Problem 8 - Precalculus: Functions and Graphs

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$x^{2}-2 y^{2}=8$$

Step-by-Step Solution

Answer

Vertices: \((\pm 2\sqrt{2}, 0)\), Foci: \((\pm 2\sqrt{3}, 0)\), Asymptotes: \(y = \pm \frac{\sqrt{2}}{2}x\).

Step 1: Identify the Hyperbola Equation

The given equation is \( x^{2} - 2y^{2} = 8 \). We need to express this in the standard form of a hyperbola. A standard hyperbola with a horizontal transverse axis is described by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). First, divide the entire equation by 8: \( \frac{x^2}{8} - \frac{y^2}{4} = 1 \). This is the standard form.

Step 2: Determine the Vertices

In the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are located at \((\pm a, 0)\). Here \( a^2 = 8 \), so \( a = \sqrt{8} = 2\sqrt{2} \). Thus, the vertices are at \((\pm 2\sqrt{2}, 0)\).

Step 3: Find the Foci

For the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), foci are located at \((\pm c, 0)\) where \( c^2 = a^2 + b^2 \). Since \( b^2 = 4 \), \( c^2 = 8 + 4 = 12 \), so \( c = \sqrt{12} = 2\sqrt{3} \). Therefore, the foci are at \((\pm 2\sqrt{3}, 0)\).

Step 4: Determine Asymptotes Equations

The asymptotes for a hyperbola with a horizontal transverse axis are given by \( y = \pm \frac{b}{a}x \). Here, \( a = \sqrt{8} \) and \( b = 2 \), so \( \frac{b}{a} = \frac{2}{\sqrt{8}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \). Thus, the asymptote equations are \( y = \pm \frac{\sqrt{2}}{2}x \).

Step 5: Sketch the Graph

To sketch the hyperbola, plot the vertices at \((\pm 2\sqrt{2}, 0)\) and draw the asymptotes through the origin with slopes \( \pm \frac{\sqrt{2}}{2} \). Locate the foci at \((\pm 2\sqrt{3}, 0)\) and draw the hyperbola opening horizontally. Ensure the curves approach the asymptotes but never intersect them.

Key Concepts

Vertices

When working with hyperbolas, understanding the concept of vertices is crucial. Vertices are the points on the hyperbola that are closest to each other, located on the major axis.

• The distance between the vertices is known as the transverse axis, which is a principal feature of the hyperbola.
• In the standard form of a hyperbola equation, \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with a horizontal transverse axis, the vertices are found at \((\pm a, 0)\).
• From our exercise, it was determined that \( a^2 = 8 \), leading to \( a = 2\sqrt{2} \).
• Therefore, the vertices of our given hyperbola are located at \((\pm 2\sqrt{2}, 0)\).

Locating these vertices aids in sketching the hyperbola and understanding its overall shape.

Foci

The foci of a hyperbola are points inside the curves that help define its shape. Unlike the vertices, which are merely points on the hyperbola, the foci play a crucial role in determining its geometry.

• In the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the foci are located at \((\pm c, 0)\).
• Here, \( c \) is determined by the equation \( c^2 = a^2 + b^2 \).
• Given that \( a^2 = 8 \) and \( b^2 = 4 \), we find \( c^2 = 12 \), so \( c = 2\sqrt{3} \).
• This indicates that the foci for our hyperbola are at \((\pm 2\sqrt{3}, 0)\).

The foci are particularly important when sketching the hyperbola, as the curves are stretched out towards these points, but never actually pass through them.

Asymptotes

Asymptotes are lines that the hyperbola gets closer to but never touches or intersects. They provide a sense of direction to the shape of the hyperbola.

• In the equation of a horizontal hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), asymptotes can be expressed as \( y = \pm \frac{b}{a}x \).
• From our exercise, \( a = \sqrt{8} \) and \( b = 2 \), thus \( \frac{b}{a} = \frac{\sqrt{2}}{2} \).
• The equations of the asymptotes are \( y = \pm \frac{\sqrt{2}}{2}x \).

Visible on the sketch, these lines provide a visual guide for the hyperbola's curves, showing how the hyperbola forms around these straight lines without crossing them.

Standard Form

Understanding the standard form of a hyperbola is essential for problem-solving and graph sketching. The standard form for a hyperbola with a horizontal transverse axis is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

• This form allows you to easily identify important parameters such as the distance between the vertices (\(a\)) and the direction of the asymptotes (\(b/a\)).
• In order to identify the standard form, the given equation \( x^2 - 2y^2 = 8 \) was rearranged by dividing every term by 8.
• The resulting equation \( \frac{x^2}{8} - \frac{y^2}{4} = 1 \) fits into the standard form, highlighting that this is a horizontal hyperbola.

Using the standard form allows for clear and concise calculations of a hyperbola's most essential features, facilitating their application in graph sketching.

Graph Sketching

Graph sketching involves accurately plotting the hyperbola, its vertices, foci, and asymptotes to visualize its geometric properties.

• Begin by marking the vertices at \((\pm 2\sqrt{2}, 0)\) on the Cartesian plane.
• Next, draw the asymptotes, which pass through the origin and have slopes \( \pm \frac{\sqrt{2}}{2} \).
• The foci, located at \((\pm 2\sqrt{3}, 0)\), are plotted inside the curve and guide the shape of the hyperbola.
• Finally, sketch the hyperbola, ensuring the curves open horizontally, towards the direction of the foci, and approach but never touch the asymptotes.

By following these steps, one can create a precise and informative visual representation of the hyperbola, aiding in the understanding of its behavior and properties.