Problem 30 - Precalculus with Limits

Question

In Exercises 29-34, find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. Vertices: \((\pm4, 0); \quad\) foci: \((\pm6, 0)\)

Step-by-Step Solution

Answer

The standard form of the equation of the given hyperbola is \[\frac{x^2}{16} - \frac{y^2}{20} = 1\]

Step 1: Identify Parameters \(a\) and \(c\)

We can identify the parameters \(a\) and \(c\) directly from the given vertices and foci. The distance from the center to a vertex represents \(a\) and the distance from the center to a focus represents \(c\). As the vertices are given by \(\pm4, 0\), we know that \(a = 4\). Similarly, as the foci are given by \(\pm6, 0\), we know that \(c = 6\).

Step 2: Calculate Parameter \(b\)

Now, we can use the relationship between the parameters \(a\), \(b\), and \(c\) in an hyperbola, \(c^2 = a^2 + b^2\), to find the value of \(b\). We substitute \(a = 4\) and \(c = 6\) in the equation, obtaining \(b = \sqrt{c^2 - a^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5}\).

Step 3: Write the Equation of the Hyperbola

The standard form of the equation of a hyperbola centered at the origin is given by \[\frac{x^2}{{a^2}} - \frac{y^2}{{b^2}} = 1\] Since the vertices and foci are on the x-axis, and the given \(a\) and \(b\) are 4 and \(2\sqrt{5}\) respectively, the equation becomes: \[\frac{x^2}{{16}} - \frac{y^2}{{20}} = 1\]

Key Concepts

Vertices of a Hyperbola

Vertices are a crucial part of understanding hyperbolas. For a hyperbola centered at the origin, the vertices are points where the hyperbola intersects its transverse axis. This axis is a line running through both foci of the hyperbola.

In our exercise, the given vertices are at \(\pm 4, 0\). These coordinates tell us that the hyperbola is horizontally oriented because the transverse axis lies on the x-axis. Specifically, the transverse axis of this hyperbola runs from \(-4, 0\) to \(4, 0\), with the center of the hyperbola as the midpoint, located at the origin \(0, 0\). The distance from the center to either vertex is \(a = 4\). This parameter \(a\) is essential as it helps determine the shape of the hyperbola, especially how "stretched" it appears horizontally.

Remember, the vertices help guide the direction of the hyperbola, whether alongside the x-axis or y-axis. In a vertical hyperbola, the vertex coordinates would be \(0, \pm a\).

Foci of the Hyperbola

Foci (singular: focus) are special points within a hyperbola that determine its shape and orientation. The standard form of a hyperbola has foci symmetrically placed around the center, lying on its transverse axis.

In the given exercise, the foci are positioned at \(\pm 6, 0\). Like the vertices, these coordinates indicate that the foci are aligned horizontally along the x-axis. The parameter \(c\) represents the distance from the center of the hyperbola to a focus. Thus, for our hyperbola, \(c = 6\).

The significance of the foci lies in how the hyperbola branches open around them. While the vertices provide a boundary, the foci ensure that the hyperbola's shape remains consistent, maintaining equal distance differences between the foci and any point on its branches. Ultimately, the foci assist in deriving the equation of the hyperbola and contribute to calculating the parameter \(b\).

Parameters in a Hyperbola

Understanding the parameters \(a\), \(b\), and \(c\) is essential in comprehending the shape and equation of a hyperbola. These parameters uniquely define the geometry and equation of a hyperbola, enabling us to discern its orientation and structure.

• \(a\): This parameter represents the distance from the center to each vertex. It helps establish how widespread the hyperbola is along the transverse axis. In the exercise, \(a = 4\).
• \(b\): The parameter \(b\) is linked with the distance determining the shape's curve on its conjugate axis. It's calculated using the relationship \(c^2 = a^2 + b^2\). For this hyperbola, \(b = 2\sqrt{5}\).
• \(c\): This parameter signifies the distance from the center to each focus—an indicator of how "open" the branches of the hyperbola are. Here, \(c = 6\).

The equation of the hyperbola can be formulated using these parameters: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). As given in our exercise, this results in \(\frac{x^2}{16} - \frac{y^2}{20} = 1\).

These parameters together encapsulate all necessary aspects of the hyperbola's geometry, helping students facilitate calculations and understand the underlying geometrical interpretations.