Section 6-4: Hyperbolas
Definition of the hyperbola
A hyperbola is the set
of all points P(x, y) in the plane such that
| PF1 - PF2
| = 2a
Again F1
and F2 are focus points.
This time the difference of these distances remain a constant at 2a.
The explanation is similar to that of the ellipse. Since the ellipse
is the sum of the distances and the hyperbola is the difference of the
distances, the equations are very similar. They differ only in the
sign and the longest side for a hyperbola is c. (Remember for the
ellipse it was a)
Drawing a Hyperbola (Manipula Math)
The equation of the above
hyperbola would have the form:
The hyperbola opens left
and right. Notice it comes in two parts. Different than an
ellipse which is a closed figure. Hyperbolas can also open up and
down. I am sure you can guess at the equation!!
This hyperbola has the
form:
To get the correct shape
of the hyperbola, we need to find the asymptotes
of the hyperbola. The asymptotes are lines that are approached but
not touched or crossed. These asymptotes are boundaries of the hyperbola.
This is one difference between a hyperbola and a parabola. For the
hyperbolas that open right/left, the asymptotes are:
and for hyperbolas opening
up/down, the asymptotes are:
To form the asymptotes
easily on the graph, all we need do is form a rectangle using a and b.
Sample Problems
1) Graph the hyperbola
x2/16 - y2/4
= 1 Find the vertices, foci and equations of the asymptotic lines.
Solution: This hyperbola opens right/left
because it is in the form x - y. a2
= 16, b2 = 4, c2
= 16 + 4 = 20. Therefore, a = 4, b = 2 and c = 4.5
Vertices: (4, 0) and (-4, 0)
Foci: (4.5, 0) and (-4.5, 0)
Equations of asymptotic lines: y = .5x and y = - .5x
To graph the hyperbola, go 2 units up/down from center point and 4 units
left/right from center point.
2) Graph the hyperbola
y2/25 - x2/9
= 1. Give the vertices, foci and equations of asymptotic lines.
Solution: This hyperbola opens up and
down because it is in the form y - x. a2
= 25, b2 = 9 and c2
= 25 + 9 = 34. Thus, a = 5, b = 3, c = 5.8
Vertices: (0, 5) and (0, -5)
Foci: (0, 5.8) and (0, -5.8)
Equations of asymptotic lines: y = (5/3)x and y = (-5/3)x
The box is formed by going 5 units up/down from center and 3 units left/right
from center.
3) Find an equation
of a hyperbola with center at the origin, one vertex at (7, 0) and a focus
at (12, 0).
Solution: The vertex and focus are on
the x-axis, so the hyperbola opens right/left. a = 7, c = 12.
That makes a2
= 49, c2 = 144 and
b2
= 144 - 49 = 95. Therefore the equation is:
x2/49
- y2/95 = 1
Translations of Hyperbolas
If the hyperbola opens
right/left the translation is:
with the equations of
the asymptotic lines as:
y - k = + (b/a)(x
- h)
If the hyperbola opens
up/down the translation is:
with the equations of
the asymptotic lines as:
y - k = + (a/b)(x
- h)
Sample Problems
1) Graph the equation: 
Find the center, vertices, foci and the equations of the asymptotic lines.
Solution: Since it is y - x it opens
up/down. a2
= 36, b2 = 25 and
c2
= 36 + 25 = 61. Thus, a = 6, b = 5 and c = 7.8
Center: (1, 2)
Vertices: (1, 8) and (1, -4) ( six units up and down from center)
Foci: (1, 9.8) and (1, -5.8) ( 7.8 units up/down from center)
Equations of asymptotic lines: y - 2 = (+6/5)(x - 1)
The box is formed by going 6 units up/down and 5 units right/left from
center.
2) Find the equation
of a hyperbola with center (1, 1), vertex (3, 1) and focus at (5, 1).
Solution: The vertex and foci are on
the same horizontal line. This makes the hyperbola open right/left.
a = 2 (distance from vertex to center), c = 4 (distance from focus to center).
Thus a2
= 4, c2 = 16 and
b2
= 16 - 4 = 12.
The equation is: 
Now for parabolas!! 
Let's back up and regroup! 