Problem 43 - Calculus Early Transcendentals

Question

Sketch a graph of the following hyperbolas. Specify the coordinates of the vertices and foci, and find the equations of the asymptotes. Use a graphing utility to check your work. $$\frac{x^{2}}{3}-\frac{y^{2}}{5}=1$$

Step-by-Step Solution

Answer

Answer: The given hyperbola is a horizontal hyperbola centered at (0,0). The vertices are at (\(\pm \sqrt{3}\), 0), the foci are at (\(\pm \sqrt{8}\), 0), and the equations of the asymptotes are \(y = \pm \frac{\sqrt{5}}{\sqrt{3}}x\).

Step 1: Identify the type of hyperbola

The hyperbola provided has the equation of the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), which indicates that it is a horizontal hyperbola.

Step 2: Find the center of the hyperbola

The center of the hyperbola is given by \((h, k)\). Comparing the given equation with the standard form, we have \(a^2 = 3\) and \(b^2 = 5\), with the center at \((h, k) = (0,0)\).

Step 3: Determine the coordinates of the vertices and foci

For a horizontal hyperbola, the vertices are located at \((h \pm a, k)\) and the foci at \((h \pm c, k)\), where \(c\) can be found using the relation \(c^2 = a^2 + b^2\). In our case, since \(a^2 = 3\) and \(b^2 = 5\), we have \(c^2 = 3 + 5 = 8\). Thus, the vertices are at \((\pm \sqrt{3}, 0)\), and the foci are at \((\pm \sqrt{8}, 0)\).

Step 4: Find the equations of the asymptotes

The equations of the asymptotes for a horizontal hyperbola are given by \(y - k = \pm \frac{b}{a}(x-h)\). In our case, since \(a^2 = 3\), \(b^2 = 5\), and \(h = k = 0\), the equations of the asymptotes are \(y = \pm \frac{\sqrt{5}}{\sqrt{3}}x\). Verify the work with a graphing utility: Plot the hyperbola using the given equation, the vertices and foci, and the asymptotes to verify that the graph obtained follows our calculations.

Key Concepts

Hyperbola

One of the fascinating conic sections is the hyperbola. A hyperbola is a type of curve that looks like two mirrored, open arcs going in opposite directions. It is created when a cone is intersected by a plane in such a way that the angle between the plane and the axis of the cone is smaller than that between the plane and one line of the cone. This unique shape is part of why hyperbolas can easily be identified in mathematical equations.
The standard form of a hyperbola's equation depends on its orientation. For a **horizontal hyperbola**, the equation is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]This indicates the hyperbola opens left and right. In contrast, a **vertical hyperbola** opens up and down and follows the equation:\[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]
For the given equation \( \frac{x^{2}}{3}-\frac{y^{2}}{5}=1 \), it's a horizontal hyperbola centered at the origin \((0, 0)\). Understanding these orientations helps in graphing and solving other properties of hyperbolas like vertices, foci, and asymptotes.

Vertices

Vertices are critical points on a hyperbola that represent the closest or farthest points to/from the center on each "arm" of the curve. Knowing the vertices helps in sketching the hyperbola accurately. In a hyperbola, we find these points along the transverse axis, which is the line segment that passes through both vertices.
For a horizontal hyperbola with a center at \((h, k)\), the vertices can be found using the formula:

• \((h \pm a, k)\) for a horizontal hyperbola

Here, \(a\) is the distance from the center to a vertex. For our hyperbola \(\frac{x^{2}}{3}-\frac{y^{2}}{5}=1\), \(a\) is equal to \(\sqrt{3}\), so the vertices are located at \((\pm \sqrt{3}, 0)\).
Recognizing and using the vertices helps in both graphing and verifying the symmetry of the hyperbola.

Foci

The foci (singular: focus) of a hyperbola are two fixed points located along the transverse axis. They are crucial because the difference in distances from any point on the hyperbola to these two foci is a constant. This geometric property defines the hyperbola. Finding the foci aids in deeply understanding the shape and nature of the curve.
For a hyperbola centered at \((h, k)\), the foci positions for a horizontal hyperbola are calculated as:

• \((h \pm c, k)\)

where \(c\) is determined by the formula \(c^2 = a^2 + b^2\). In this problem, with \(a^2 = 3\) and \(b^2 = 5\), we compute \(c\) as \(\sqrt{8}\). Therefore, the foci are located at \((\pm \sqrt{8}, 0)\).
This understanding of the foci is essential to confirm the hyperbola's full geometric profile.

Asymptotes

Asymptotes of a hyperbola provide lines that the curve approaches but never actually meets. They provide a boundary, guiding the hyperbola's arms. Understanding asymptotes is key when sketching a hyperbola and assists in seeing how wide the branches should appear.
For a hyperbola oriented horizontally, asymptotes follow the formula:

• \(y - k = \pm \frac{b}{a}(x-h)\)

Using the given equation \(\frac{x^{2}}{3}-\frac{y^{2}}{5}=1\), the asymptotes are lines determined by replacing \(a\) and \(b\):
\(y = \pm \frac{\sqrt{5}}{\sqrt{3}}x\). Asymptotes help us understand and draw the extended "arms" of the hyperbola, pointing out the direction in which they extend to infinity.