Question 572756
<pre>
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 ({{{-3-2sqrt(5)}}}, 3) and ({{{-3+2sqrt(5)}}}, 3)
 
First we plot the vertices:
 
{{{drawing(400,400,-10,7,-7,10,
 
graph(400,400,-10,7,-7,10), circle(-4,2,.1), circle(1,2,.1),
circle(-7,2,.1),circle(4,2,.1)
 )}}}
 
We see that the hyperbola opens right and left, that is, 
it looks something like this: )(
 
So we know its standard equation is this:
 
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}
 
We connect the vertices to find the transverse axis:

{{{drawing(400,400,-10,7,-7,10,
 
graph(400,400,-10,7,-7,10), circle(-4,2,.1),circle(1,2,.1),
circle(-7,2,.1),circle(4,2,.1), green(line(-4,2,1,2))
 )}}}

 
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={{{5/2}}}
 
The center of the hyperbola is the midpoint of the transverse axis,
and we can see that the midpoint of the transverse axis is ({{{-3/2}}},2), so
we have (h,k) = ({{{-3/2}}},2).  So we plot the center:

{{{drawing(400,400,-10,7,-7,10,
 
graph(400,400,-10,7,-7,10), circle(-3/2,2,.1), circle(-4,2,.1),circle(1,2,.1),
circle(-7,2,.1),circle(4,2,.1), green(line(-4,2,1,2))
 )}}}

To find a, we subtract the x-coordinate of 
the center from the x-coordinate of the right vertex, and get
  
1 - ({{{-3/2}}}) = {{{2/2}}} + {{{3/2}}} = {{{5/2}}}

So a={{{5/2}}}. 

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 - ({{{-3/2)}}} = {{{8/2}}} + {{{3/2}}} = {{{11/2}}}
 
Next we find b from the Pythagorean relationship common to all
hyperbolas, which is
 
c² = a² + b²
 
Substituting for c and a
 
{{{(11/2)^2}}} = ({{{5/2}}})² + b²

{{{121/4}}} = {{{25/4}}} + b²

{{{121/4}}} - {{{25/4}}} = b²

{{{96/4}}} = b²

24 = b²

{{{sqrt(24)}}} = b

{{{2sqrt(6)}}} = b

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

{{{(x-h)^2/a^2-(y-k)^2/b^2}}} = 1

{{{(x-(-3/2))^2/(5/2)^2-(y-2)^2/24}}} = 1

{{{(x+3/2)^2/(25/4)-(y-2)^2/24}}} = 1


Next we draw in the conjugate axis which is 2b units
or {{{2sqrt(6)}}} or about 4.9 units long with the center as its midpoint.
That is, we draw a vertical line {{{2sqrt(6)}}}, about 2.45 units
upward and the same number of units downward from the center:

{{{drawing(400,400,-10,7,-7,10,
 
graph(400,400,-10,7,-7,10), circle(-4,2,.1),circle(1,2,.1),
circle(-7,2,.1),circle(4,2,.1), green(line(-4,2,1,2),
line(-3/2,2+2sqrt(6),-3/2,2-2sqrt(6))
)
 )}}}

 
Next we draw the defining 2a×2b rectangle which has the transverse axis 
and the conjugate axis as perpendicular bisectors of its sides:
 
{{{drawing(400,400,-10,7,-7,10,
 
graph(400,400,-10,7,-7,10), circle(-4,2,.1),circle(1,2,.1),
circle(-7,2,.1),circle(4,2,.1), green(line(-4,2,1,2),
line(-3/2,2+2sqrt(6),-3/2,2-2sqrt(6))
, rectangle(-4,2-2sqrt(6),1,2+2sqrt(6))) )}}}
 
Next we draw the extended diagonals of the defining rectangle:
 
{{{drawing(400,400,-10,7,-7,10,
blue(line(-12,-18.5751384,15,34.3332646),
line(-15,28.45448922,20,-40.13122358)), 
graph(400,400,-10,7,-7,10), circle(-4,2,.1),circle(1,2,.1),
circle(-7,2,.1),circle(4,2,.1), green(line(-4,2,1,2),
line(-3/2,2+2sqrt(6),-3/2,2-2sqrt(6))
, rectangle(-4,2-2sqrt(6),1,2+2sqrt(6))) )}}}
 
We can now sketch in the hyperbola:

{{{drawing(400,400,-10,7,-7,10,
blue(line(-12,-18.5751384,15,34.3332646),
line(-15,28.45448922,20,-40.13122358)), 
graph(400,400,-10,7,-7,10,2+sqrt(3.84(x+1.5)^2-24)),

graph(400,400,-10,7,-7,10,2-sqrt(3.84(x+1.5)^2-24)),






 circle(-4,2,.1),circle(1,2,.1),
circle(-7,2,.1),circle(4,2,.1), green(line(-4,2,1,2),
line(-3/2,2+2sqrt(6),-3/2,2-2sqrt(6))
, rectangle(-4,2-2sqrt(6),1,2+2sqrt(6))) )}}}
 
But we still have to find the equations of those two blue
asymptotes.

We know one point they go through, namely the center ({{{-3/2}}},2)

We need to know the slopes of the two asymptotes. They are 
{{{rise/run}}} = ±{{{b/a}}} = ±{{{2sqrt(6)/(5/2)}}} = ±{{{2sqrt(6)}}}·{{{2/5}}} = ±{{{4sqrt(6)/5}}}

Now we use the point-slope form.

y - y<sub>1</sub> = m(x - x<sub>1</sub>)
y - 2 = ±{{{4sqrt(6)/5}}}(x - ({{{-5/2}}}))
y - 2 = ±{{{4sqrt(6)/5}}}(x + {{{5/2}}})
    y = 2 ± {{{4sqrt(6)/5}}}(x + {{{5/2}}})

One asymptote has the equation with the positive slope,
and the other has the equation with the negative slope.
 
Edwin</pre>