CONIC SECTIONS: THE HYPERBOLA
Example 1:
Find the coordinates of the center, vertices, and foci of the hyperbola given by
. The equation is already in standard form
where h = 2, k = 5, a = 4 (from denominator of positive term!), and b = 5
Now, we can find the requested information as follows:
Coordinates of the Center (h, k)
(2, 5)
Coordinates of the Vertices (h + a, k) and (h
a, k)
(2 + 4, 5) and (2
4, 5)
then (6, 5) and (
2, 5)
Coordinates of the Foci (h + c, k) and (h
c, k), where
Since a = 4 and b = 5, then
.
and
Example 2:
Find the coordinates of the center, vertices, and foci of the hyperbola given by
.
First, we will convert the equation of the hyperbola to standard form. Since there is a positive y2-term raised to the second power we know that this hyperbola is of the form
.
NOTE: The center (h, k) of this hyperbola is at (0, 0) because there is only a y2-term and an x2-term. Therefore, we can write
where h = 0, k = 0, a = 5 (from denominator of positive term!), and b = 4
Now, we can find the requested information as follows:
Coordinates of the Center (h, k)
(0, 0)
Coordinates of the Vertices (h, k + a) and (h, k
a)
(0, 0 + 5) and (0, 0
5)
then (0, 5) and (0,
5)
Coordinates of the Foci (h, k + c) and (h, k
c), where
Since a = 5 and b = 4, then
.
and
then
and
Example 3:
Find the coordinates of the center, vertices, foci, and the equations of the asymptotes of the hyperbola given by
.
First, we will convert the equation of the hyperbola to standard form.
It is either
or
, where a = b or a
b
Let's divide both sides of the equation by
12 to get
and
or
We can now write the standard form of the given hyperbolic equation as follows:
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where h =
1, k = 0, a = 2 (from denominator of positive term!), and b =
Now, we can find the requested information as follows:
Coordinates of the Center (h, k)
(
1, 0)
Coordinates of the Vertices (h, k + a) and (h, k
a)
(
1, 0 + 2) and (
1, 0
2)
(
1, 2) and (
1,
2)
Coordinates of the Foci (h, k + c) and (h, k
c), where
Since a = 2 and b =
, then
.
and
and
Example 4:
Find the coordinates of the center, vertices, and foci of the hyperbola given by
.
First, we will convert the equation of the hyperbola to standard form. Since there is a positive x2-term raised to the second power we know that this hyperbola is of the form
.
The given equation is almost in standard form. All we have to do is give the x2-term a denominator of 1 as follows:
where h = 1, k = 2, a = 1 (from denominator of positive term!), and b = 3
Now, we can find the requested information as follows:
Coordinates of the Center (h, k)
(1, 2)
Coordinates of the Vertices (h + a, k) and (h
a, k)
(1 + 1, 2) and (1
1, 2)
then (2, 2) and (0, 2)
Coordinates of the Foci (h + c, k) and (h
c, k), where
Since a = 1 and b = 3, then
.
(1 + 2, 2) and (1
2, 2)
then (3, 2) and (
1, 2)