4x²-y²-8x+20 = 0         |        x²-4y²-4x+24y-36 = 0

Rearrange the terms in this order x²,x,y²,y, and constants
on the right:

   4x²-8x-y² = -20                      x²-4x-4y²+24y = 36 

 4(x²-2x)-y² = -20                   (x²-4x)-4(y²-6y) = 36

Complete the square in the parentheses. In the equation on
the left since there is no y-term write y as (y-0)

4(x²-2x+1)-(y-0)² = -20+4         (x²-4x+4)-4(y²-6y+9) = 36+4=36

    4(x-1)²-(y-0)² = -16                (x-2)²-4(y-3)² = 4

Get 1 on the right by dividing through by the constant term:

                        

               

               

Switch the terms on the left so that the positive term comes first:

               

The equation on the left has its y-term first and therefore represents a
hyperbola that looks like this  whereas the equation on the right 
has its x-term first and therefore it represents a hyperbola that looks
like this )(

               

  

For the hyperbola on the left:

center=(h,k)=(1,0)
vertices=(h,k±a)=(1,4)&(1,-4)
focus=(h,k±c)
c²=a²+b²
c²=16+4
c²=20
c=√20
c=√4*5
c=2√5
focus=(h,k±c)=(1,±2√5)

asymptotes are the line through center (1,0)
with slope  =  = ±2
Use point slope form.
They are y = 2x-2 and y = -2x+2 

For the hyperbola on the right:

center=(h,k)=(2,3)
vertices=(h±a,k)=(4,3)&(0,3)
focus=(h±c,k)
c²=a²+b²
c²=4+1
c²=5
c=2√5
focus=(h±c,k)=(2±√5,3)

asymptotes are the line through center (2,3)
with slope  = 
Use point slope form.
They are y = x+2 and y = x+4 


Edwin