Problem 64 - College Algebra

Question

For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=\frac{2}{3} x\) and \(y=-\frac{2}{3} x,\) and its closest distance to the center fountain is 12 yards.

Step-by-Step Solution

Answer

The equation is \(\frac{x^2}{144} - \frac{y^2}{64} = 1\).

Step 1: Understand the Asymptotes Form

The asymptotes of the hyperbola are given by the equations \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\). Since these are in the form \(y = \pm \frac{b}{a}x\), where \(b = 2\) and \(a = 3\), this means \(\frac{b}{a} = \frac{2}{3}\). This tells us that the hyperbola is centered at the origin and is symmetric about both axes.

Step 2: Determine the Form of the Hyperbola Equation

For a hyperbola centered at the origin with horizontal and vertical asymptotes \(y = \pm \frac{b}{a}x\), the equation is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) (if the hyperbola opens left-right) or \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\) (if it opens up-down). From the asymptotes equation, this hyperbola opens left-right.

Step 3: Use Closest Distance Criteria

The closest distance of the hyperbola to the center (origin) is equal to the distance to the vertex, which is given by \(a\) in the equation. Here, it's given as 12 yards. Hence, \(a = 12\).

Step 4: Calculate \(b\) Using Asymptotes Ratio

Since \(\frac{b}{a} = \frac{2}{3}\) and \(a = 12\), we solve for \(b\) as follows: \(b = \frac{2}{3} \times 12 = 8\).

Step 5: Write the Equation of the Hyperbola

By substituting \(a = 12\) and \(b = 8\) into the equation for the hyperbola, we get: \[\frac{x^2}{12^2} - \frac{y^2}{8^2} = 1\]. Simplifying, this becomes \[\frac{x^2}{144} - \frac{y^2}{64} = 1\].

Step 6: Graph the Hyperbola

To sketch the graph, plot the center of the hyperbola at the origin (0, 0). Draw the asymptotes using the lines \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\). Mark the vertices at ±12 on the x-axis. Draw the branches of the hyperbola opening left-right, approaching but not touching the asymptotes.

Key Concepts

Asymptotes

Asymptotes are lines that a curve approaches but never actually meets. In the context of a hyperbola, the asymptotes provide valuable information about the graph's overall shape and orientation. For the hyperbola in this problem, the asymptotes are given by the equations \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\).

These lines intersect at the origin, giving us a crucial clue about the hyperbola's center location. Since the slopes of the asymptotes are \(\pm \frac{b}{a}\), with \(b = 2\) and \(a = 3\), we know the hyperbola is centered at the origin. The hyperbola in question is symmetric about the x and y axes.

The role of asymptotes in hyperbolas is to define the "path" or "direction" in which the branches of the hyperbola open. Here, the hyperbola branches will approach these lines as they extend towards infinity, but they will never intersect them.

Equation of a Hyperbola

The general equation for a hyperbola depends on its orientation and the relationship between the x and y components. For a hyperbola centered at the origin with horizontal asymptotes, the standard equation is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

In this problem, we are given the asymptotes \(y = \pm \frac{2}{3}x\), which indicates a hyperbola opening left-right. The ratio of the asymptotes, \(\frac{b}{a} = \frac{2}{3}\), helps determine the specific values of \(a\) and \(b\).

Using the closest distance to the center (which is one of the vertices of the hyperbola), we identify \(a = 12\). By solving \(b = \frac{2}{3} \, a\), we find that \(b = 8\).

Finally, inserting \(a = 12\) and \(b = 8\) into the standard form of the hyperbola's equation gives us:

• \(\frac{x^2}{144} - \frac{y^2}{64} = 1\).

This equation fully defines the hyperbola's shape.

Center of the Hyperbola

The center of a hyperbola is the point around which the structure is symmetrical. For this hyperbola, the center is located at the origin, denoted as (0, 0).

Identifying the center is crucial because it serves as the fixed point from which other key features of the hyperbola are defined, such as vertices, foci, and the asymptotes.

In this particular case, since the asymptotes intersect at the origin and the hyperbola is symmetric about both axes, the conclusion that the center is (0, 0) follows naturally. The origin acts as a pivotal point ensuring coherence in the graph's geometry.

Distance and Hyperbola Properties

The properties of a hyperbola involve distances, focusing mainly on the distances between the center, vertices, and foci. For the hyperbola in this exercise, one direct property is the distance to the vertex, given as 12 yards.

This distance corresponds to \(a\) in the hyperbola equation. Vertices are vital in determining how 'stretched' the hyperbola is along its major axis. Remember, the distance from the center to a vertex along the axis on which the hyperbola opens provides the \(a\) value directly, which helps in shaping the equation.

Understanding these distances gives insight into the hyperbola's overall structure and behavior. The vertices mark the closest points of the branches of the hyperbola to the center, reiterating the nature of the hyperbola as a geometric shape distinct from ellipses and circles.