Problem 27 - Thomas Calculus

Question

In Exercises \(23-30\) , 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

Eccentricity: \(\sqrt{5}\), Foci: (\(\pm \sqrt{10}, 0\)), Directrices: \(x = \pm \frac{\sqrt{10}}{5}\).

Step 1: Rewrite the Equation

To handle the hyperbola equation, we first rewrite the given equation to standard form. Divide every term of the equation by 16 to simplify it: \[ \frac{8x^2}{16} - \frac{2y^2}{16} = 1 \] This reduces to: \[ \frac{x^2}{2} - \frac{y^2}{8} = 1 \].

Step 2: Identify the Elements

The standard form of a hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) for horizontal hyperbolas. From \(\frac{x^2}{2} - \frac{y^2}{8} = 1\), we identify that \(a^2 = 2\) and \(b^2 = 8\).

Step 3: Calculate the Eccentricity

The eccentricity (\(e\)) of a hyperbola is calculated using \(e = \sqrt{1 + \frac{b^2}{a^2}}\). 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 Foci

The foci (c) of a hyperbola are found using \(c = ae\). Thus, \(c = \sqrt{2} \times \sqrt{5} = \sqrt{10}\). Since it's a horizontal hyperbola, the foci are at (\(\pm \sqrt{10}, 0\)).

Step 5: Find the Directrices

The directrices can be located at \(\pm \frac{a^2}{c}\). Here, \(a^2 = 2\) and \(c = \sqrt{10}\), so the directrices are at \(x = \pm \frac{2}{\sqrt{10}}\), which simplifies to \(x = \pm \frac{\sqrt{10}}{5}\).

Step 6: Sketch the Graph

Plot the hyperbola with its vertices along the x-axis. Include the foci at (\(\pm \sqrt{10}, 0\)) and draw the lines parallel to the y-axis through the directrices at \(x = \pm \frac{\sqrt{10}}{5}\).

Key Concepts

Eccentricity

The eccentricity of a hyperbola is a measure that indicates how much the shape of the hyperbola deviates from being circular. This concept helps us understand the "stretch" or "flatness" of the hyperbola in comparison to a circle. It is represented by the symbol \(e\). For a hyperbola, the eccentricity is always greater than 1.
To calculate the eccentricity \(e\), we use the formula:

• \( e = \sqrt{1 + \frac{b^2}{a^2}} \)

In this formula, \(a^2\) and \(b^2\) are derived from the standard form of the hyperbola equation, where \(a\) represents the distance from the center to each vertex along the major axis, and \(b\) is related to the distance along the conjugate axis.
In the example provided, \(a^2 = 2\) and \(b^2 = 8\), so plugging these values into the expression gives us \(e = \sqrt{5}\). This shows that our hyperbola is significantly more stretched out compared to a circle.

Foci of a Hyperbola

In any hyperbola, the foci are crucial points because they help to define its shape. These are two points located along the interior side of each curve arm — imaginary lines that create the structure of a hyperbola.
The foci distance \(c\) from the center is calculated using the formula:

• \( c = a \times e \)

We already know that \(e = \sqrt{5}\) and \(a = \sqrt{2}\). So, the foci \(c\) can be calculated as \( c = \sqrt{2} \times \sqrt{5} = \sqrt{10} \).
For a horizontal hyperbola, the foci are located at the coordinates \((\pm c, 0)\), meaning that in this case, the foci are placed at \( (\pm \sqrt{10}, 0) \). These serve as anchors to the hyperbola's shape, making it uniquely distinct from other conic sections like ellipses.

Standard Form of Hyperbola

The standard form of a hyperbola is crucial for identifying and calculating its various characteristics. This form simplifies complex equations, making it straightforward to find features such as the center, vertices, axes, and more.
For horizontal hyperbolas, the standard form is:

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

This equation represents hyperbolas oriented along the x-axis. In our specific problem, the equation is given as \( \frac{x^2}{2} - \frac{y^2}{8} = 1 \). We can simplify this by identifying \(a^2 = 2\) and \(b^2 = 8\); these values help us in calculating other elements like eccentricity and positions of foci.
In cases where the y-term comes first

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

this represents vertical hyperbolas. Thus, knowing the standard form aids in initial understanding and setup for further calculations.

Directrices of a Hyperbola

Directrices serve as reference lines that help in understanding the geometric relationship of a hyperbola with its foci. In a hyperbola, there are two directrices, and they are positioned at equal distances on either side of the center.
The position of the directrices is determined by:

• \( x = \pm \frac{a^2}{c} \)

Given \(a^2 = 2\) and \(c = \sqrt{10}\) from our exercise, we find the directrices at \(x = \pm \frac{2}{\sqrt{10}}\). Simplifying, we have them located at \(x = \pm \frac{\sqrt{10}}{5}\).
These lines are not part of the hyperbola itself but help illustrate the pathway of light or sound waves reflecting off the curve in physics or engineering contexts.