Question 115143
 I need to find the center,vertices, and foci of the ellipse
{{{4x^2+25y^2-8x+100y+4=0}}}
<pre><font size = 4><b>
You first have to put it in standard form, which is either

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

if it turns out that the larger number is under
the term on the left which contains x, and the 
graph will be an ellipse with a horizontal major
axis and a vertical minor axis, where

1. The center is (h, k) 
2. The length of the semi-major axis is " a ".
3. The ends of the major axis are the points (h-a,k) and (h+a,k)
4. The ends of the minor axis are the points (h,k-b) and (h,k+b)
5. Foci are (h-c,k), (h+c,k) where c is calculated from c² = a²-b²  

or

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

if it turns out that the larger number is under
the term on the left which contains y, and the 
graph will be an ellipse with a horizontal major
axis and a vertical minor axis, where and the 
graph will be an ellipse with a vertical major 
axis and a horizontal minor axis

where

1. The center is (h, k) 
2. The length of the semi-major axis is " a ".
3. The ends of the major axis are the points (h,k-a) and (h,k+a)
4. The ends of the minor axis are the points (h-b,k) and (h+b,k)
5. Foci are (h,k-a), (h,k+a) where c is calculated from c² = a²-b²  

We can't tell which it is unti we get it into standard form:

4x² + 25y² - 8x + 100y + 4 = 0

Rearrange getting x² term then the x term then y²and y terms
That is, swap the second and third terms:

4x² - 8x + 25y² + 100y + 4 = 0

Subtract 4 from both sides:

    4x² - 8x + 25y² + 100y = -4

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

  4(x² - 2x) + 25(y² + 4y) = -4

Complete the square in the first parentheses:

The coefficient of x is -2
Multiply it by {{{1/2}}}, getting -1
Square -1, getting +1.
Add +1 inside the first parentheses.
But the parentheses has a 4 in front.
So this amounts to adding 4·1 or 4 to
the left side, so we must add +4 to
the right side:

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

Complete the square in the second parentheses:
The coefficient of y is 4
Multiply it by {{{1/2}}}, getting 2
Square -2, getting +4.
Add +4 inside the first parentheses.
But the parentheses has a 25 in front.
So this amounts to adding 25·4 or 100 to
the left side, so we must add +100 to
the right side:

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

Factor both parentheses on the left and combine the 
numbers on the right

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

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

To get a 1 on the right side, divide every term
by 100
    
            4(x - 1)²   25(y + 2)²    100
           ---------  + ---------- = -----
               100         100        100

Cancel             
            1           1              1  
            <s>4</s>(x - 1)²   <s>25</s>(y + 2)²    <s>100</s>
           ---------  + ---------- = -----
               <s>100</s>         <s>100</s>        <s>100</s>
               25           4          1  
 

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


Since 25 is greater than 4, and since a² is always the
larger or these in an ellipse, then a² is under the
term which contains x, we compare that to:

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

so it is an ellipse with a horizontal major axis and
a vertical minor axis. Comparing letters in this with
the numbers in the equation we got, we see that:

h = 1, k = -2, a² = 25, b² = 4 so a = 5 and b = 2

This means
1. The center is (h, k) = (1, -2)
2. The length of the semi-major axis is a = 5
3. The ends of the major axis (vertices) are the points (h-a,k), (h+a,k)
   or (1-5,-2) and (1+5,-2) or (-4,-2) and (6,-2)
4. The ends of the minor axis are the points (h,k-b), (h,k+b)
   or (1,-2-2), (1,-2+2) or (1,-4) and (1,0)
5. Foci are (h-c,k), (h+c,k) where c is gotten from c² = a²-b²  
   We calculate c:
   c² = a² - b²
   c² = 5² - 2²
   c² = 25 - 4
   c² = <u>21</u>
   c = <font face = "symbol">Ö</font>21                                 __             __         
   So foci are (h-c,k) and (h+c,k), or (1-<font face = "symbol">Ö</font>21,-2) and (1+<font face = "symbol">Ö</font>21), 
   or about (-3.6,-2) and (5.6,-2)

We draw the major and minor axes and plot the foci:   

{{{drawing(400,370,-5,7,-5,7, line(-4,-2,6,-2),line(1,-4,1,0), 
   graph(400,375,-5,7,-5,7),   
   locate(1-sqrt(21)-.1,-1.695,o),locate(1+sqrt(21)-.1,-1.695,o) 
)}}}

Then we sketch in the ellipse:

{{{drawing(400,370,-5,7,-5,7, line(-4,-2,6,-2),line(1,-4,1,0), 
   graph(400,375,-5,7,-5,7,sqrt(100-4(x-1)^2)/5 - 2),
   graph(400,375,-5,7,-5,7,-sqrt(100-4(x-1)^2)/5 - 2),
   locate(1-sqrt(21)-.1,-1.695,o),locate(1+sqrt(21)-.1,-1.695,o) 
)}}}

The answers to your problem are:

center = (1, -2)
vertices = <u>en</u>ds of major a<u>xi</u>s = (-4,-2) and (6,-2)
foci = (1-<font face = "symbol">Ö</font>21,-2) and (1+<font face = "symbol">Ö</font>21,-2), 

Edwin</pre>