Problem 21 - Thomas Calculus

Question

In Exercises \(17-24\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$8 x^{2}-2 y^{2}=16$$

Step-by-Step Solution

Answer

The eccentricity is \(\sqrt{5}\), foci are \((\pm \sqrt{10}, 0)\), and directrices are \(x = \pm \frac{2}{\sqrt{10}}\).

Step 1: Write the Equation in Standard Form

To start solving the equation, first write it in the standard form of a hyperbola. Divide each term by 16 to simplify: \( \frac{x^2}{2} - \frac{y^2}{8} = 1 \). Now, it is in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \(a^2 = 2\) and \(b^2 = 8\).

Step 2: Identify the Values of a and b

Find the values of \(a\) and \(b\) from \(a^2\) and \(b^2\):\[ a = \sqrt{2} \] \[ b = \sqrt{8} = 2\sqrt{2} \].

Step 3: Calculate the Eccentricity e

For a hyperbola, the eccentricity \(e\) is calculated by \(e = \sqrt{1 + \frac{b^2}{a^2}}\). Substitute \(a^2 = 2\) and \(b^2 = 8\) into the formula: \[ e = \sqrt{1 + \frac{8}{2}} = \sqrt{1 + 4} = \sqrt{5} \].

Step 4: Find the Coordinates of the Foci

In a hyperbola \(x^2/a^2 - y^2/b^2 = 1 \), the foci are \((\pm ae, 0)\). Calculate \(ae = a\sqrt{5}= \sqrt{2}\sqrt{5}=\sqrt{10}\). The foci are at \((\pm \sqrt{10}, 0)\).

Step 5: Identify the Equations of the Directrices

The directrices of the hyperbola are found using the equation \(x = \pm \frac{a^2}{c}\), where \(c = a \sqrt{5}\). Substitute in the values: \[ x = \pm \frac{2}{\sqrt{10}} = \pm \frac{\sqrt{10}}{5} \], rationalized: \[ \approx \pm 0.632 \].

Step 6: Graph the Hyperbola, Foci, and Directrices

Graph the hyperbola based on the equation \(\frac{x^2}{2} - \frac{y^2}{8} = 1\) with vertices at \((\pm \sqrt{2}, 0)\). Locate the foci at \(( \pm \sqrt{10}, 0)\) and sketch the vertical lines representing the directrices at \(x = \pm \frac{\sqrt{10}}{5}\). Ensure all points and lines are clearly marked to visualize the structure of the hyperbola.

Key Concepts

Standard Form of a Hyperbola

When dealing with hyperbolas, the standard form of the equation is crucial for identifying key characteristics of the hyperbola. The standard form for a hyperbola oriented with its transverse axis along the x-axis is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). For hyperbolas oriented with the transverse axis along the y-axis, the standard form is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).

This form allows students to easily see the orientation and the dimensions of the hyperbola. The terms \( a^2 \) and \( b^2 \) help determine the distances from the center to the vertices and the co-vertices, respectively.

In the given exercise, converting the original equation \( 8x^2 - 2y^2 = 16 \) to its standard form \( \frac{x^2}{2} - \frac{y^2}{8} = 1 \) revealed that \( a^2 = 2 \) and \( b^2 = 8 \). This implies that the hyperbola is oriented along the x-axis, with its vertices located at \( (\pm \sqrt{2}, 0) \).

Eccentricity of Conic Sections

The eccentricity \( e \) of a hyperbola gives insight into its shape, specifically how much it deviates from a perfect circle. The general formula to find eccentricity for a hyperbola is \( e = \sqrt{1 + \frac{b^2}{a^2}} \).

For the given hyperbola \( \frac{x^2}{2} - \frac{y^2}{8} = 1 \), substituting the values \( a^2 = 2 \) and \( b^2 = 8 \) into this formula gives \( e = \sqrt{1 + \frac{8}{2}} = \sqrt{5} \).

An eccentricity greater than 1 always characterizes hyperbolas, which means they are not circular but instead two separate curved sections moving away from one another. The larger the value of \( e \), the "flatter" or more "stretched" the hyperbola will appear.

Hyperbola Graphing

Graphing a hyperbola involves plotting its key features, such as the vertices, foci, and directrices.

To graph the hyperbola given by \( \frac{x^2}{2} - \frac{y^2}{8} = 1 \), start by identifying and plotting the vertices at \( (\pm \sqrt{2}, 0) \). Vertices mark the closest points of the hyperbola to its center.

Next, locate the foci at \( (\pm \sqrt{10}, 0) \), since they are essential for understanding how the hyperbola "opens." As hyperbolas extend infinitely, foci help in sketching its path. Finally, sketch the asymptotes by extending lines through the origin with slopes \( \frac{b}{a} \) and \( -\frac{b}{a} \), providing a guideline for the curve.

Mark the directrices to visualize their influence on the graph's structure, completing the graph's holistic view.

Directrix of Hyperbola

In the context of hyperbolas, the directrix is a straight line that, along with the foci, helps define the curve. Typically, a hyperbola has two directrices corresponding to its foci.

The position of the directrix for a hyperbola in standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by \( x = \pm \frac{a^2}{c} \), where \( c \) is the distance from the center to a focus, calculated as \( c = a \cdot e \).

In our example, with \( a^2 = 2 \) and \( c = \sqrt{10} \), this results in the directrices being located at \( x = \pm \frac{2}{\sqrt{10}} \). Rationalizing this expression gives \( \pm 0.632 \), indicating their proximity to the y-axis is not overly distant, therefore closely influencing the curvature of the hyperbola.

Understanding directrices in hyperbolas unveils the dynamic relationship between points on these lines and the foci, further emphasizing the geometric nature of conic sections.