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# Hyperbolas Help (page 2)

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## Matching Hyperbola Equations with Graphs

In the next problem, we will match equations of hyperbolas with their graphs. Being able to identify the vertices will not be enough. We will also need to use the equations of the asymptotes to find b (we will know a from the vertices). Because the center of each hyperbola will be at (0, 0), the asymptotes will either be and .

#### Examples

Match the equation with its graph in Figures 12.24–12.27.

Fig. 12.24

Fig. 12.25

Fig. 12.26

Fig. 12.27

• The vertices are (−2, 0) and (2, 0). The slopes of the asymptotes are −1 and 1. The graph is in Figure 12.25 .

• The vertices are (−2, 0) and (2, 0). The slopes of the asymptotes are and . The graph is in Figure 12.26 .

• The vertices are (0, −2) and (0, 2). The slopes of the asymptotes are −1 and 1. The graph is in Figure 12.27 .

• The vertices are (0, −2) and (0, 2). The slopes of the asymptotes are −2 and 2. The graph is in Figure 12.24 .

## Finding the Equation for a Hyperbola

We can find the equation for a hyperbola when we know some points or a point and the asymptotes. If we have the vertices and foci, then finding an equation for a hyperbola will be similar to finding an equation for an ellipse. If we are given the vertices and asymptotes or foci and asymptotes, we will need to use the slope of one of the asymptotes to find either a or b (we will know one but not the other from the vertices or foci). The first thing we need to decide is which formulas to use—those in Figures 12.21 or Figure 12.22. If the vertices or foci are on the same horizontal line (the y -coordinates are the same), we will use Figure 12.21. If they are on the same vertical line (the x -coordinates are the same), we will use Figure 12.22.

Fig. 12.21

Fig. 12.22

#### Examples

Find an equation for the hyperbola.

• The vertices are (3, −1) and (3, 7) and is an asymptote. The vertices are on the same vertical line, so we need to use the information in Figure 12.22. The vertices are ( h , ka ) = (3, −1) and ( h , k + a ) = (3, 7). This gives us h = 3, ka = −1 and k + a = 7.

The center is (3, 3) and a = 4. Once we have b , we will be done. The slope of one of the asymptotes in Figure 12.22 is , so we have , so b = 3. The equation is

• The vertices are (−8, 5) and (4, 5), and the foci are (−12, 5) and (8, 5). The vertices and foci are on the same horizontal line, so we need to use the information in Figure 12.21. The vertices are ( ha , k ) = (−8, 5) and ( h + a , k ) = (4, 5). Now we know k = 5 and we have the system ha = −8 and h + a = 4.

−2 − a = −8 Let h = −2 in ha = − 8

a = 6

A focus is ( hc , k ) = (−2 − c , 5) = (−12, 5), which gives us −2 − c = –12. Now that we see that c = 10, we can put this and a = 6 in to find b .

The equation is

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