Example 4

Again, let's pull as much information out of the equation as possible. We see that the center of the hyperbola is . (Remember that

and this is why the y--coordinate of the center is -1.) Next, note that

and

Since the ``positive'' term in the equation involves the variable x, we know that the transverse axis of this hyperbola is horizontal (parallel with the x--axis). Moreover, we know that the transverse axis has length 14 and the vertices occur at points which are 7 units in either direction of the center. This all implies that the vertices are at and , which could also be written as and . As a sidelight, we also know that the endpoints of the conjugate axis are exactly 5 units above and below the center, which places them at the points and . Moreover, we can easily determine the equations for the asymptotes. They are given by

and

From this information, we can easily plot the hyperbola in question. For the sake of completion, let's quickly determine the location of the foci. Again using the relationship

we know that

Thus, the foci are exactly units to the left and right of the center of the hyperbola. We can also then determine that the eccentricity of this hyperbola is given by

Finally, a sketch of the graph is given in Figure H7.