Question 455974
<pre><font face = "lucida console" color = "indigo" size = 2><b>
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)

{{{drawing(400,400,-4,7,-3.5,7.5,
graph(400,400,-4,7,-3.5,7.5), circle(1,2,.1) )}}}

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:

{{{drawing(400,400,-4,7,-3.5,7.5,
graph(400,400,-4,7,-3.5,7.5), circle(1,2,.1),
green(line(0,2,2,2))

 )}}}

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:

{{{drawing(400,400,-4,7,-3.5,7.5,
graph(400,400,-4,7,-3.5,7.5), circle(1,2,.1),
green(line(0,2,2,2),line(1,0,1,4))

 )}}}

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:

{{{drawing(400,400,-4,7,-3.5,7.5,
graph(400,400,-4,7,-3.5,7.5), circle(1,2,.1),
green(line(0,2,2,2),line(1,0,1,4),
line(0,0,2,0),line(2,0,2,4),line(2,4,0,4),line(0,4,0,0)) )}}}
      
Next we draw the extended diagonals of the defining rectangle.
They will be the asymptotes of the hyperbola:

{{{drawing(400,400,-4,7,-3.5,7.5,
graph(400,400,-4,7,-3.5,7.5), circle(1,2,.1),
green(line(0,2,2,2),line(1,0,1,4),
line(0,0,2,0),line(2,0,2,4),line(2,4,0,4),line(0,4,0,0),
line(-10,-20,14,28), line(-8,20,17,-30)) )}}}

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:

{{{drawing(400,400,-4,7,-3.5,7.5,
graph(400,400,-4,7,-3.5,7.5,2+2sqrt((x-1)^2-1)), circle(1,2,.1),

graph(400,400,-4,7,-3.5,7.5,2-2sqrt((x-1)^2-1)), 
green(line(0,2,2,2),line(1,0,1,4),
line(0,0,2,0),line(2,0,2,4),line(2,4,0,4),line(0,4,0,0),
line(-10,-20,14,28), line(-8,20,17,-30)) )}}}

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 = &#8730;5

Then the foci are the points which are c units right
and left of the vertices.  They are
    _            _ 
(1+&#8730;5,2) and (1-&#8730;5,2)

I'll plot the two foci:

{{{drawing(400,400,-4,7,-3.5,7.5,
graph(400,400,-4,7,-3.5,7.5,2+2sqrt((x-1)^2-1)), circle(1,2,.1),
circle(1+sqrt(5),2,.1), circle(1-sqrt(5),2,.1),
graph(400,400,-4,7,-3.5,7.5,2-2sqrt((x-1)^2-1)), 
green(line(0,2,2,2),line(1,0,1,4),
line(0,0,2,0),line(2,0,2,4),line(2,4,0,4),line(0,4,0,0),
line(-10,-20,14,28), line(-8,20,17,-30)) )}}}

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  {{{m = (y[2]-y[1])/(x[2]-x[1])=(2-0)/(1-0)=2/1=2}}}

Substitute in {{{y-y[1]=m(x-x[1])}}}

{{{y-0=2(x-0)}}}

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  {{{m = (y[2]-y[1])/(x[2]-x[1])=(2-0)/(1-2)=2/(-1)=-2}}}

Substitute in {{{y-y[1]=m(x-x[1])}}}

{{{y-2=-2(x-0)}}}

{{{y-2=-2x}}}

y = -2x + 2

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

Edwin</pre>