Problem 29 - Precalculus Mathematics for Calculus

Question

Use a graphing device to graph the hyperbola. $$\frac{y^{2}}{2}-\frac{x^{2}}{6}=1$$

Step-by-Step Solution

Answer

Graph the vertical hyperbola centered at the origin with vertices along the y-axis, foci \((0, 2\sqrt{2})\), and asymptotes \(y = \pm \frac{1}{\sqrt{3}}x\).

Step 1: Identify the Equation Type

This equation is a hyperbola. It is written in the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), where \(a^2 = 2\) and \(b^2 = 6\). This tells us that the hyperbola is vertical because the \(y\) term is positive.

Step 2: Determine the Center of the Hyperbola

For a hyperbola in this form, the center is at the origin \((0, 0)\). We identify these values based on the absence of \(h\) and \(k\) in the equation, which would translate \((x-h)^2\) and \((y-k)^2\) terms.

Step 3: Calculate the Vertices

The vertices of a vertical hyperbola are located \(a\) units above and below the center. Here, \(a = \sqrt{2}\). Therefore, the vertices are at \((0, \sqrt{2})\) and \((0, -\sqrt{2})\).

Step 4: Compute Foci

The foci are \(c\) units from the center, calculated using \(c = \sqrt{a^2 + b^2}\). Substituting \(a^2 = 2\) and \(b^2 = 6\), we find \(c = \sqrt{8} = 2\sqrt{2}\). Thus, the foci are located at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\).

Step 5: Define the Asymptotes

For the hyperbola \(\frac{y^2}{2} - \frac{x^2}{6} = 1\), the asymptotes are given by \(y = \pm \frac{a}{b}x\). Therefore, the asymptotes are \(y = \pm \frac{\sqrt{2}}{\sqrt{6}}x = \pm \frac{1}{\sqrt{3}}x\).

Step 6: Graph the Hyperbola

Using a graphing device, plot the hyperbola, marking the vertices at \((0, \sqrt{2})\) and \((0, -\sqrt{2})\) and drawing the asymptotes \(y = \pm \frac{1}{\sqrt{3}}x\). The hyperbola will open upwards and downwards between these asymptotes. Ensure the foci are placed at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\).

Key Concepts

Vertices

In the context of a hyperbola, the vertices are points that represent the closest distance between the two arms of the hyperbola. A hyperbola, by definition, opens either horizontally or vertically. In our specific case, we deal with a vertical hyperbola.

To find the vertices of the hyperbola, we need to identify the value of \(a\) from the standard form of the equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). In the given equation, \(a^2 = 2\), so \(a = \sqrt{2}\).

This tells us how far the vertices are from the center, which for our equation is located at the origin \((0, 0)\). Hence, we get:

• One vertex at \((0, \sqrt{2})\)
• Another at \((0, -\sqrt{2})\)

These vertices indicate how stretched the hyperbola will be along the vertical axis, which helps in visualizing and drawing it accurately.

Asymptotes

Asymptotes are crucial for understanding how a hyperbola behaves as it extends infinitely. They are imaginary lines that the arms of the hyperbola approach but never touch. In some way, they act as guidelines showing the path the hyperbola follows.

For a hyperbola of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the equations for the asymptotes are determined by the expression \(y = \pm \frac{a}{b}x\). In this exercise:

• \(a = \sqrt{2}\)
• \(b = \sqrt{6}\)

This results in the asymptotes being expressed as:

• \(y = \pm \frac{\sqrt{2}}{\sqrt{6}}x\)
• Simplifying gives us \(y = \pm \frac{1}{\sqrt{3}}x\)

By determining and plotting these asymptotes, one can draw a more accurate hyperbola, as they will guide how the tails of the graph move.

Foci

The foci of a hyperbola are two important points that define its overall shape. Unlike in an ellipse where the sum of distances from the foci to any point on the curve remains constant, in a hyperbola, it is the difference that remains constant.

To find the foci, we use the formula \(c = \sqrt{a^2 + b^2}\). For our given hyperbola:

• \(a^2 = 2\)
• \(b^2 = 6\)

Substituting these into the formula gives:

• \(c = \sqrt{8} = 2\sqrt{2}\)

The foci are located \(c\) units away from the center, along the hyperbola’s axis. Since the hyperbola is vertical, the foci for this specific example will be at:

• \((0, 2\sqrt{2})\)
• \((0, -2\sqrt{2})\)

These points are crucial for plotting the hyperbola’s accurate trajectory and understanding its structural properties.