Problem 5 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ 4 x^{2}-16 y^{2}=64 $$

Step-by-Step Solution

Answer

Center: (0, 0), Foci: (±2√5, 0), Vertices: (±4, 0), Asymptotes: y = ±1/2x, Eccentricity: √5/2.

Step 1: Write the Equation in Standard Form

The given hyperbola is: \ \[ 4x^{2} - 16y^{2} = 64 \]. \ Divide the entire equation by 64 to form the standard equation: \ \[ \frac{x^{2}}{16} - \frac{y^{2}}{4} = 1 \]. This is in the form: \ \[ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \].

Step 2: Identify the Center

For the hyperbola equation \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), the center is at the origin, \( (0, 0) \).

Step 3: Find the Vertices

The vertices of a hyperbola in the form \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \) are at \( (\pm a, 0) \). \ Here, \( a = \sqrt{16} = 4 \). \ Thus, the vertices are \( (4, 0) \) and \( (-4, 0) \).

Step 4: Determine the Foci

The foci of a hyperbola are calculated using the equation \( c^{2} = a^{2} + b^{2} \). \ Substitute \( a^{2} = 16 \) and \( b^{2} = 4 \) to find \( c \): \ \[ c = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]. \ The foci are \( (\pm 2\sqrt{5}, 0) \).

Step 5: Calculate the Asymptotes

For the hyperbola \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), \( y = \pm \frac{b}{a}x \) are the equations of the asymptotes. \ \( b = \sqrt{4} = 2 \) and \( a = 4 \). \ Thus, the equations are \( y = \pm \frac{2}{4}x = \pm \frac{1}{2}x \).

Step 6: Compute Eccentricity

The eccentricity \( e \) is given by \( e = \frac{c}{a} \). \ Here, \( c = 2\sqrt{5} \) and \( a = 4 \). \ Thus, \( e = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2} \).

Step 7: Graph the Hyperbola

Plot the center at \( (0, 0) \). \ Plot the vertices at \( (4, 0) \) and \( (-4, 0) \). \ Plot the foci at approximately \( (2\sqrt{5}, 0) \) and \( (-2\sqrt{5}, 0) \). \ Draw the asymptotes with equations \( y = \pm \frac{1}{2}x \). Draw the hyperbola opening along the x-axis.

Key Concepts

Standard Form of Hyperbola

The standard form of a hyperbola is crucial for identifying its key features such as vertices, foci, and asymptotes. A hyperbola in its standard form is written as \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \) if it opens horizontally, or \( \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1 \) if it opens vertically. This form is analogous to a hyperbola's blueprint, revealing its structure once the equation is properly arranged.
For instance, starting from an equation like \( 4x^{2} - 16y^{2} = 64 \), you would divide every term by 64 to simplify it to \( \frac{x^{2}}{16} - \frac{y^{2}}{4} = 1 \). This transformation into the standard form allows us to easily identify the necessary parts for further analysis. By comparing it with \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), it's clear that here, \( a^{2} = 16 \) and \( b^{2} = 4 \).

Vertices of Hyperbola

The vertices of a hyperbola are the points where the hyperbola intersects the transverse axis—its main axis of symmetry. For a horizontally oriented hyperbola in the form \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), the vertices are located at \((\pm a, 0)\).
From our standard form \( \frac{x^{2}}{16} - \frac{y^{2}}{4} = 1 \), we have \( a = \sqrt{16} = 4 \). Thus, the vertices of this particular hyperbola are at the coordinates \((4, 0)\) and \((-4, 0)\). These points provide the outer boundaries along the axis that the hyperbola opens to.

Foci of Hyperbola

The foci of a hyperbola are two fixed points used in its formal definition. These points lie along the transverse axis, further from the center than the vertices are. To calculate the distance \( c \) from the center to each focus, you use the equation \( c^{2} = a^{2} + b^{2} \).
In our problem, we have \( a^{2} = 16 \) and \( b^{2} = 4 \), leading to \( c = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \).
Therefore, the foci are situated at \((\pm 2\sqrt{5}, 0)\), meaning each focus is approximately \(4.47\) units from the center along the x-axis. These points are crucial because they help illustrate how the hyperbola is stretched.

Asymptotes of Hyperbola

Asymptotes of a hyperbola are the lines that the hyperbola approaches but never touches. They help define the "direction" of the open arms of the hyperbola. For a hyperbola in the form \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), the equations of the asymptotes are derived as \( y = \pm \frac{b}{a}x \).
Here, \( b = \sqrt{4} = 2 \) and \( a = 4 \), so the asymptotes are represented by \( y = \pm \frac{2}{4}x \) or \( y = \pm \frac{1}{2}x \).
These lines are key to sketching the hyperbola accurately because they show the slant along which the hyperbola opens away from the center point.

Eccentricity of Hyperbola

Eccentricity is a measure of how "stretched" a hyperbola is and is denoted by \( e \). For a hyperbola, the eccentricity is calculated as \( e = \frac{c}{a} \). A larger eccentricity value indicates the hyperbola is more open and stretched farther from the center.
Within the context of our example, the eccentricity is computed with \( c = 2\sqrt{5} \) and \( a = 4 \), giving \( e = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2} \).
Eccentricities of hyperbolas are always greater than 1, representing how they differ structurally from ellipses, which have eccentricities less than 1. This eccentricity value tells us that our hyperbola is moderately stretched along its axis.