Problem 13 - Precalculus: Mathematical for Calculus

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: Identify the General Form

We start with the hyperbola equation: \(x^2 - 4y^2 - 8 = 0\). The standard form of a hyperbola centered at the origin is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Our task is to rewrite the equation in this form.

Step 2: Rewrite in Standard Form

First, add 8 to both sides of the equation to get its standard form: \(x^2 - 4y^2 = 8\). Next, divide every term by 8: \(\frac{x^2}{8} - \frac{y^2}{2} = 1\). The equation is now in the standard form: \(\frac{x^2}{8} - \frac{y^2}{2} = 1\).

Step 3: Identify Parameters

From the standard form \(\frac{x^2}{8} - \frac{y^2}{2} = 1\), we can identify the parameters: \(a^2 = 8\) and \(b^2 = 2\). Solve for \(a\) and \(b\): \(a = \sqrt{8} = 2\sqrt{2}\), \(b = \sqrt{2}\).

Step 4: Find the Vertices

Since the hyperbola is oriented along the x-axis (horizontal), the vertices will be at \((\pm a, 0)\). Thus, the vertices are \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\).

Step 5: Find the Foci

To find the foci, we need \(c\), where \(c^2 = a^2 + b^2\). Calculate \(c^2 = 8 + 2 = 10\), so \(c = \sqrt{10}\). The foci are at \((\pm \sqrt{10}, 0)\).

Step 6: Determine the Asymptotes

The asymptotes for a hyperbola oriented along the x-axis are given by the equation \(y = \pm \frac{b}{a}x\). Substituting \(b = \sqrt{2}\) and \(a = 2\sqrt{2}\), the equations are: \(y = \pm \frac{1}{2}x\).

Step 7: Sketch the Graph

Draw the coordinate axes. Plot the vertices at \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\). Draw the asymptotes: \(y = \pm \frac{1}{2}x\). The hyperbola opens left and right from the vertices, approaching but never touching the asymptotes.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are crucial in defining its shape and orientation. They lie on the main axis, determining the direction of the hyperbola's "open ends." For a hyperbola in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), when it is oriented along the x-axis (horizontal), the vertices are at \( (\pm a, 0) \). In this exercise, the value of \( a \) has been calculated as \( 2\sqrt{2} \), hence the vertices are located at \( (2\sqrt{2}, 0) \) and \( (-2\sqrt{2}, 0) \). This means the hyperbola stretches along the x-axis, opening to the left and right from the vertices, which are equidistant from the origin.

Foci of Hyperbola

The foci of a hyperbola are points around which the curves bend, located further out along the main axis than the vertices. They are essential in the hyperbola's definition via the constant difference of distances from any point on the hyperbola to the foci.
For this hyperbola, the distance from the center to each focus, \( c \), is found using the relationship \( c^2 = a^2 + b^2 \). Here, \( a^2 = 8 \) and \( b^2 = 2 \), giving us \( c = \sqrt{10} \). Thus, the foci are at \( (\pm \sqrt{10}, 0) \). The foci are further out along the x-axis than the vertices, since \( \sqrt{10} > 2\sqrt{2} \).

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never touches. They guide the opening branches of the hyperbola. For a hyperbola with a horizontal transverse axis in standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the asymptotes are defined by the equations \( y = \pm \frac{b}{a}x \).
In this case, with \( a = 2\sqrt{2} \) and \( b = \sqrt{2} \), the equations simplify to \( y = \pm \frac{1}{2}x \). These asymptotes pass through the origin and have slopes of \( \pm \frac{1}{2} \), showing us how the hyperbola spreads and the direction in which it opens. As the hyperbola's arms extend outwards, they stretch towards these lines without ever intersecting them.

Graphing Hyperbolas

Graphing hyperbolas involves plotting essential points and lines, which include the vertices, foci, and asymptotes. This process provides a visual understanding of the hyperbola's structure. First, identify and plot the vertices \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\) on the graph, which marks the extent of the hyperbola along its transverse axis.
Then, sketch the asymptotes, which are \( y = \pm \frac{1}{2}x \). These will appear as straight, diagonal lines crossing through the origin. The hyperbola will open from the vertices towards the direction of these asymptotes. Although it comes very close, the hyperbola will never actually touch these lines. This graphical representation is crucial in visualizing the specific characteristics that define this hyperbola.

Hyperbola Standard Form

The standard form of a hyperbola makes it simpler to identify key features, such as the vertices and asymptotes. It provides a clear structure, represented by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) for hyperbolas oriented along the x-axis. For the given equation \( x^2 - 4y^2 - 8 = 0 \), we rearrange it to standard form by first adding 8 to both sides, resulting in \( x^2 - 4y^2 = 8 \).
Then, divide each term by 8 to get \( \frac{x^2}{8} - \frac{y^2}{2} = 1 \). This form reveals that \( a^2 = 8 \) and \( b^2 = 2 \), helping us find the vertices, foci, and asymptotes. Understanding this standard form is essential before graphing a hyperbola, as it dictates how and where the curves are placed on the coordinate plane.