91Ó°ÊÓ

In hyperbolas, the vertices are key points from which we can gather a lot of information to construct the equation. For this exercise, we have the vertices at

• (9, -3)
• (-5, -3)

The vertices are on the transverse axis and are symmetrically placed about the center. They establish the length of the transverse axis. By using them, we can easily find the distance from the center to each vertex, which is the value of "a" in the hyperbola equation. We essentially measure from the center to one of these vertices to define the horizontal "spread" of the hyperbola.

The foci of a hyperbola are also important. They lie on the transverse axis, just like the vertices, but are further from the center. In this hyperbola's case, the foci are given as:

• (2 + \(\sqrt{53}\), -3)
• (2 - \(\sqrt{53}\), -3)

The foci tell us about the hyperbola's shape and help determine the distance "c" from the center. Here, the calculation reveals "c" as \(\sqrt{53}\). Both foci being aligned on the x-axis with the same y-coordinate confirm a horizontal hyperbola. The relationship involving the vertices, foci, and axes leads us to the distance formula for hyperbolas, which is vital for understanding their dimensions and orientation.

The transverse axis is a critical line segment for hyperbolas. It extends between the vertices and through the center. In a hyperbola with horizontal orientation, like this example, it primarily lies along the x-axis. The length of the transverse axis aligns with the distance between the vertices.To find the transverse axis, we calculated the distance between the two vertices \((9, -3)\) and \((-5, -3)\), resulting in a distance of 14. This automatically tells us a key property: the full length of the transverse axis is 14. Since the transverse axis length is given by \(2a\), we find \(a = \frac{14}{2} = 7\). This length of 7 is used in calculations concerning hyperbola dimensions.

Calculating the center of a hyperbola is straightforward when we have both vertices. The center is the midpoint of the line segment connecting the vertices. We apply the midpoint formula to find it. Given vertices

• (9, -3)
• (-5, -3)

Applying the formula, \[\left(\frac{9 + (-5)}{2}, \frac{-3 + (-3)}{2}\right) = \left(\frac{4}{2}, \frac{-6}{2}\right) = (2, -3)\]This center point (2, -3) is essential for forming the hyperbola's equation, serving as the reference point for both vertices and foci.

The distance formula helps measure the space between two points in a plane. It's an essential tool in determining various attributes of a hyperbola. In this exercise, the formula was vital for:

• Finding the distance between the vertices.
• Ensuring the accuracy of the transverse axis length.

Given two points, say \((x_1, y_1)\) and \((x_2, y_2)\), the distance formula is \[\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]For our vertices \((9, -3)\) and \((-5, -3)\), it becomes \[\sqrt{(9 - (-5))^2 + ((-3) - (-3))^2} = \sqrt{14^2} = 14\]This 14 also represents the length of the transverse axis, confirming it matches the theoretical dimensions. The formula consistently aids in identifying and verifying all these critical hyperbola elements.