SOLUTION: Graph the hyperbola (x+3)^2-9(y-4)^2=9 Endpoints of traverse axis? Endpoints of conjugate axis? I know how to find the center, foci and asymptotes but am confused on how to

Instead of doing yours for you, I'll do one
exactly like it, which you can use as a model:




 
, , 
 
, so 
 
, so 
 
The center (h,k) = (-2,7)
 
We start out plotting the center C(h,k) = C(-2,7)
 

 
Next we draw the left semi-transverse axis,
which is a segment a=5 units long horizontally 
left from the center.  This semi-transverse
axis ends up at one of the two vertices (-7,7).
We'll call it V1(-7,7):
 

 
Next we draw the right semi-transverse axis,
which is a segment a=5 units long horizontally 
right from the center. This other semi-transverse
axis ends up at the other vertex (3,7).
We'll call it V2(3,7):
 

 
That's the whole transverse ("trans"="across",
"verse"="vertices", the line going across from
one vertex to the other. It is 2a in length,
so the length of the transverse axis is 2a=2(5)=10
 
Next we draw the upper semi-conjugate axis,
which is a segment b=8 units long verically 
upward from the center.  This semi-conjugate
axis ends up at (-2,15).
 

 
Next we draw the lower semi-conjugate axis,
which is a segment b=8 units long verically 
downward from the center.  This semi-conjugate
axis ends up at (-2,-1). 
 

 
That's the complete conjugate axis. It is 2b in length,
so the length of the transverse axis is 2b=2(8)=16
 
Next we draw the defining rectangle which has the
ends of the transverse and conjugate axes as midpoints
of its sides:
 

 
Next we draw and extend the two diagonals of this defining
rectangle:
 

 
Now we can sketch in the hyperbola:
 


The foci are points inside the hyperbola, which are the distance c
from the center, where c is calculated by

  (just like the Pythagorean theorem, from whence
it comes):







So the two foci are  units 
right and left of the center, which is
(h,k) = (-2,7)

Therefore the foci are:

(,7) and (,7) 

They are approximately the points:

(-11.4,7) and (7.4,7).  I won't bother plotting
them.

All that's left to do is find the equations of the two asymptotes.
Their slopes are ± or ±
 
The asymptote that has slope  goes through the center
C(-2,7), so its equation is found using the point-slope
formula:
 



Multiply through by 5



 
The asymptote that has slope  goes through the center
C(-2,7), so its equation is found using the point-slope
formula:
 



Multiply through by 5



 -----------------------------------------------
Edwin