Hyperbolas

The Hyperbola in Standard Form

A hyperbolaThe set of points in a plane whose distances from two fixed points, called foci, has an absolute difference that is equal to a positive constant. is the set of points in a plane whose distances from two fixed points, called foci, has an absolute difference that is equal to a positive constant. In other words, if points F1 and F2 are the foci and d is some given positive constant then (x,y) is a point on the hyperbola if d=|d1−d2| as pictured below:

In addition, a hyperbola is formed by the intersection of a cone with an oblique plane that intersects the base. It consists of two separate curves, called branchesThe two separate curves of a hyperbola.. Points on the separate branches of the graph where the distance is at a minimum are called vertices.Points on the separate branches of a hyperbola where the distance is a minimum. The midpoint between a hyperbola’s vertices is its center. Unlike a parabola, a hyperbola is asymptotic to certain lines drawn through the center. In this section, we will focus on graphing hyperbolas that open left and right or upward and downward.

The asymptotes are drawn dashed as they are not part of the graph; they simply indicate the end behavior of the graph. The equation of a hyperbola opening left and right in standard formThe equation of a hyperbola written in the form (x−h)2a2−(y−k)2b2=1. The center is (h,k), a defines the transverse axis, and b defines the conjugate axis. follows:

(x−h)2a2−(y−k)2b2=1

Here the center is (h,k) and the vertices are (h±a,k). The equation of a hyperbola opening upward and downward in standard formThe equation of a hyperbola written in the form (y−k)2b2−(x−h)2a2=1. The center is (h,k), b defines the transverse axis, and a defines the conjugate axis. follows:

(y−k)2b2−(x−h)2a2=1

Here the center is (h,k) and the vertices are (h,k±b).

The asymptotes are essential for determining the shape of any hyperbola. Given standard form, the asymptotes are lines passing through the center (h,k) with slope m=±ba. To easily sketch the asymptotes we make use of two special line segments through the center using a and b. Given any hyperbola, the transverse axisThe line segment formed by the vertices of a hyperbola. is the line segment formed by its vertices. The conjugate axisA line segment through the center of a hyperbola that is perpendicular to the transverse axis. is the line segment through the center perpendicular to the transverse axis as pictured below:

The rectangle defined by the transverse and conjugate axes is called the fundamental rectangleThe rectangle formed using the endpoints of a hyperbolas, transverse and conjugate axes.. The lines through the corners of this rectangle have slopes m=±ba. These lines are the asymptotes that define the shape of the hyperbola. Therefore, given standard form, many of the properties of a hyperbola are apparent.

Equation	Center	a	b	Opens
(x−3)225−(y−5)216=1	(3,5)	a=5	b=4	Left and right
(y−2)236−(x+1)29=1	(−1,2)	a=3	b=6	Upward and downward
(y+2)23−(x−5)2=1	(5,−2)	a=1	b=3	Upward and downward
x249−(y+4)28=1	(0,−4)	a=7	b=22	Left and right

The graph of a hyperbola is completely determined by its center, vertices, and asymptotes.

Example 1

Graph: (x−5)29−(y−4)24=1.

Solution:

In this case, the expression involving x has a positive leading coefficient; therefore, the hyperbola opens left and right. Here a=9=3 and b=4=2. From the center (5,4), mark points 3 units left and right as well as 2 units up and down. Connect these points with a rectangle as follows:

The lines through the corners of this rectangle define the asymptotes.

Use these dashed lines as a guide to graph the hyperbola opening left and right passing through the vertices.

Answer:

Example 2

Graph: (y−2)24−(x+1)236=1.

Solution:

In this case, the expression involving y has a positive leading coefficient; therefore, the hyperbola opens upward and downward. Here a=36=6 and b=4=2. From the center (−1,2) mark points 6 units left and right as well as 2 units up and down. Connect these points with a rectangle. The lines through the corners of this rectangle define the asymptotes.

Use these dashed lines as a guide to graph the hyperbola opening upward and downward passing through the vertices.

Answer:

Note: When given a hyperbola opening upward and downward, as in the previous example, it is a common error to interchange the values for the center, h and k. This is the case because the quantity involving the variable y usually appears first in standard form. Take care to ensure that the y-value of the center comes from the quantity involving the variable y and that the x-value of the center is obtained from the quantity involving the variable x.

As with any graph, we are interested in finding the x- and y-intercepts.

Example 3

Find the intercepts: (y−2)24−(x+1)236=1.

Solution:

To find the x-intercepts set y=0 and solve for x.

(0−2)24−(x+1)236=11−(x+1)236=1−(x+1)236=0(x+1)2=0x+1=0x=−1

Therefore there is only one x-intercept, (−1,0). To find the y-intercept set x=0 and solve for y.

(y−2)24−(0+1)236=1(y−2)24−136=1(y−2)24=3736(y−2)2=±376y−2=±373y=2±373=6±373

Therefore there are two y-intercepts, (0,6−373)≈(0,−0.03) and (0,6+373)≈(0,4.03). Take a moment to compare these to the sketch of the graph in the previous example.

Answer: x-intercept: (−1,0); y-intercepts: (0,6−373) and (0,6+373).

Consider the hyperbola centered at the origin,

9x2−5y2=45

Standard form requires one side to be equal to 1. In this case, we can obtain standard form by dividing both sides by 45.

9x2−5y245=45459x245−5y245=4545x25−y29=1

This can be written as follows:

(x−0)25−(y−0)29=1

In this form, it is clear that the center is (0,0), a=5, and b=3. The graph follows.

Try this! Graph: y225−(x−5)29=1.

Answer:

(click to see video)