Question 1033120
<pre><font size = 4 color = "indigo"><b>

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

{{{((x+2)^2)/25 - ((y-7)^2)/64 = 1}}}

{{{((x-h)^2)/a^2 - ((y-k)^2)/b^2 = 1}}}
 
{{{h = -2}}}, {{{k=7}}}, 
 
{{{a^2=25}}}, so {{{a=5}}}
 
{{{b^2=64}}}, so {{{b=8}}}
 
The center (h,k) = (-2,7)
 
We start out plotting the center C(h,k) = C(-2,7)
 
{{{drawing(400,600,-12,8,-10,20,
line(-2.1,7,-1.9,7), line(-2,6.9,-2,7.1), locate(-4,8,"C(-2,7)"),
graph(400,600,-12,8,-10,20) )}}}
 
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):
 
{{{drawing(400,600,-12,8,-10,20,
graph(400,600,-12,8,-10,20),
line(-2.1,7,-1.9,7), line(-2,6.9,-2,7.1), locate(-4,8,"C(-2,7)"),
line(-7,7,-2,7), locate(-10.5,8,"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):
 
{{{drawing(400,600,-12,8,-10,20,
 
line(-2.1,7,-1.9,7), line(-2,6.9,-2,7.1), locate(-4,8,"C(-2,7)"),
line(-2,7,3,7), locate(3.5,8,"V2(3,7)"),
line(-7,7,-2,7), locate(-10.5,8,"V1(-7,7)"),
graph(400,600,-12,8,-10,20)
 
 )}}}
 
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).
 
{{{drawing(400,600,-12,8,-10,20,
 
line(-2.1,7,-1.9,7), line(-2,6.9,-2,7.1), locate(-4,8,"C(-2,7)"),
line(-2,7,3,7), locate(3.5,8,"V2(3,7)"),
line(-7,7,-2,7), locate(-10.5,8,"V1(-7,7)"),
line(-2,7,-2,15), locate(-4,16,"(-2,15)"),
 
graph(400,600,-12,8,-10,20) )}}}
 
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). 
 
{{{drawing(400,600,-12,8,-10,20,
locate(-4,16,"(-2,15)"),
line(-2.1,7,-1.9,7), line(-2,6.9,-2,7.1), locate(-4,8,"C(-2,7)"),
line(-2,7,3,7), locate(3.5,8,"V2(3,7)"),
line(-7,7,-2,7), locate(-10.5,8,"V1(-7,7)"),
line(-2,-1,-2,7), line(-2,7,-2,15), locate(-4,-1,"(-2,-1)"),
graph(400,600,-12,8,-10,20) )
 
 )}}}
 
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:
 
{{{drawing(400,600,-12,8,-10,20,
locate(-4,16,"(-2,15)"),
line(-2.1,7,-1.9,7), line(-2,6.9,-2,7.1), locate(-4,8,"C(-2,7)"),
line(-2,7,3,7), locate(3.5,8,"V2(3,7)"),
line(-7,7,-2,7), locate(-10.5,8,"V1(-7,7)"),
line(-2,-1,-2,7), line(-2,7,-2,15), locate(-4,-1,"(-2,-1)"),
graph(400,600,-12,8,-10,20),line(-7,-1,3,-1), line(3,-1,3,15),
line(3,15,-7,15),line(-7,15,-7,-1)
 )}}}
 
Next we draw and extend the two diagonals of this defining
rectangle:
 
{{{drawing(400,600,-12,8,-10,20,locate(-4,-1,"(-2,-1)"),
locate(-4,16,"(-2,15)"),
line(-2.1,7,-1.9,7), line(-2,6.9,-2,7.1), locate(-4,8,"C(-2,7)"),
line(-2,7,3,7), locate(3.5,8,"V2(3,7)"),
line(-7,7,-2,7), locate(-10.5,8,"V1(-7,7)"),
line(-2,-1,-2,7), line(-2,7,-2,15),
rectangle(-7,-1,3,15),
line(8,23,-17,-17),
line(13,-17,-17,31),
graph(400,600,-12,8,-10,20 )
 
 )}}}
 
Now we can sketch in the hyperbola:
 
{{{drawing(400,600,-12,8,-10,20,locate(-4,-1,"(-2,-1)"),
locate(-4,16,"(-2,15)"),
 
line(-2.1,7,-1.9,7), line(-2,6.9,-2,7.1), locate(-4,8,"C(-2,7)"),
line(-2,7,3,7), locate(3.5,8,"V2(3,7)"),
line(-7,7,-2,7), locate(-10.5,8,"V1(-7,7)"),
line(-2,-1,-2,7), line(-2,7,-2,15),
rectangle(-7,-1,3,15),
line(8,23,-17,-17),
line(13,-17,-17,31),
graph(400,600,-12,8,-10,20, 7-sqrt(  (64(x+2)^2-1600)/25 ) ),
graph(400,600,-12,8,-10,20, 7+sqrt(  (64(x+2)^2-1600)/25 ) )
 
 )}}}

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

{{{c^2=a^2+b^2}}}  (just like the Pythagorean theorem, from whence
it comes):

{{{c^2=a^2+b^2}}}
{{{c^2=5^2+8^2}}}
{{{c^2=25+64}}}
{{{c^2=89}}}
{{{c=sqrt(89)}}}

So the two foci are {{{sqrt(89)}}} units 
right and left of the center, which is
(h,k) = (-2,7)

Therefore the foci are:

({{{-2-sqrt(89)}}},7) and ({{{-2+sqrt(89)}}},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 ±{{{b/a}}} or ±{{{8/5}}}
 
The asymptote that has slope {{{8/5}}} goes through the center
C(-2,7), so its equation is found using the point-slope
formula:
 
{{{y-y[1]=m(x-x[1])}}}
{{{y-(7)=(8/5)(x-(-2))}}}
{{{y-7=(8/5)(x+2)}}}
Multiply through by 5
{{{5y-35=8(x+2)}}}
{{{5y-35=8x+16}}}
{{{-8x+5y=51}}}
 
The asymptote that has slope {{{-8/5}}} goes through the center
C(-2,7), so its equation is found using the point-slope
formula:
 
{{{y-y[1]=m(x-x[1])}}}
{{{y-(7)=(-8/5)(x-(-2))}}}
{{{y-7=(-8/5)(x+2)}}}
Multiply through by 5
{{{5y-35=-8(x+2)}}}
{{{5y-35=-8x-16}}}
{{{8x+5y=19}}}
 -----------------------------------------------
Edwin</pre>