uc3rmw_0804e08

Hi there. In this example, we want to find the equation of a hyperbola centered at h comma k, given information about its graphs that include its asymptotes. The problem states, find an equation for the hyperbola whose graphs are shown. In part A, we have a hyperbola with branches opening to the left and to the right. The center of the hyperbola is at 0 comma 1. It's got a vertex here at 2 comma 1, and a focus at 4 comma 1. In part B, we have a hyperbola with branches opening upward and downward. There are two asymptotes shown here. One is given by the equation y equals 3/2 x minus 5/2. And the other is given by the equation y equals minus 3/2 x plus 1/2. The vertex is at 1 comma 2, and the center of this hyperbola is at 1 comma minus 1. First let's recall the equations for the asymptotes of a hyperbola centered at h comma k. The main properties of hyperbolas centered at h comma k are shown here. We have the standard equation given by x minus h squared divided by a squared minus y minus k squared divided by b squared equals 1, where a is greater than 0, and b is greater than 0. This is the standard equation for a hyperbola with transverse axis along the line y equals k. The length of the transverse axis is 2a. The conjugate axis is along the line x equals h. And the length of the conjugate axis is 2b. The center is h comma k and therefore, the vertices have coordinates h minus a comma k and h plus a comma k. And the endpoints of the conjugate axis are given by h common k minus b and h comma k plus b. The foci are h minus c comma k and h plus c comma k. We have the equation involving a, b, and c given by c squared equals a squared plus b squared. And the asymptotes have the form y minus k equals plus or minus b over a times, in parentheses, x minus h. Now the standard form for the equation of a hyperbola whose transverse axis is along the line x equals h is given by y minus k squared divided by a squared minus x minus h squared divided by b squared equals 1. The length of the transverse axis is the same for this equation where the transverse axis is along the line y equals k. It's 2a. The conjugate axis in this case is y equals k. And the length of the conjugate axis is 2b. The center is at h comma k, and the vertices are, therefore, h comma k minus a, and h comma k plus a. The endpoints of the conjugate axis are h minus b comma k, and h plus b comma k. And foci are at h comma k minus c, and h comma k plus c. We have the equation involving a, b, and c as c squared equals a squared plus b squared, just as it is for the other equation. And the asymptotes, in this case, are y minus k equals plus or minus a over b times, in parentheses, x minus h. In part A, since the foci are on the x-axis, the transverse axis is on the x-axis. From the graph, we see that the center is at hk equals 0 comma 1. So h is equal to 0, and k is equal to 1. The distance a between the center at 0 comma 1 and the vertex 2 comma 1 is 2. Therefore, a is equal 2. The distance c is between the center 0 comma 1 and the focus at 4 comma 1. And so c is equal to 4. The transverse axis, we see lies on the horizontal line y equals 1, as shown here. So we have the value for c. We've determine the value for a. We now want to find the value for b. We can use the relationship between a, b, and c. That is the equation c squared equals a squared plus b squared to solve for b squared. We plug in the fact that c equals 4. So we have 4 squared. And that's equal to, plugging in that a equals 2, we have 2 squared plus b squared. We now want to solve this equation for b squared. We have that b squared equals 4 squared minus 2 squared, which simplifies to 16 minus 4, or 12. So b squared equals 12. Now that we have a and b, we can plug that into the standard form for the hyperbola centered at 0 comma 1 to find the equation of the hyperbola with the specific information given. We have that the standard form for the equation of a hyperbola with transverse axis on the horizontal line y equals 1 is given by x minus h squared over a squared minus y minus k squared over b squared equals 1. We know the center h comma k, we know what a is, and we know what b is. So now we can plug in all that information to find the equation specific to this graph. We have x minus h, where h is 0, squared over a squared-- we know that a is 2, so that's going to be 2 squared-- minus y minus k-- k is 1-- squared over b squared. We determine that b squared is 12. And that equals 1. We can simplify this a bit. We have x squared over 4 minus, in parentheses, y minus 1 squared over 12 equals 1. So here's the equation of the hyperbola given the information in the graph. In part B, the center h comma k is equal to 1 comma minus 1. So we have that h is equal to 1, and k is equal to minus 1. The distance a between the center at 1 comma minus 1 and the vertex 1 comma 2, as shown in the graph, is 3 units. And so a is equal to 3. We see that the slope of the asymptote with the positive slope given by this line is 3/2. The transverse axis is on the vertical line x equal 1, as shown here. The form for asymptotes for hyperbolas with transverse axis on the vertical line x equal 1 are given by y equals plus or minus a over b times x. We just found that the slope of our asymptotes are plus or minus 3/2. And so we can relate our asymptotes based on the information given in the graph to the standard form to find a and b. We have that the slopes of our asymptotes are plus or minus 3/2. And therefore, a over b is equal to 3/2. Now we can relate b and a. We have that b is equal to 2/3 a. Now that we have b in terms of a and we know that a is equal to 3, we can find b. b is equal to 2/3 times a where a is 3. So we have 2/3 times 3, or 2. So we have a equal to 3 and b equal to 2. And we can plug that into the standard form for the equation of a hyperbola with transverse axis along the line x equal 1. That standard form is y minus k squared over a squared minus x minus h squared over b squared equal 1. Again, the center is at 1 comma minus 1. And we just found a and b-- a is equal to 3; b is equal to 2. We're going to plug in all that information to find the equation specific to the hyperbola graphed here. We have the y minus k, which is equal to y minus minus 1 squared over a squared-- a is equal to 3, so that's going to be 3 squared-- minus x minus h, where h is equal to 1, squared over b squared. We found that b is equal to 2. So that's going to be 2 squared. And that's all equal to 1. We'll go ahead and simplify that equation. We have an equation that is y plus 1 squared over 9 minus x minus 1 squared over 4 equal to 1. So there's the form of the equation for this hyperbola graphed here.