<span style="text-decoration: overline">5</span>)

The foci are plotted below:

{{{drawing(300,400,-2,4,-4,4,graph(300,400,-2,4,-4,4),
circle(1,0,.08),green(line(1,-3,1,3),line(-1,0,3,0)),
arc(1,0,4,6),locate(1,3.4,"(1,3)"),locate(1,-3,"(1,-3)"),
circle(1,sqrt(5),.08),circle(1,-sqrt(5),.08),
locate(1,-2.1,F(1,-sqrt(5))), locate(1,2.1,F(1,sqrt(5)))

 )}}}

The center is (1,0). The vertices are (1,±3).
the covertices are (-1,0) and (3,0),
the major axis is 6, the minor axis is 4.

Edwin</pre>