Question 298194
Write the equation. 

Hyperbola with C(-2,0)  V(-2,3)  y=±.5(x+2)

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

<font size = 2>[Notice that I had to change your asymptotes' equations, because
the equations you gave (y=±.5x) for asymptotes were not possible 
with a hyperbola with center C(-2,0), for the two asymptotes
must intersect at the center of the hyperbola, so I changed
them to have the equations y=±.5(x+2) instead, for either you
copied something wrong, or there was a typo in the book you
got the problem from.  But rather than do nothing, I went
ahead and changed the asymptotes.]</font>  

First we plot what is given, the vertex, the center, and the two
asymptotes, whose equations are {{{y=.5(x+2)}}} and {{{y=-.5(x+2)}}} 
 
{{{drawing(400,400,-13,7,-7,13,
 
green(graph(400,400,-13,7,-7,13,-.5x-1)), green(graph(400,400,-13,7,-7,13,.5x+1)), 
line(-2+.1,3,-2-.1,3), line(-2,3+.1,-2,3-.1), line(-2+.1,3+.1,-2-.1,3-.1), line(-2+.1,3-.1,-2-.1,3+.1),
line(-2+.1,0,-2-.1,0), line(-2,0+.1,-2,0-.1), line(-2+.1,0+.1,-2-.1,0-.1), line(-2+.1,0-.1,-2-.1,0+.1) )}}}
 
We see that the hyperbola opens upward and downward.
 
So we know its standard equation is this:
 
{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}, where the center is C(h,k),
and so we know that the center is (-2,0), then h=-2, and k=0.
 
The center is halway between the vertices, so you are right
that the other vertex is (-2,-3). We plot it:

{{{drawing(400,400,-13,7,-7,13,
 
green(graph(400,400,-13,7,-7,13,-.5x-1)), green(graph(400,400,-13,7,-7,13,.5x+1)), 
line(-2+.1,3,-2-.1,3), line(-2,3+.1,-2,3-.1), line(-2+.1,3+.1,-2-.1,3-.1), line(-2+.1,3-.1,-2-.1,3+.1),
line(-2+.1,0,-2-.1,0), line(-2,0+.1,-2,0-.1), line(-2+.1,0+.1,-2-.1,0-.1), line(-2+.1,0-.1,-2-.1,0+.1),
line(-2+.1,-3,-2-.1,-3), line(-2,-3+.1,-2,-3-.1), line(-2+.1,-3+.1,-2-.1,-3-.1), line(-2+.1,-3-.1,-2-.1,-3+.1)


 )}}}


We connect the vertices to find the transverse axis

{{{drawing(400,400,-13,7,-7,13,
 
green(graph(400,400,-13,7,-7,13,-.5x-1)), green(graph(400,400,-13,7,-7,13,.5x+1)), 
line(-2+.1,3,-2-.1,3), line(-2,3+.1,-2,3-.1), line(-2+.1,3+.1,-2-.1,3-.1), line(-2+.1,3-.1,-2-.1,3+.1),
line(-2+.1,0,-2-.1,0), line(-2,0+.1,-2,0-.1), line(-2+.1,0+.1,-2-.1,0-.1), line(-2+.1,0-.1,-2-.1,0+.1),
line(-2+.1,-3,-2-.1,-3), line(-2,-3+.1,-2,-3-.1), line(-2+.1,-3+.1,-2-.1,-3-.1), line(-2+.1,-3-.1,-2-.1,-3+.1),
green(line(-2,3,-2,-3))
 )}}}

 
We can see that the transverse axis is 6 units long, and since the
transverse axis is {{{2a}}} units long, then {{{2a=6}}} and {{{a=3}}}
 
Next we draw the top and bottom of the defining rectangle, which
are the horizontal line segments between the two intercepts which
passes through the vertex:
 
