Problem 32 - College Algebra

Question

Sketch a graph of the hyperbola, labeling vertices and foci. \(\frac{x^{2}}{64}-\frac{y^{2}}{4}=1\)

Step-by-Step Solution

Answer

Plot vertices at \((-8,0)\) and \((8,0)\), foci at \((\pm\sqrt{68}, 0)\), and sketch the hyperbola with asymptotes \(y=\pm\frac{1}{4}x\).

Step 1: Identify the Type of Hyperbola

The equation given is \(\frac{x^2}{64} - \frac{y^2}{4} = 1\), which is in the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). This represents a horizontal hyperbola because the \(x\)-term is positive.

Step 2: Determine the Transverse and Conjugate Axes

For this hyperbola, the transverse axis is along the \(x\)-axis with length \(2a\), and the conjugate axis is along the \(y\)-axis with length \(2b\). Here, \(a^2=64\) so \(a=8\), and \(b^2=4\) so \(b=2\).

Step 3: Locate the Center of the Hyperbola

The center of the hyperbola is at the origin \((0, 0)\) because the equation is centered at the origin.

Step 4: Calculate and Mark the Vertices

The vertices of a horizontal hyperbola are located \(a\) units from the center on the transverse axis. Thus, the vertices are at \((\pm 8, 0)\).

Step 5: Determine and Mark the Foci

The distance \(c\) from the center to each focus is found using the formula \(c = \sqrt{a^2 + b^2}\). Here, \(c = \sqrt{64 + 4} = \sqrt{68}\). The foci are at \((\pm \sqrt{68}, 0)\).

Step 6: Draw the Asymptotes

The equations of the asymptotes for a horizontal hyperbola are \(y = \pm \frac{b}{a}x\). Thus, the asymptotes are \(y = \pm \frac{2}{8}x\) or \(y = \pm \frac{1}{4}x\).

Step 7: Sketch the Hyperbola

Begin by plotting the center \((0,0)\), then mark the vertices \((\pm8,0)\) and foci \((\pm\sqrt{68},0)\). Draw dashed lines for the asymptotes and sketch the hyperbola opening left and right, approaching but never crossing the asymptotes.

Key Concepts

Horizontal Hyperbola

A hyperbola is a type of conic section that consists of two separate curves, known as branches, which open indefinitely. In the story of hyperbolas, a horizontal hyperbola stands out when the transverse axis stretches along the horizontal or x-axis of a graph. This specific branch orientation occurs when the equation of the hyperbola is written in the form

• \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

In this expression, the positive term appears with the \(x^2\) variable, thus indicating the "horizontal" nature of the hyperbola. The branches of the hyperbola run parallel to the x-axis, distinguishing it from a vertical hyperbola which would have the \(y^2\) term positive instead. This orientation impacts how both the vertices and foci position themselves, which we'll explore shortly.

Vertices of Hyperbola

Vertices are critical points on a hyperbola that lie on the transverse axis and represent the closest points on each branch of the hyperbola to each other. For a horizontal hyperbola like the one described by the equation

• \(\frac{x^2}{64} - \frac{y^2}{4} = 1\)

the vertices are located at a distance 'a' units from the center along the x-axis. To calculate this, you identify \(a\) by taking the square root of \(a^2\), which in this case, \(a^2 = 64\), leading to \(a = 8\).
Thus, the vertices are found at

• \((\pm 8, 0)\)

on the coordinate plane. These denote the horizontal stretches of the hyperbola and help define its overall shape and orientation.

Foci of Hyperbola

The foci of a hyperbola are a pair of points that sit on the transverse axis beyond the vertices. They provide essential insight into the geometric structure and eccentricity of the hyperbola. These points are located using the formula

• \(c = \sqrt{a^2 + b^2}\)

where \(c\) is the distance from the center to each focus. For our example, we determined

• \(c = \sqrt{64 + 4} = \sqrt{68}\)

which approximately equals \(8.25\). Hence, the foci of the hyperbola are positioned at

• \((\pm \sqrt{68}, 0)\)

demonstrating how they extend further than the vertices along the x-axis.

Asymptotes of Hyperbola

Asymptotes are lines that a hyperbola gets infinitely close to but never actually intersects. They serve as boundary guidelines the branches of the hyperbola adhere to. For a horizontal hyperbola, the equations of the asymptotes are

• \(y = \pm \frac{b}{a}x\)

In our specific equation, we have:

• \(b = 2\)
• \(a = 8\)

Thus resulting in the asymptote equations:

• \(y = \pm \frac{1}{4}x\)

These asymptotes help in sketching the hyperbola, showing how the branches will extend and align closely with these linear paths while never crossing them, indicating infinite extension.