It is in the standard form



We can tell that the hyperbola opens upward and downward because
the term in y comes first in the standard equation.

where the center is  = 
The length of the semi-transverse axis is  =  = 7

The vertices are 7 units above and below the center (9,-3), so
we add to and subtract 7 from the y-ccordinate of the center. So the
vertices are 

(9,-3+7) and (9,-3-7) or simplifying:

(9,4) and (9,-10)

-------------------------

But since you could have been asked more than that I'll go into
some more detail that may help you in other hyperbola problems. 

The length of the semi-conjugate axis is  =  = 8
 

First let's plot the center:

 

Now let's draw the transverse and conjugate axes by drawing
1.  a vertical line a=7 units above the center.
2.  a vertical line a=7 units below the center.
3.  a horizontal line b=8 units left of the center.
3.  a horizontal line b=8 units right of the center.
  


 

The green vertical line is the complete transverse axis which goes from
vertex to vertex. The horizontal line is the complete conjugate axis.

Next we draw the defining rectangle which has the transverse and conjugate
axes bisecting the sides of the rectangle:

 

Next we draw the two asymptotes, by drawing and extending the
diagonals of the defining rectangle:



Now we can sketch in the hyperbola:



You could have been asked to graph the hyperbola.  You could
also have been asked to find the equations of the asymptotes, which
you could do because you have points that they pass through. You
could also be asked to find the foci.  Then you'd have to find c
from the equation c² = a² + b².

Edwin