Problem 27 - Algebra and Trigonometry

Question

Find an equation for the hyperbola that satisfies the given conditions. Foci \(( \pm 5,0),\) vertices \(( \pm 3,0)\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \).

Step 1: Identify the Hyperbola Orientation

Since the given foci and vertices have x-coordinates \( \pm 5 \) and \( \pm 3 \) respectively, while the y-coordinates are 0, the hyperbola is centered on the x-axis. This means that the transverse axis is horizontal, and the general equation of the hyperbola will be \((x^2/a^2) - (y^2/b^2) = 1\).

Step 2: Determine the Values of a and c

The vertices of the hyperbola \( ( \pm 3, 0) \) indicate that the distance from the center to each vertex \( a = 3 \). The foci \( ( \pm 5, 0) \) indicate that the distance from the center to each focus \( c = 5 \).

Step 3: Use the Relationship c^2 = a^2 + b^2

In a hyperbola, the relationship between a, b, and c is given by \{ c^2 = a^2 + b^2 \}. Using \( a = 3\) and \( c = 5\), we calculate \( b^2 \) as follows:\\[ c^2 = a^2 + b^2 \]\[ 5^2 = 3^2 + b^2 \]\[ 25 = 9 + b^2 \]\[ b^2 = 16 \].

Step 4: Formulate the Equation of the Hyperbola

Now that we have \( a^2 = 9\) and \( b^2 = 16\), we can plug them into the standard form of the equation for a hyperbola with a horizontal transverse axis: \((x^2/a^2) - (y^2/b^2) = 1\). Thus, the equation is: \[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \].

Key Concepts

Foci

In the context of a hyperbola, the foci are two critical points located along the transverse axis. They help define the shape of the hyperbola. Each hyperbola has two branches, and the foci are positioned on the outside of these branches.
The purpose of the foci is to maintain a constant difference in distances to any point on the hyperbola.For our exercise, the foci are at the coordinates \(( \pm 5, 0)\). This placement confirms that our hyperbola is horizontally oriented. Knowing where the foci are helps us determine other features of the hyperbola, such as the transverse axis and the standard form of its equation.
It’s essential to calculate the distance from the center of the hyperbola to each focus, denoted as \(c\). In our case, \(c = 5 \).

Vertices

Vertices are the points where each branch of the hyperbola is closest to its center. They form an essential part of the hyperbola as they essentially "anchor" it visually and mathematically.For our exercise, the vertices are located at \(( \pm 3, 0)\). This means the hyperbola stretches away from the center to these points. The distance from the center to each vertex is termed as \(a\).In our example, this measure is \(a = 3\). The vertices help us understand how "wide" the hyperbola looks at its nearest points to the center. Determining the vertices is a crucial step in forming the hyperbola's equation.

Transverse Axis

The transverse axis is a line that segments through the midline of a hyperbola, passing through the foci. It is the direction of the main curve of the hyperbola and helps identify the hyperbola's orientation.For the given hyperbola, the transverse axis is horizontal, lying along the x-axis. It indicates that the equation we will use is in the form:\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] Horizontal placement means that both vertices and foci are neatly aligned along the same line. The transverse axis allows us to discern whether the hyperbola's stretch is horizontal or vertical, which is pivotal in choosing the correct equation form.

Hyperbola Standard Form

The equation of a hyperbola in its standard form provides an algebraic representation that describes its shape and position based on certain parameters like vertices and foci.Given a horizontal transverse axis, the equation takes the form:\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]Where \(a^2\) and \(b^2\) are derived from the distances to the vertices and the auxiliary distances (calculated as shown in the original solution). For our exercise, we computed:- \( a^2 = 9 \)- \( b^2 = 16 \)Inserting these into the standard form gives:\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \] This equation defines the precise boundaries and nature of our specific hyperbola. Understanding how to derive and apply the standard form is key to mastering the mathematics of hyperbolas.