Divide terms into x and y variables: 
(x^{2} − 4x ) − (16y^{2} + 32y ) − 28 = 0 
By applying the method of completing the square formula (x + a)^{2} = x^{2} + 2ax + a^{2} we get: 

(x − 2)^{2} − 4 − 16(y + 1)^{2} + 16 − 28 = 0 

(x − 2)^{2} − 16(y + 1)^{2} = 16 
After dividing by 16 we get: 

The center of this hyperbola is at (2 , − 1) h = −2 and k = 1.
The transvers axis half length (a) is equal to 4.
and the conjugate axis half length (b) is equal to 1.
Because a > b this hyperbola is horizontal so the transverse axis is along the x axis.
The foci distance is calculated from the equation: 

In order to find the coordinates of the foci we will take the center of the hyperbola at (2 , −1) and add and subtract the value of c in the x direction.
(2 + c , −1) (2 − c , −1) = (2 + 4.12 , −1) (2 − 4.12 , −1) = (6.12 , −1) (−2.12 , −1)
To find the coordinate of the vertices we perform the same process as for the foci but with the value of a.
(2 + a , −1) (2 − a , −1) = (2 + 4 , −1) (2 − 4 , −1) = (6 , −1) (−2 , −1)
Repeat the same method as before but with + sign instead of minus x^{2} − 16y^{2} − 4x − 32y + 28 = 0
(x^{2} − 4x ) − (16y^{2} + 32y ) + 28 = 0 
(x − 2)^{2} − 4 − 16(y + 1)^{2} + 16 + 28 = 0 
(x − 2)^{2} − 16(y + 1)^{2} = −40 
Divide by −40 and again by 16 we get: 

and finally, 

We can see that changing the sign of the last term changed the value of the free term to negative and hence the hyperbola changed to vertical also the values of a and b had been changed.
