Hyperbolas

Let F1and F2 be fixed points in the plane. A hyperbola is the collection of points in the plane such that the difference of the distances from the point to F1and F2 is a fixed constant. The points F1and F2 are called the foci of the hyperbola.
The picture illustrates the definition. The point P is a typical point on the hyperbola. The difference of the distances from P to each of the foci is a constant. If we denote this constant by 2a, a > 0, then | |PF1| - |PF2| | = 2a for any point P on the hyperbola. Note that the curve is made up of two continuous curves known as branches of the hyperbola.

Other than the foci there are other special points associated with a hyperbola which we have pointed out in the diagram. The line through the foci cuts the hyperbola at two points, labeled A and B in the diagram. The points A and B are called the vertices. The line segment joining the vertices is called the transverse axis. The point midway between the foci and lying on the transverse axis is called the center of the hyperbola. This is the point labeled O in the diagram.

If we position a hyperbola in the plane with its center at the origin and its foci along the x axis we can obtain a nice equation for a hyperbola. The derivation of the equation is almost identical to the derivation of the equation for the ellipse. We will not go through all the details here but set up the machinery for you to obtain the equation. As with the ellipse, we let the positive constant be 2a and let c be a positive number so that the foci are located at (-c,0) and (c,0). We have

|PF1| - |PF2| = 2a
|PF1| = 2a + |PF2|
|PF1|2 = (2a + |PF2|)2
|PF1|2 = 4a2 4a|PF2| +|PF2|2
You should now continue with the algebra and the simplification to show that the equation of the hyperbola is
x2/a2-y2/b2 = 1
where b2 = c2-a2. More details on the algebra can be found by clicking here. The two lines drawn in the diagram are called asymptotes and their equations are y = bx/a. We shall say more about the asymptotes shortly. The asymptotes are a great help in sketching the graph of a hyperbola.

To understand the importance of the asymptotes let's rewrite the equation of the hyperbola as x2/a2 = y2/b2 + 1. Now for very large values of x the addition of 1 on the right side of the equation becomes insignificant. That is, for very large values of x

x2/a2 y2/b2
and so x/a y/b and solving for y gives y bx/a which are the equations of the asymptotes. In other words, the asymptotes provide approximations to the behavior of y for large values of x. As we see in the picture, the hyperbola gets "closer and closer" to the asymptotes for large values of x. In terms of graphing, if we draw in the asymptotes they provide a guideline for the shape of the hyperbola.

The numbers a and b do have geometric meaning. Referring to the diagram, we know that at the points A and B, y=0. Then, substituting y=0 in the equation of the hyperbola gives x2/a2 = 1 which simplifies to x2=a2. This equation has solutions x=a or -a. Since a > 0, then A has coordinates (-a,0) and B has coordinates (a,0). Thus the vertices of the hyperbola are at (-a,0) and (a,0). Now draw in points, C and D, with coordinates (0,b) and (0,-b) respectively. The line segment CD passes through the center, O, and is perpendicular to the transverse axis. It is called the conjugate axis. Note that the diagonals of the rectangle containing A, B, C and D are line segments of the asymptotes. This gives a convenient way to draw in the asymptotes - namely draw vertical lines through the vertices, draw horizontal lines through the points (0,b) and (0,-b). The points of intersection of these four lines form the vertices of a rectangle. Draw in the lines that contain the diagonals of the rectangle and you have drawn the asymptotes. Note that the length of CD is 2b and the length of AB is 2a (compare with the minor and major axis of the ellipse). Note that since b2 = c2-a2 then b < c and similarly a < c.

In a similar manner, we could have placed the foci on the y axis and the center at the origin In this case, the equation of the hyperbola comes out to be

y2/a2 - x2/b2 = 1.
The foci are located at (0,c) and (0,-c) and the vertices at (0,a) and (0,-a). The equations of the asymptotes are y = ax/b.

The equations we have just established are known as standard equations of a hyperbola in standard position. Standard position always implies the center is at the origin and the foci are on one of the axes.

Hyperbolas in standard position
In this demonstration you can alter the location of the foci and the value of a by moving the sliders. Recall that 2a is the difference of the distances of a point on the hyperbola to each foci. Initially the foci are on the x axis but you can also elect to place them on the y axis.

Example Find the foci of the hyperbola
x2/16-y2/9 = 1.


The foci are located on the x axis since the x term is positive. We have a2 = 16 and b2 = 9. Then c2 = 25 and so c = 5. Therefore, the foci are located at (-5,0) and (5,0).

Example Find the equation of the hyperbola in standard position with a focus at (0,13) and with transverse axis of length 24.

The other focus is located at (0,-13) and since the foci are on the y axis we are looking to find an equation of the form y2/a2-x2/b2 = 1. The value of a is one-half the length of the transverse axis and so a = 12. Also, b2 = c2 - a2 = 169 - 144 = 25. Hence b = 5. So the equation of the hyperbola is
y2/144-x2/25 = 1.


Equation of a hyperbola
In this exercise set, you are required to find the equation of a hyperbola from given information. You will be given a focus and the length of the transverse or conjugate axis or you will be given the lengths of the axes and axis that contains the foci. After you have entered your answer, press "CHECK".

If we translate a hyperbola in standard position so that its center is moved to (h, k) then the equation of the hyperbola is given as follows:

The equation of a hyperbola translated from standard position so that its center is at (h,k) is given by
(x-h)2/a2-(y-k)2/b2 = 1.
if its foci lie on a line parallel to the x axis and
(y-k)2/a2-(x-h)2/b2 = 1.
if its foci lie on a line parallel to the y axis.

This form of the equation of a hyperbola is called the standard equation. However, if we take a standard equation (x-h)2/u2-(y-k)2/v2 = 1 and expand it we get an equation of the form Ax2 +By2 +Cx +Dy +E=0, where A, B, C, D and E are constants. The question we have is if we are given such an equation can we recognize it as the equation of a hyperbola? The answer to this question can be determined through the process of completing the square (just as we did for the ellipse and the parabola).

Example Determine if 16x2 -9y2 -64x+18y-89=0 is the equation of a hyperbola. If so, find the coordinates of the center and the foci.

Our first task is to group together the x terms and group together the y terms. We have
16x2 -9y2 -64x+18y-89=0
(16x2 -64x) -(9y2 -18y)-89=0
16(x2 -4x) -9(y2 -2y)-89=0
Now complete the square on the x terms and the y terms.
16(x2 -4x+4-4) -9(y2 -2y+1-1)-89=0
16((x-2)2 -4) -9((y-1)2 -1)-89=0
Next we remove the outermost parentheses in the grouped terms.
16(x-2)2 -64 -9(y-1)2 +9-89=0
16(x-2)2 -9(y-1)2 -144=0
16(x-2)2 -9(y-1)2 =144
(x-2)2/9 -(y-1)2/16 =1

This we should recognize as a hyperbola with center (2,1). The transverse axis is parallel to the x axis. Now, c2 = 9 + 16 = 25, so that c = 5. The foci of the hyperbola in standard position (and foci along the x axis) are at (c,0) and (-c,0) so the foci of the translated hyperbola are at (c+2,1) and (-c+2,1). That is, at (7,1) and (-3,1).

Finding the Center and Foci of a Hyperbola
In this exercise you are given the equation of a hyperbola in expanded form. On paper you should determine the location of the center and the foci. We suggest you first rewrite the equation in standard form. To accomplish this, you will probably need to complete the square on the x terms and the y terms. Once you have your answer try to sketch the graph of the hyperbola. Finally, press the "Graph" button to see the answers.