{{{drawing(400,400,-13,7,-7,13,
 
green(graph(400,400,-13,7,-7,13,-.5x-1)), green(graph(400,400,-13,7,-7,13,.5x+1)), 
line(-2+.1,3,-2-.1,3), line(-2,3+.1,-2,3-.1), line(-2+.1,3+.1,-2-.1,3-.1), line(-2+.1,3-.1,-2-.1,3+.1),
line(-2+.1,0,-2-.1,0), line(-2,0+.1,-2,0-.1), line(-2+.1,0+.1,-2-.1,0-.1), line(-2+.1,0-.1,-2-.1,0+.1),
line(-2+.1,-3,-2-.1,-3), line(-2,-3+.1,-2,-3-.1), line(-2+.1,-3+.1,-2-.1,-3-.1), line(-2+.1,-3-.1,-2-.1,-3+.1),
green(line(-2,3,-2,-3)), green(line(4,3,-8,3)), green(line(4,-3,-8,-3))
 )}}}

It appears that the corners of the defining rectangle are at 
(-8,3), (4,3), (-8,-3), and (4,-3).  But we can't just look and see.

We know the y-coordinate of the corners of the defining rectangle
are 3 and -3.  So we substitute 3 and -3 for y into each of the asymptotes' 
equations:

{{{system(y=.5(x+2),
3=.5(x+2),
3/.5=x+2,
6=x+2,4=x)}}}{{{system(y=-.5(x+2),
3=-.5(x+2),
3/(-.5)=x+2,
-6=x+2,
-8=x)}}}{{{system(y=.5(x+2),
-3=-.5(x+2),
-3/.5=x+2,
-6=x+2,
-8=x)}}}{{{system(y=-.5(x+2),
-3=-.5(x+2),
(-3)/(-.5)=x+2,
6=x+2,
4=x)}}}

We finish drawing in the defining rectangle, by
drawing its two vertical sides:

{{{drawing(400,400,-13,7,-7,13,
 
green(graph(400,400,-13,7,-7,13,-.5x-1)), green(graph(400,400,-13,7,-7,13,.5x+1)), 
line(-2+.1,3,-2-.1,3), line(-2,3+.1,-2,3-.1), line(-2+.1,3+.1,-2-.1,3-.1), line(-2+.1,3-.1,-2-.1,3+.1),
line(-2+.1,0,-2-.1,0), line(-2,0+.1,-2,0-.1), line(-2+.1,0+.1,-2-.1,0-.1), line(-2+.1,0-.1,-2-.1,0+.1),
line(-2+.1,-3,-2-.1,-3), line(-2,-3+.1,-2,-3-.1), line(-2+.1,-3+.1,-2-.1,-3-.1), line(-2+.1,-3-.1,-2-.1,-3+.1),
green(line(-2,3,-2,-3)), green(line(4,3,-8,3)), green(line(4,-3,-8,-3)),
green(line(-8,-3,-8,3)), green(line(4,-3,4,3))

 )}}}

Now we can sketch in the hyperbola:
{{{drawing(400,400,-13,7,-7,13,
green(graph(400,400,-13,7,-7,13,-.5x-1)), green(graph(400,400,-13,7,-7,13,.5x+1)), 
line(-2+.1,3,-2-.1,3), line(-2,3+.1,-2,3-.1), line(-2+.1,3+.1,-2-.1,3-.1), line(-2+.1,3-.1,-2-.1,3+.1),
line(-2+.1,0,-2-.1,0), line(-2,0+.1,-2,0-.1), line(-2+.1,0+.1,-2-.1,0-.1), line(-2+.1,0-.1,-2-.1,0+.1),
line(-2+.1,-3,-2-.1,-3), line(-2,-3+.1,-2,-3-.1), line(-2+.1,-3+.1,-2-.1,-3-.1), line(-2+.1,-3-.1,-2-.1,-3+.1),
green(line(-2,3,-2,-3)), green(line(4,3,-8,3)), green(line(4,-3,-8,-3)),
green(line(-8,-3,-8,3)), green(line(4,-3,4,3)),
green(graph(400,400,-13,7,-7,13,sqrt(9+(x+2)^2/4))),
green(graph(400,400,-13,7,-7,13,-sqrt(9+(x+2)^2/4)))  

 )}}}



The conjugate axis is along the x-axis, and extends from (-8,0) to (4,0)
Tha conjugate axis has length 12, and since b = one-half of the
conjugate axis's length, b = 6.

Therefore the equation of the hyperbola is

{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}, where h=-2, and k=0, a=3, b=6

{{{(y-0)^2/3^2-(x-(-2))^2/6^2=1}}}, or

{{{y^2/9-(x+2)^2/36=1}}}.

Edwin</pre>