Problem 37 - Thomas Calculus

Question

Exercises \(35-38\) give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation from the information given. $$ \begin{array}{l}{\text { Vertices: }( \pm 3,0)} \\ {\text { Asymptotes: } y=\pm \frac{4}{3} x}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Identify Characteristics

The vertices indicate that the hyperbola is centered at the origin with vertices at \(( ext{major axis}, 0)\) suggesting the transverse axis is horizontal. The equation of the asymptotes \(y = \pm \frac{4}{3}x\) also indicates a horizontal transverse axis.

Step 2: Write the General Equation

The standard form of a hyperbola with a horizontal transverse axis is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \((\pm a,0)\) are the vertices.

Step 3: Use Vertex Information

From the vertices \((\pm 3,0)\), it follows that \(a = 3\). Thus, \(a^2 = 9\). The equation becomes \(\frac{x^2}{9} - \frac{y^2}{b^2} = 1\).

Step 4: Use Asymptote Equation

The slope of the asymptotes is \(\frac{b}{a}\). From the equation \(y = \pm \frac{4}{3}x\), the slope \(\frac{4}{3} = \frac{b}{a}\). Given \(a = 3\), solve for \(b: b = 4\). Thus, \(b^2 = 16\).

Step 5: Write the Hyperbola Equation

Substitute \(a^2 = 9\) and \(b^2 = 16\) into the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). The equation of the hyperbola is \(\frac{x^2}{9} - \frac{y^2}{16} = 1\).

Key Concepts

Standard Form Equation

The standard form equation of a hyperbola provides the basic structure that describes all the hyperbolas, giving us a mathematical way to portray their unique shape. When the transverse axis is horizontal, the standard form of the equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). This format tells us how the hyperbola opens and where its major and minor axes lie.

• \(a\) represents the distance from the center of the hyperbola to each vertex on the transverse axis.
• \(b\) corresponds to the distance that shapes the asymptotes but is not directly visible in the graph as a specific point like the vertices.

Thus, the standard form equation not only sets the boundaries for the hyperbola but also describes its geometric properties comprehensively.

Vertices

Vertices are crucial points where the hyperbola changes direction and represent the tips of its main curves. For hyperbolas centered at the origin with a horizontal transverse axis, the vertices are placed symmetrically about the origin at \((\pm a, 0)\).
Given in the original problem, the vertices at \((\pm 3, 0)\) indicate that the hyperbola stretches along the x-axis.

• The distance between the vertices directly provides us with the value of \(a\). Here, since the distance is 3 units, we set \(a = 3\) and subsequently, \(a^2 = 9\).

Understanding the placement of vertices is essential, as they give us insights into the orientation and scale of the hyperbola.

Asymptotes

Asymptotes are lines that the hyperbola approaches but never actually reaches. They play a key role in defining the openness and direction of the hyperbola's branches. In this exercise, asymptotes are given as \(y = \pm \frac{4}{3}x\).
The slope of these lines, \(\frac{b}{a}\), provides further information about the hyperbola's shape. Here, it suggests that \(\frac{4}{3} = \frac{b}{3}\) since \(a = 3\). By solving this equation, we find \(b = 4\) and thus \(b^2 = 16\).

• Asymptotes help to determine how steeply the arms of the hyperbola open and give a boundary within which the hyperbola lies.

They are not visible in the hyperbola's graph but are critical for its mathematical representation.

Transverse Axis

The transverse axis of a hyperbola is the line segment that stretches between the vertices and it directs the primary spread of the hyperbola's branches. In a hyperbola centered at the origin, a horizontal transverse axis indicates that the hyperbola opens left and right along the x-axis.

• The length of the transverse axis is \(2a\), which is simply twice the distance from the center to a vertex.
• In our case, since \(a = 3\), the transverse axis measures 6 units in length.

The direction and length of the transverse axis are fundamental in establishing the orientation of the hyperbola and are drawn through its center across the widest part.