Problem 45 - Calculus Early Transcendentals

Question

Find an equation of the following hyperbolas, assuming the center is at the origin. Sketch a graph labeling the vertices, foci, and asymptotes. Use a graphing utility to check your work. A hyperbola with vertices (±4,0) and foci (±6,0)

Step-by-Step Solution

Answer

Answer: The equation of the hyperbola is: \[ \frac{x^2}{16} - \frac{y^2}{20} = 1 \] The graph is a horizontal hyperbola with vertices at (±4,0), foci at (±6,0), and asymptotes y = ±(sqrt(5)/2)x.

Step 1: Identify the given values and find a, b, and c

We know that the center of the hyperbola is at the origin (0,0). The vertices are located at (±4,0), so the distance a from the center is 4 units. The foci are located at (±6,0), so the distance c from the center is 6 units. We can use the relationship between a, b, and c to find b: \[c^2 = a^2 + b^2\] Plug the values of a and c: \[6^2 = 4^2 + b^2\] Solving for b, we get: \[b^2 = 20\] So, b = sqrt(20).

Step 2: Write the equation of the hyperbola

Now that we have the values of a and b, we can write the equation of the hyperbola. Since the vertices and foci are along the x-axis, we have a horizontal hyperbola. The equation is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2}= 1 \] Plug in the values of a and b: \[ \frac{x^2}{4^2} - \frac{y^2}{(sqrt(20))^2}= 1 \] This simplifies to: \[ \frac{x^2}{16} - \frac{y^2}{20}= 1 \]

Step 3: Sketch and label the graph

To sketch the graph, we need to label the vertices, foci, and asymptotes. The vertices are (±4,0), and the foci are (±6,0). To find the asymptotes, we use the equation: \[ y = ± \frac{b}{a}x \] Plug in the values of a and b: \[ y = ± \frac{sqrt(20)}{4}x \] So, the asymptotes are y = ±(sqrt(5)/2)x. Now, sketch the graph, making sure to plot the vertices, foci, and draw the asymptotes. The graph should resemble an open sideways hyperbola with the vertices and foci on the x-axis.

Step 4: Use a graphing utility to check your work

Use a graphing utility, like Desmos or a graphing calculator, to check your work. Input the equation: \[ \frac{x^2}{16} - \frac{y^2}{20}= 1 \] and also input the asymptotes: \[ y = (sqrt(5)/2)x \] \[ y = -(sqrt(5)/2)x \] Make sure your sketch matches the output of the graphing utility.

Key Concepts

Vertices of a Hyperbola

The vertices of a hyperbola are the points where the hyperbola intersects the line connecting its two branches. To find the vertices, we look at the values given in the problem. For a hyperbola centered at the origin, with vertices at

• (±4,0)

this means that the vertices lie along the x-axis. The distance from the center (0,0) to a vertex is denoted as \(a\), which is the semi-major axis's length. In this case, \(a = 4\).

Vertices are important because they give us the "width" of the hyperbola's branches and help in drawing the graph accurately.

In context, these points mark the boundary before the hyperbola begins to curve inward, creating that typical "open" shape. The vertices are intrinsic to the hyperbola's equation, as they factor into how wide or narrow the branches appear.

Foci of a Hyperbola

The foci refer to two fixed points on the interior of each hyperbola branch, which help define the hyperbola's shape. They play a crucial role because the difference in distances from any point on the hyperbola to each focus is constant. For the hyperbola in the example, with foci at

• (±6,0)

we note that the foci are located further along the x-axis than the vertices, indicating a horizontal orientation.

The distance from the center of the hyperbola to each focus is labeled \(c\). In this hyperbola, \(c = 6\). Knowing \(c\) is essential because it connects with \(a\) and \(b\) using the equation \[ c^2 = a^2 + b^2 \].

Understanding where the foci are helps in sketching the hyperbola and ensuring it's visually accurate. The closer the foci are to the vertices, the more "narrow" the hyperbola appears; the further away, the wider it opens.

Asymptotes of a Hyperbola

Asymptotes are lines that the hyperbola approaches as it goes towards infinity. They aren’t part of the hyperbola but provide a guideline for its shrinking and expanding direction. The asymptotes of a hyperbola are found using the formula \[ y = \pm \frac{b}{a}x \].

For this hyperbola example, substituting \(a = 4\) and \(b = \sqrt{20}\), we calculate the slopes of the asymptotes to be \[ \pm \frac{\sqrt{20}}{4} \]. Simplifying, this becomes \[ y = \pm \frac{\sqrt{5}}{2}x \].

These asymptotes help in drawing the hyperbola because they represent the "directions" towards which the arms extend.

• A horizontal hyperbola has asymptotes sloping similarly on both sides.
• The branches open out along these invisible lines as they extend further.

Drawing these lines first when sketching can greatly aid in ensuring your hyperbola's arms extend correctly and in the right direction.