9.5 Graph and Write Equations of Hyperbolas

Hyperbola

The set of all points P such that the difference of the distances between P and two fixed points, called the foci, is a constant

Foci

Two fixed points in a hyperbola

Vertices

The points of intersection of a hyperbola and the line through the foci

Transverse Axis

The line segment that connects the vertices of a hyperbola

Center

The midpoint of the transverse axis

STANDARD EQUATION OF A HYPERBOLA WITH CENTER AT THE ORIGIN

Equation

Transverse Axis

Asymptotes

Vertices

Horizontal

( ±_a_, 0)

Vertical

b

=

±

y

a

x

(0, ±_a_)

The foci lie on the transverse axis, c units from the center, where c² = _a² + b².

Example 1

Graph an equation of a hyperbola

Graph 36y² - 9x² = 324. Identify the vertices, foci, and asymptotes of the hyperbola.

Solution

1. Rewrite the equation in standard form.

2. Identify the vertices, foci, and asymptotes. Note that a² = _9_ and b² = _36_ , so a = _3_ and b = _6_ . The y²-term is _positive_ , so the transverse axis is _vertical_ and the vertices are (0, ± _3_). Find the foci.

c² = a² + b² = _3² + 6²_ = _45_ , so c = The foci are at (0, ± ) » (0, ±6.7).

The asymptotes are y = ± x, or y =

3. Draw the hyperbola. Draw a rectangle centered at the origin that is 2a = _6_ units high and 2b = _12_ units wide. The asymptotes pass through opposite corners of the rectangle. Then, draw the hyperbola passing through the vertices and approaching the asymptotes.

Example 2

Write an equation of a hyperbola

Write an equation of the hyperbola with foci at (-5, 0) and (5, 0) and vertices at (-4, 0) and (4, 0).

The foci and vertices lie on the _x_-axis equidistant from the origin, so the transverse axis is _horizontal_ and the center is the origin. The foci are each _5_ units from the center, so c = _5_. The vertices are each _4_ units from the center, so a = _4_ .

Because c² = a² + b², you have b² = c² - a². Find b². Find b².

b² = c² - a² = _5² - 4²_ = _9_

Because the transverse axis is horizontal, the standard form of the equation is as follows:

Substitute 4 for a and 9 for b².

Simplify.

Example 3

Solve a multi-step problem

Lamp The diagram shows the hyperbolic cross section of a lamp. Write an equation for the cross section of the lamp. The lamp is 10 inches high. How wide is the base?

Solution

1. From the diagram, a = _2_and b = _4_ .

Because the transverse axis is _horizontal_, an equation for the cross section of the lamp

is

or

2. Find the x-coordinate at the lamp's bottom edge. Because the lamp is 10 inches tall, substitute y = _5_ into the equation and solve.

x² = _10.25_

x » _3.20_

So, the lamp has a width of 2x or 2(_3.20_) = _6.40_ inches.