SOLUTION: Give the equation in standard form of the hyperbola with vertices (-4,2) and (1,2) and foci (-7,2) and (4,2). Give the center and the asymptotes.

Give the equation in standard form of the hyperbola with vertices (-4,2)
and (1,2) and foci (-7,2) and (4,2). Give the center and the asymptotes. 
 
In a message dated 6/24/2011 2:53:25 P.M. Eastern Daylight Time, AnlytcPhil@aol.com writes:
what is the equation of a hyperbola with vertices (-5,3) (-1,3) and foci (, 3) and (, 3)
 
First we plot the vertices:
 

 
We see that the hyperbola opens right and left, that is, 
it looks something like this: )(
 
So we know its standard equation is this:
 

 
We connect the vertices to find the transverse axis:



 
We can see that the transverse axis is 5 units long, and since the
transverse axis is 2a units long, then 2a=5 and a=
 
The center of the hyperbola is the midpoint of the transverse axis,
and we can see that the midpoint of the transverse axis is (,2), so
we have (h,k) = (,2).  So we plot the center:



To find a, we subtract the x-coordinate of 
the center from the x-coordinate of the right vertex, and get
  
1 - () =  +  = 

So a=. 

We are given the foci (-7,2) and (4,2)
 
The number of units from each of the foci to the center is the value c.
 
To find that distance c, we subtract the x-coordinate of the center 
from the x-coordinate of the right focus, and get
 
c = 4 - ( =  +  = 
 
Next we find b from the Pythagorean relationship common to all
hyperbolas, which is
 
c² = a² + b²
 
Substituting for c and a
 
 = ()² + b²

 =  + b²

 -  = b²

 = b²

24 = b²

 = b

 = b

Now we can give the standard equation of the hyperbola, since we now
know h, k, a, and b, a² and b²:

 = 1

 = 1

 = 1


Next we draw in the conjugate axis which is 2b units
or  or about 4.9 units long with the center as its midpoint.
That is, we draw a vertical line , about 2.45 units
upward and the same number of units downward from the center:



 
Next we draw the defining 2a×2b rectangle which has the transverse axis 
and the conjugate axis as perpendicular bisectors of its sides:
 

 
Next we draw the extended diagonals of the defining rectangle:
 

 
We can now sketch in the hyperbola:


 
But we still have to find the equations of those two blue
asymptotes.

We know one point they go through, namely the center (,2)

We need to know the slopes of the two asymptotes. They are 
 = ± = ± = ±· = ±

Now we use the point-slope form.

y - y1 = m(x - x1)
y - 2 = ±(x - ())
y - 2 = ±(x + )
    y = 2 ± (x + )

One asymptote has the equation with the positive slope,
and the other has the equation with the negative slope.
 
Edwin