Problem 13 - Algebra and Trigonometry

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. \(x^{2}-4 y^{2}-8=0\)

Step-by-Step Solution

Answer

Vertices: \((2\sqrt{2}, 0), (-2\sqrt{2}, 0)\); Foci: \((\sqrt{10},0), (-\sqrt{10},0)\); Asymptotes: \(y=\pm\frac{1}{2}x\).

Step 1: Rewrite the Hyperbola Equation

First, bring the given hyperbola equation to the standard form. The equation is \(x^2 - 4y^2 - 8 = 0\). Add 8 to both sides to get: \(x^2 - 4y^2 = 8\). Then, divide every term by 8 to normalize it: \(\frac{x^2}{8} - \frac{y^2}{2} = 1\). Now, the equation is in the standard form of a hyperbola.

Step 2: Identify the Orientation and Parameters

The standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) indicates this is a horizontally-oriented hyperbola. Here, \(a^2 = 8\) and \(b^2 = 2\). Calculate \(a\) by \(a = \sqrt{8} = 2\sqrt{2}\) and \(b = \sqrt{2}\).

Step 3: Find the Center of the Hyperbola

Since the hyperbola's equation is centered at the origin, its center is at \((0,0)\).

Step 4: Calculate the Vertices

For the horizontally-oriented hyperbola, vertices are located \(a\) units left and right from the center. Thus, the vertices are \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\).

Step 5: Calculate the Foci

The formula for the distance to the foci is \(c = \sqrt{a^2 + b^2}\). Substitute and solve: \(c = \sqrt{8 + 2} = \sqrt{10}\). Therefore, the foci are located \((\sqrt{10},0)\) and \((-\sqrt{10}, 0)\).

Step 6: Determine the Asymptotes

For the horizontally-oriented hyperbola, asymptotes have the equations \(y = \pm\frac{b}{a}x\). Substitute \(a\) and \(b\) to get: \(y = \pm\frac{\sqrt{2}}{2\sqrt{2}}x = \pm\frac{1}{2}x\). Thus, the asymptotes are \(y = \frac{1}{2}x\) and \(y = -\frac{1}{2}x\).

Step 7: Sketch the Graph

Using the points from the steps: plot the center at \((0,0)\), place the vertices at \((2\sqrt{2},0)\) and \((-2\sqrt{2},0)\), and draw the asymptotes. Construct the hyperbola opening left and right, approaching but never touching the asymptotes.

Key Concepts

Vertices

In the context of a hyperbola, vertices are key points that help define its shape. They are the points where the hyperbola crosses its transverse axis, which is the line segment that goes through both the center and the vertices. For the given hyperbola equation \[ \frac{x^2}{8} - \frac{y^2}{2} = 1 \]the hyperbola is horizontally oriented.
This means the vertices are located in relation to the horizontal axis. The distance from the center to each vertex is \(a\), where \(a\) is calculated as \(a = \sqrt{8} = 2\sqrt{2}\).

• The center is at \((0,0)\).
• The vertices are \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\).

Understanding the positioning of the vertices aids greatly in sketching and analyzing the hyperbola's overall structure.

Foci

The foci of a hyperbola, similar to ellipses, are key points that lie along the transverse axis and help in forming the hyperbola's shape. In hyperbolas, the foci are located outside the vertices, one on each side, and they are essential in defining the hyperbola’s strict curve.
The distance to the foci from the center is given by \(c\), which is calculated using the formula:\[ c = \sqrt{a^2 + b^2} \]For this hyperbola:

• Calculate \(c\) as \(c = \sqrt{8 + 2} = \sqrt{10}\).
• So, the foci are at \((\sqrt{10}, 0)\) and \((-\sqrt{10}, 0)\).

The foci can be thought of as the guiding points that affect the path of the hyperbolic arms, making them flare outwards.

Asymptotes

In hyperbolas, asymptotes are crucial as they provide a boundary framework that the hyperbola approaches but never actually touches. The asymptotes cross each other at the center and form an 'X' shape, clearly showing the general direction of the hyperbola’s arms.
For a horizontally-oriented hyperbola like ours, the asymptotes are found using the slope \(\frac{b}{a}\):

• The slope here is \(\pm\frac{\sqrt{2}}{2\sqrt{2}} = \pm\frac{1}{2}\).
• Thus, the equations for the asymptotes are \(y = \frac{1}{2}x\) and \(y = -\frac{1}{2}x\).

Asymptotes provide visual guides that help portrait the hyperbola's wings' behavior as they extend outwards.

Standard Form

The standard form of a hyperbola is essential for understanding its properties and graphing it accurately. It gives us immediate information about its orientation and critical values that determine its shape and position. The standard form of the hyperbola is:\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]For the hyperbola in our exercise, we restructured the original equation,\( x^2 - 4y^2 - 8 = 0 \), into this form:\[ \frac{x^2}{8} - \frac{y^2}{2} = 1 \]

• It indicates a horizontal hyperbola because the \(x^2\) term is positive.
• This helps us find \(a^2 = 8\) and \(b^2 = 2\), leading to crucial calculations for vertices, foci, and asymptotes.

Understanding the standard form not only facilitates a better grasp of the hyperbola's structure but also simplifies the calculation of other vital elements such as axes, eccentricity, and graphing.