Problem 21 - Precalculus: Mathematics for Calculus

Question

An equation of a hyperbola is given. (a) Find the vertices, foci, and asymptotes of the hyperbola. (b) Determine the length of the transverse axis. (c) Sketch a graph of the hyperbola. $$x^{2}-4 y^{2}-8=0$$

Step-by-Step Solution

Answer

Vertices: \((\pm 2\sqrt{2}, 0)\); Foci: \((\pm \sqrt{10}, 0)\); Asymptotes: \(y=\pm\frac{1}{2}x\); Transverse axis: \(4\sqrt{2}\).

Step 1: Rewrite the Equation

Begin by rewriting the given equation in standard form. Given the equation \(x^2 - 4y^2 - 8 = 0\), move the constant to the other side: \ x^2 - 4y^2 = 8. \

Step 2: Standard Form of Hyperbola

Divide every term in the equation by \(8\) to express it in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). This results in: \ \frac{x^2}{8} - \frac{y^2}{2} = 1. \ Now, we have \(a^2 = 8\) and \(b^2 = 2\).

Step 3: Identify Vertices

In the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are at \((\pm a, 0)\). Calculate \(a\): \ a = \sqrt{8} = 2\sqrt{2}. \ Therefore, the vertices are \((\pm 2\sqrt{2}, 0)\).

Step 4: Find Foci

For the hyperbola, the distance to the foci \(c\) satisfies \(c^2 = a^2 + b^2\). Calculate \(c\) using \(a^2 = 8\) and \(b^2 = 2\): \ c^2 = 8 + 2 = 10, \ c = \sqrt{10}. \ Hence, the foci are \((\pm \sqrt{10}, 0)\).

Step 5: Asymptotes

The asymptotes of the hyperbola have equations \(y = \pm \frac{b}{a}x\). Calculate the slopes with \(b = \sqrt{2}\) and \(a = \sqrt{8} = 2\sqrt{2}\): \ \frac{b}{a} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2}. \ Therefore, the asymptotes are \(y = \pm \frac{1}{2}x\).

Step 6: Length of the Transverse Axis

The length of the transverse axis in a hyperbola is \(2a\). Using \(a = 2\sqrt{2}\), calculate: \ 2a = 4\sqrt{2}. \

Step 7: Sketch the Graph

To sketch the hyperbola, plot the vertices \((\pm 2\sqrt{2}, 0)\), foci \((\pm \sqrt{10}, 0)\), and draw the asymptotes \(y = \pm \frac{1}{2}x\). Ensure the hyperbola opens left and right, tracing a curve approaching but never touching the asymptotes.

Key Concepts

Vertices

Vertices of a hyperbola are the points where the hyperbola intersects its transverse axis. This gives us a clear indication of the extreme edges of the hyperbola's shape. In the standard form of a hyperbola \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]the vertices are located at \((\pm a, 0)\). From the solution, we see:

• \(a = \sqrt{8} = 2\sqrt{2}\)
• Thus, the vertices are \((\pm 2\sqrt{2}, 0)\)

These points represent the highest and lowest x-values of the hyperbola on the x-axis.
They act as crucial reference points when drawing or interpreting the graph of a hyperbola.

Foci

The foci of a hyperbola are two fixed points located along the transverse axis, inside the curve.These points are pivotal as they define the hyperbola's shape and direction.
Mathematically, the foci are determined using:\[ c^2 = a^2 + b^2 \]For the hyperbola in question, we have:

• \(a^2 = 8\)
• \(b^2 = 2\)
• Thus, \(c^2 = 10\) and \(c = \sqrt{10}\)
• The foci are \((\pm \sqrt{10}, 0)\)

These foci lie beneath the vertices on the x-axis and contribute to the uniqueness of the hyperbola, influencing its opening and spanning both sides.

Asymptotes

Asymptotes are the lines that a hyperbola approaches but never actually touches. They give directionality and width to a hyperbola's curve, presenting the wayit stretches into infinity.
The equations of the asymptotes for a horizontal hyperbola in standard form are:\[ y = \pm \frac{b}{a}x \]In our hyperbola:

• \(b = \sqrt{2}\)
• \(a = 2\sqrt{2}\)
• The slopes of the asymptotes become \( \pm \frac{1}{2} \)
• The equations are \(y = \pm \frac{1}{2}x\)

These lines dictate the framework of the hyperbola's shape, ensuring it never intersects them.
It's helpful to sketch these lines first when graphing the hyperbola, setting the basic structure for the curve's path.

Transverse Axis

The transverse axis is the line segment that runs through the center and vertices of the hyperbola.
Understanding its length helps to grasp the overall size and spread of the hyperbola.
For a horizontal hyperbola like the one here, the transverse axis lies along the x-axis.The length of this axis can be expressed as:\[ 2a \]In the given hyperbola:

• \(a = 2\sqrt{2}\)
• The length is \(2 \, a = 4\sqrt{2}\)

The transverse axis is essential for determining the size and extent of the hyperbola.
The vertices lie at either end of the transverse axis, providing a span of the hyperbola over the coordinate plane.

Graph of Hyperbola

Graphing a hyperbola involves piecing together several elements, ensuring the curve is correctly positioned and shaped.
Key steps include:

• Identifying vertices: \((\pm 2\sqrt{2}, 0)\)
• Locating foci: \((\pm \sqrt{10}, 0)\)
• Sketching asymptotes: \(y = \pm \frac{1}{2}x\)

Start by plotting the vertices and foci on the coordinate plane.
Then draw the asymptotes, extending through the origin with the correct slope.
Finally, sketch the hyperbola's two arms, beginning at each vertex, curving outward toward the asymptotes.The graph should clearly show that the hyperbola's arms open left and right, tracing paths closer to the asymptotes as they extend.
This visual representation allows for a deeper understanding of the hyperbola's nature and form.