Problem 62 - College Algebra

Question

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 \(y=2 x\) and \(y=-2 x\) , and its closest distance to the center fountain is 6 yards.

Step-by-Step Solution

Answer

The hyperbola equation is \(\frac{x^2}{36} - \frac{y^2}{144} = 1\) with asymptotes at \(y = 2x\) and \(y = -2x\); vertices at \((6,0)\) and \((-6,0)\).

Step 1: Identify the characteristics of the hyperbola

In a hyperbola centered at the origin, the asymptotes can be represented as the lines \(y = \pm \frac{b}{a}x\). Here, the given asymptotes are \(y = 2x\) and \(y = -2x\), indicating that \(\frac{b}{a} = 2\). Additionally, the hyperbola is centered at the origin.

Step 2: Use the closest distance to find 'a'

The closest distance from the center (0,0) to the hyperbola is the length of the transverse axis divided by 2, which is equal to \(a\). We are given that this distance is 6 yards, so \(a = 6\).

Step 3: Calculate 'b' using the asymptote equation

Using \(\frac{b}{a} = 2\) and \(a = 6\), we can solve for \(b\). Therefore, \(b = 2 \times a = 12\).

Step 4: Write the equation of the hyperbola

The standard form of a hyperbola centered at the origin with horizontal transverse axis is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Substituting \(a = 6\) and \(b = 12\), the equation becomes \(\frac{x^2}{36} - \frac{y^2}{144} = 1\).

Step 5: Sketch the hyperbola

To sketch the graph, plot the center, asymptotes, and mark the vertices and co-vertices. The vertices are at \((6,0)\) and \((-6,0)\) because the transverse axis is horizontal. The co-vertices, although not needed for sketching asymptotes, would be \((0,12)\) and \((0,-12)\). The asymptotes guide the shape of the hyperbola.

Key Concepts

Understanding Asymptotes in Hyperbolas

Asymptotes are critical in understanding hyperbolas. They are straight lines that your hyperbola will approach but never actually meet. You can think of them as invisible boundaries that shape the hyperbola’s path. In graphs, asymptotes help guide the curve’s direction and provide a framework for how the hyperbola opens.
For a hyperbola centered at the origin, the equations for the asymptotes are given by the lines \(y = \pm \frac{b}{a} x\). In our problem, the asymptotes are \(y = 2x\) and \(y = -2x\). This tells us that the slope of the asymptotes, \(\frac{b}{a}\), is 2.
These asymptotes are crucial for sketching the graph because they tell us the angles at which the branches of the hyperbola open. The closer the hyperbola is to these lines, the more exact your sketch will be. Remember, the role of these asymptotes is purely guiding, with the curve never intersecting them.

The Role of the Transverse Axis

The transverse axis is a line segment that contains both the center and vertices of the hyperbola. It plays a big part in determining how wide or narrow the hyperbola will be. In simple terms, it's the main axis around which the hyperbola forms.
In the case of a horizontally oriented hyperbola, like in our exercise, the length of the transverse axis is equivalent to \(2a\), where \(a\) represents half the distance between the vertices. Since the problem indicates that the closest distance the hedge gets to the fountain is 6 yards, \(a\) equals 6. Thus, the full transverse axis is 12 yards long.
This distance is not just a physical length. It informs us graphically how far apart the peaks of the hyperbola extend horizontally. Knowing \(a\) assists in understanding the exact location for the vertices of the hyperbola.

Locating the Center of a Hyperbola

The center of a hyperbola is the midpoint between its vertices and co-vertices. It's an essential point since it serves as a reference for the entire hyperbola. Being aware of the center's location allows you to plot both the hyperbola and its asymptotes accurately.
In many problems, including this one, the center was given as the origin, or the point (0,0). When the hyperbola is centered at the origin, the formulas for its standard equation become more straightforward. It simplifies the process of determining its characteristics and ensures you plot the graph correctly.
To sum up, whether you're asked to sketch, analyze, or rotate a hyperbola, identifying its center offers a solid foundation for handling the problem effectively.