Problem 65 - 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 \(y=\frac{3}{4} x\) and \(y=-\frac{3}{4} x,\) and its closest distance to the center fountain is 20 yards.

Step-by-Step Solution

Answer

The equation of the hyperbola is \( \frac{x^2}{400} - \frac{y^2}{225} = 1 \), with vertices at (20, 0) and (-20, 0).

Step 1: Understanding the Hyperbola's Asymptotes Equation

The equations of the asymptotes for the hyperbola are given as \( y = \frac{3}{4}x \) and \( y = -\frac{3}{4}x \). For a hyperbola centered at the origin with these asymptotes, the slopes suggest that \( \frac{b}{a} = \frac{3}{4} \). Therefore, the relationship can be expressed as \( b = \frac{3}{4}a \).

Step 2: Using the Closest Distance to Find 'a'

The closest distance of the hyperbola to the center, known as the distance of the vertices from the center, corresponds to \( a = 20 \) yards. This value of 'a' gives the horizontal distance from the center to each vertex.

Step 3: Calculating 'b' Using the Slope Equation

Using the relationship \( b = \frac{3}{4}a \) and the value of \( a = 20 \), calculate \( b \): \[ b = \frac{3}{4} \times 20 = 15. \] Thus, \( b = 15 \).

Step 4: Formulating the Equation of the Hyperbola

With \( a = 20 \) and \( b = 15 \), the equation of the hyperbola will be \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Substituting in the values, we have \( \frac{x^2}{400} - \frac{y^2}{225} = 1 \).

Step 5: Sketching the Hyperbola

To sketch the hyperbola, plot the vertices at (20, 0) and (-20, 0) since it opens horizontally. The asymptotes have slopes \( \frac{3}{4} \) and \( -\frac{3}{4} \), guiding the orientation of the hyperbola. The hyperbola approaches these lines as it moves away from the center.

Key Concepts

Asymptotes of Hyperbola

Asymptotes of a hyperbola are straight lines that the curve approaches but never actually meets as it extends to infinity. They are crucial for understanding the general shape and orientation of the hyperbola. In our problem, the given asymptotes are equations of the lines: \( y = \frac{3}{4}x \) and \( y = -\frac{3}{4}x \). These lines tell us that the hyperbola will have an opening towards the horizontal axis with steepness indicated by the slopes \( \pm \frac{3}{4} \). This slope helps us establish the relationship between \( a \) and \( b \), because for a hyperbola centered at the origin (0,0), if the equation of the asymptotes is \( y = \pm \frac{b}{a}x \), then \( \frac{b}{a} = \frac{3}{4} \). This leads to the equation \( b = \frac{3}{4}a \), which we use to calculate the necessary parameters of the hyperbola.

Distance to Center

The distance from the hyperbola to its center is a significant parameter. It gives us the value of 'a', which is half the length of the transverse axis of the hyperbola. For this exercise, the closest distance to the center, also identified as the vertices' location, is given as 20 yards. Therefore, \( a = 20 \). This means that the vertices, which are key positions on the hyperbola, are located at distances of 20 yards from the center along the x-axis. Understanding this distance helps in formulating the hyperbola equation and indeed, influences the dimensions of the hyperbola considerably.

Hyperbola Sketch

To sketch a hyperbola, consider its asymptotes, center, and vertices. First, plot the asymptotes on a graph, which in this problem are lines with slopes \( \frac{3}{4} \) and \( -\frac{3}{4} \). This step helps in determining the spread direction of the hyperbola branches. Next, place the center at the origin \((0, 0)\). Since \( a = 20 \), mark the vertices at \((20, 0)\) and \((-20, 0)\) on the horizontal axis. The hyperbola will open horizontally, moving closer to the asymptotes as it extends away from the center. These lines act as guides determining the shape of the hyperbola’s branches and illustrating its infinite nature beyond the sketched region.

Vertices of Hyperbola

The vertices of a hyperbola are the points where the curve is closest to its center. They define the width of the hyperbola along the x-axis and are symmetric about the center. Specifically in our given problem, since \( a = 20 \), the vertices occur at points \( (20, 0) \) and \( (-20, 0) \). These vertices mark the endpoints of the hyperbola's transverse axis. Understanding the vertices' positions is crucial as it determines the overall layout and spread of the hyperbola on a coordinate plane. Moreover, positioning the vertices aids in constructing the correct visual representation of the hyperbola as part of sketching the entire graph.