Question 196080
<pre><font size = 4 color = "indigo"><b>
{{{((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 vertically 
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 vertically 
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 conjugate 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),rectangle(-7,-1,3,15)
 )}}}

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 ) )

 )}}}

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>