Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation, indicating vertices, foci, and asymptotes (if it is a hyperbola). $$ \frac{x^{2}}{16}-\frac{y^{2}}{4}=1 $$

Step-by-Step Solution

Answer

Sketch a horizontal hyperbola centered at (0,0) with vertices at (4,0) and (-4,0), foci at approximately (4.47,0) and (-4.47,0), and asymptotes at lines y=±1/2x.

Step 1: Identify the Type of Conic Section

The given equation \( \frac{x^{2}}{16}-\frac{y^{2}}{4}=1 \) is in the standard form \( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \), indicating that it represents a hyperbola. In this case, the transverse axis is horizontal.

Step 2: Identify the Vertices

For this hyperbola, the center is at the origin, \((0,0)\). The vertices are located along the x-axis at \(()a,0\) and \((-a,0)\). The value of \(a\) is 4, so the vertices are at \((4,0)\) and \((-4,0)\).

Step 3: Identify the Foci

The foci of the hyperbola are given by \((\pm c, 0)\), where \(c\) is calculated as \(c=\sqrt{a^{2}+b^{2}} = \sqrt{16+4} = \sqrt{20} \approx 4.47\). Thus, the foci are approximately at \((4.47, 0)\) and \((-4.47, 0)\).

Step 4: Identify the Asymptotes

The hyperbola has asymptotes given by the lines \(y=\pm\frac{b}{a}x\). With \( b=2 \) and \( a=4 \), the equations of the asymptotes are \(y=\pm\frac{1}{2}x\).

Step 5: Sketch the Graph

First, plot the vertices at \((4,0)\) and \((-4,0)\). Next, plot the approximate positions of the foci at \((4.47, 0)\) and \((-4.47, 0)\). Then, draw the asymptotes \(y=\pm\frac{1}{2}x\). Finally, sketch the hyperbola opening horizontally, approaching the asymptotes as \(x\rightarrow\pm\infty\).

Key Concepts

Vertices of a hyperbola

In hyperbola equations, vertices play an essential role in defining the shape and position of the curve. The equation \( \frac{x^{2}}{16}-\frac{y^{2}}{4}=1 \) is a horizontal hyperbola, meaning it opens left and right.
The vertices are key points on the hyperbola that touch the axis along which the hyperbola opens.
For this equation, the center is at the origin, \((0,0)\). The term beneath \(x^{2}\), which is \(a^{2}=16\), dictates the positioning of the vertices along the x-axis.

• Calculate \(a\) by taking the square root of \(a^{2}: a=\sqrt{16}=4\).
• The vertices are then located at \((\pm a, 0)\), specifically at \((4,0)\) and \((-4,0)\).

These vertices essentially determine the limits of the hyperbola in its horizontal extent.

Foci of a hyperbola

The foci are points located within the curves of a hyperbola that help define its shape and determine how 'stretched' the hyperbola appears. For the hyperbola given by \( \frac{x^{2}}{16} - \frac{y^{2}}{4} = 1 \), they lie along the transverse axis (which in this case is horizontal).
To find the foci, we need to determine \(c\), where \(c\) is the distance from the center to each focus. The formula is:

• \( c = \sqrt{a^{2} + b^{2}} \)

Here, \( a^{2} = 16 \) and \( b^{2} = 4 \). Calculate \(c\) as follows:

• \( c = \sqrt{16 + 4} = \sqrt{20} \approx 4.47 \)
• Thus, the foci are at \((\pm c, 0)\), approximately positioned at \((4.47, 0)\) and \((-4.47, 0)\).

The foci are crucial for constructing the precise shape of the hyperbola, indicating how wide or narrow it is.

Asymptotes of a hyperbola

Asymptotes are imaginary lines that the hyperbola approaches but never actually touches. They provide a framework that helps in guiding the drawing of the hyperbola. For the equation \( \frac{x^{2}}{16} - \frac{y^{2}}{4} = 1 \), the asymptotes can be found using the slopes derived from coefficients of the equation.
The asymptotes of a hyperbola with a horizontal transverse axis are given by:

• \( y = \pm \frac{b}{a}x \)

Here, \( a = 4 \) and \( b = 2 \). Therefore, calculate the slopes:

• \( \frac{b}{a} = \frac{2}{4} = \frac{1}{2} \)
• The equations of the asymptotes are \( y = \frac{1}{2}x \) and \( y = -\frac{1}{2}x \).

When sketched, the asymptotes form guiding lines along which the hyperbola extends and approaches, providing a boundary for the hyperbolic curves.