SOLUTION: I have to find the center, vertices, co-vertices, foci, and asymptotes for the hyperbola {{{4x^2-8x-y^2+4y-4=0}}}. Can someone help me please?

4x² - 8x - y² + 4y - 4 = 0

Factor the coefficient of x², which is 4, out of 
the two x-terms 

4(x² - 2x) - y² + 4y - 4 = 0

Factor the coefficient of y², which is -1, out of 
the two y-terms

4(x² - 2x) - (y² - 4y) - 4 = 0

Get the term -4 off the left side by adding
4 to both sides:
 
    4(x² - 2x) - (y² - 4y) = 4

Multiply the coefficient of x inside the parentheses,
which is -2 by ½ getting -1.  Then square -1 getting +1.
Add +1 inside the first parentheses, and offset it by
adding +4 to the right side, since adding +1 inside
the first parentheses amounts to adding +4 to the left side
because of the 4 outside the first parentheses:

4(x² - 2x + 1) - (y² - 4y) = 4 + 4

Combine the 4 + 4 on the right as 8

4(x² - 2x + 1) - (y² - 4y) = 8

Multiply the coefficient of y inside the second parentheses,
which is -4 by ½ getting -2.  Then square -2 getting +4.
Add +4 inside the second parentheses, and offset it by
adding -4 to the right side, since adding +4 inside
the second parentheses amounts to adding -4 to the left side
because of the - outside the second parentheses:

4(x² - 2x + 1) - (y² - 4y + 4) = 8 - 4

Combine the 8 - 4 on the right as 4

4(x² - 2x + 1) - (y² - 4y + 4) = 4

Factor x² - 2x + 1 as (x - 1)(x - 1) and then as (x - 1)²
Factor y² - 4y + 4 as (y - 2)(y - 2) and then as (y - 2)²

          4(x - 1)² - (y - 2)² = 4

Next we must get a 1 on the right.
So we divide all the terms by 4

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

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

Next we compare that to

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

which means that it is a hyperbolka that looks like this )(

We see that h = 1, k = 2, a² = 1 so a = 1 and b² = 4 so b = 2

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

Plot the center (1,2)



Draw the transverse axis which is 2a or 2 units long and
is horizontal, with the center at its midpoint.  That 
means we draw 1 units left of the center and 1 units
right of the center to get the complete transverse
axis, drawn in green below:



Next draw the conjugate axis which is 2b or 4 units long and
is vertical, with the center at its midpoint.  That 
means we draw 2 units up from the center and 2 units
down from the center to get the complete transverse
axis, also drawn in green below:



Next we draw the defining rectangle, which is a
rectangle with the ends of the transverse and
conjugate axes as midpoints of the sides.  I'll
draw it in green, too:


      
Next we draw the extended diagonals of the defining rectangle.
They will be the asymptotes of the hyperbola:



Now we can sketch in the hyperbola to have its vertices as
the ends of the transverse axis and to approach the \
asymptotes.  I'll draw the hyperbola in red:



vertices, co-vertices, foci, and asymptotes

The vertices are the endpoints of the transverse axis.
They are (0,2) and (2,2)

The co-vertices are the endpoints of the conjugate axis.
They are (1,0) and (1,4)

To find the foci we calculate c from the formula

c² = a² + b²
c² = 1² + 2²
c² = 1 + 4
c² = 5_
 c = √5

Then the foci are the points which are c units right
and left of the vertices.  They are
    _            _ 
(1+√5,2) and (1-√5,2)

I'll plot the two foci:



Now all we need are the equations of the two asymptotes:

The asymptote that slants uphill to the right passes through (0,0)
and (1,2),

We use the slope formula  

Substitute in 



y = 2x

That's the equation of the asymptote that slants 
uphill to the right.

The asymptote that slants downhill to the right
passes through (2,0) and (1,2),

Using the slope formula  

Substitute in 





y = -2x + 2

That's the equation of the asymptote that slants
downhill to the right.

Edwin