Hyperbolas

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: y = (b/a) and y = - (b/a) and for hyperbolas opening up/down, the asymptotes are: y = (a/b) and y = - (a/b)

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: (x-h)²/a² - (y-k)²/b² = 1
with the equations of the asymptotic lines as:
y - k = + (b/a)(x - h)


If the hyperbola opens up/down the translation is: (y-k)²/a² - (x-h)²/b² = 1
with the equations of the asymptotic lines as:
y - k = + (a/b)(x - h)

Sample Problems

1) Graph the equation: (y-2)²/36 - (x-1)²/25 = 1. 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: (x-1)²/4 - (y-1)²/12 = 1