I'll do this one instead.  It's done exactly the same as yours step by step:



Rearrange to get the terms in x together and the terms in y together.



Factor 25 out of the 1st and 2nd terms on the left, and factor -16 out of the
3rd and 4th terms on the left:



Skip a big space at the end of each parentheses because we're going to insert
a couple of terms in each one to complete the square.



In the first parentheses:
Multiply the coefficient of x, which is -8, by 1/2 getting -4.
Square -4, getting (-4)2 or +16, so we add and subtract 16 in the
first blank, that is, put +16-16 in the first blank:



In the second parentheses:
Multiply the coefficient of y, which is -6, by 1/2 getting -3.
Square -3, getting (-3)2 or +9, so we add and subtract 9 in the
second blank, that is, put +9-9 in the second blank:



Factor the first three terms in each parentheses:
            
                
 


Distribute the 25 and the -16, being careful to leave the squared parentheses intact:







Divide every term by 400 to get 1 on the right side:



cancel:



Compare to



Where the center is (h,k) = (4,3),
semi-transverse axis length = a = 4, 
semi-conjugate axis length = b = 5

Plot the center (h,k) = (4,3).
Draw the transverse axis horizontally from the center a=4 units 
right and a=4 units left.

Draw the conjugate axis vertically from the center b=5 units 
upward and b=5 units downward.



The endpoints of the transverse axes are the two vertices.
From the graph, we see they are (0,3), and (8,3).

The endpoints of the conjugate axes are the two co-vertices.
From the graph, we see they are (4,-2), and (4,8).

Next, we draw the defining rectangle around the transverse
and conjugate axes like this.  And we draw and extend the
two diagonals of the defining rectangle:



Now we can sketch in the hyperbola approching the two ansymptotes:



Edwin

I forgot to do the foci above.



The focal distance is c, which is found from the Pythagorean
relation 




  <---(I think on your problem this comes out to be an integer.)

So the foci are  or about 6.4 units right and left of the
center. so we get two points right and left of the center about 6.4 units
from the center, the foci are approximately (4+6.4,3) and (4-6.4,3), which
are approximately the points (10.4,3) and (-2.4,3), as shown on the graph
below:





Edwin