# uc3rmw_0804e08

0 (0 Likes / 0 Dislikes)

Hi there.
In this example, we want to find
the equation of a hyperbola centered
at h comma k, given information about
its graphs that include its asymptotes.
The problem states, find an equation for
the hyperbola whose graphs are shown.
In part A, we have a hyperbola with branches
opening to the left and to the right.
The center of the hyperbola is at 0 comma 1.
It's got a vertex here at 2 comma
1, and a focus at 4 comma 1.
In part B, we have a hyperbola with
branches opening upward and downward.
There are two asymptotes shown here.
One is given by the equation
y equals 3/2 x minus 5/2.
And the other is given by the equation
y equals minus 3/2 x plus 1/2.
The vertex is at 1 comma 2, and the center
of this hyperbola is at 1 comma minus 1.
First let's recall the
equations for the asymptotes
of a hyperbola centered at h comma k.
The main properties of hyperbolas
centered at h comma k are shown here.
We have the standard equation
given by x minus h squared divided
by a squared minus y minus k squared
divided by b squared equals 1, where a
is greater than 0, and b is greater than 0.
This is the standard equation for
a hyperbola with transverse axis
along the line y equals k.
The length of the transverse axis is 2a.
The conjugate axis is
along the line x equals h.
And the length of the conjugate axis is 2b.
The center is h comma k and therefore,
the vertices have coordinates
h minus a comma k and h plus a comma k.
And the endpoints of the
conjugate axis are given
by h common k minus b and h comma k plus b.
The foci are h minus c comma
k and h plus c comma k.
We have the equation involving
a, b, and c given by c squared
equals a squared plus b squared.
And the asymptotes have the form y minus
k equals plus or minus b over a times,
in parentheses, x minus h.
Now the standard form for the equation of
a hyperbola whose transverse axis is along
the line x equals h is given
by y minus k squared divided
by a squared minus x minus h squared
divided by b squared equals 1.
The length of the transverse axis
is the same for this equation where
the transverse axis is
along the line y equals k.
It's 2a.
The conjugate axis in
this case is y equals k.
And the length of the conjugate axis is 2b.
The center is at h comma k, and the
vertices are, therefore, h comma k minus a,
and h comma k plus a.
The endpoints of the conjugate axis are
h minus b comma k, and h plus b comma k.
And foci are at h comma k
minus c, and h comma k plus c.
We have the equation involving
a, b, and c as c squared
equals a squared plus b squared,
just as it is for the other equation.
And the asymptotes, in this case,
are y minus k equals plus or minus
a over b times, in parentheses, x minus h.
In part A, since the foci are on the x-axis,
the transverse axis is on the x-axis.
From the graph, we see that the
center is at hk equals 0 comma 1.
So h is equal to 0, and k is equal to 1.
The distance a between the center at 0
comma 1 and the vertex 2 comma 1 is 2.
Therefore, a is equal 2.
The distance c is between the center
0 comma 1 and the focus at 4 comma 1.
And so c is equal to 4.
The transverse axis, we see lies on the
horizontal line y equals 1, as shown here.
So we have the value for c.
We've determine the value for a.
We now want to find the value for b.
We can use the relationship
between a, b, and c.
That is the equation c squared
equals a squared plus b
squared to solve for b squared.
We plug in the fact that c equals 4.
So we have 4 squared.
And that's equal to, plugging in that a
equals 2, we have 2 squared plus b squared.
We now want to solve this
equation for b squared.
We have that b squared equals 4
squared minus 2 squared, which
simplifies to 16 minus 4, or 12.
So b squared equals 12.
Now that we have a and b, we can
plug that into the standard form
for the hyperbola centered at 0 comma
1 to find the equation of the hyperbola
with the specific information given.
We have that the standard
form for the equation
of a hyperbola with transverse
axis on the horizontal line y
equals 1 is given by x minus
h squared over a squared minus
y minus k squared over b squared equals 1.
We know the center h comma k, we know
what a is, and we know what b is.
So now we can plug in all that
information to find the equation
specific to this graph.
We have x minus h, where h is
0, squared over a squared--
we know that a is 2, so
that's going to be 2 squared--
minus y minus k--
k is 1-- squared over b squared.
We determine that b squared is 12.
And that equals 1.
We can simplify this a bit.
We have x squared over 4 minus, in
parentheses, y minus 1 squared over 12
equals 1.
So here's the equation of the hyperbola
given the information in the graph.
In part B, the center h comma
k is equal to 1 comma minus 1.
So we have that h is equal to
1, and k is equal to minus 1.
The distance a between the center at 1
comma minus 1 and the vertex 1 comma 2,
as shown in the graph, is 3 units.
And so a is equal to 3.
We see that the slope of the asymptote
with the positive slope given by this line
is 3/2.
The transverse axis is on the vertical
line x equal 1, as shown here.
The form for asymptotes for
hyperbolas with transverse axis
on the vertical line x equal 1 are given
by y equals plus or minus a over b times x.
We just found that the slope of our
asymptotes are plus or minus 3/2.
And so we can relate our asymptotes
based on the information given
in the graph to the standard
form to find a and b.
We have that the slopes of our
asymptotes are plus or minus 3/2.
And therefore, a over b is equal to 3/2.
Now we can relate b and a.
We have that b is equal to 2/3 a.
Now that we have b in terms of a and we
know that a is equal to 3, we can find b.
b is equal to 2/3 times a where a is 3.
So we have 2/3 times 3, or 2.
So we have a equal to 3 and b equal to 2.
And we can plug that into the standard
form for the equation of a hyperbola
with transverse axis
along the line x equal 1.
That standard form is y minus k
squared over a squared minus x
minus h squared over b squared equal 1.
Again, the center is at 1 comma minus 1.
And we just found a and b-- a
is equal to 3; b is equal to 2.
We're going to plug in all that
information to find the equation
specific to the hyperbola graphed here.
We have the y minus k, which is equal to
y minus minus 1 squared over a squared--
a is equal to 3, so that's
going to be 3 squared--
minus x minus h, where h is equal
to 1, squared over b squared.
We found that b is equal to 2.
So that's going to be 2 squared.
And that's all equal to 1.
We'll go ahead and simplify that equation.
We have an equation that is y plus 1 squared
over 9 minus x minus 1 squared over 4
equal to 1.
So there's the form of the equation
for this hyperbola graphed here.