Problem 19 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Eccentricity is \( \sqrt{2} \), foci are at \( (0, 4) \) and \( (0, -4) \), directrices are \( y = 2 \) and \( y = -2 \).

Step 1: Identify and Rewrite in Standard Form

The given equation of the hyperbola is \( y^2 - x^2 = 8 \). We start by rewriting it in the standard form of a hyperbola. Notice this is already in the form of \( \frac{y^2}{8} - \frac{x^2}{8} = 1 \), which matches the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Here, \( a^2 = 8 \) and \( b^2 = 8 \), thus \( a = \sqrt{8} \text{ and } b = \sqrt{8} \).

Step 2: Calculate the Eccentricity

For a hyperbola represented by \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the eccentricity \( e \) is given by \( e = \sqrt{1 + \frac{b^2}{a^2}} \). Substituting our values, we get \( e = \sqrt{1 + \frac{8}{8}} = \sqrt{2} \).

Step 3: Calculate the Foci

The foci of a hyperbola with a vertical transverse axis are located at \( (0, \pm c) \). The value of \( c \) is found using \( c = e \cdot a \). So, \( c = \sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4 \). Thus, the foci are located at \( (0, 4) \) and \( (0, -4) \).

Step 4: Find the Equations of the Directrices

For a vertical hyperbola, the directrices have the equation \( y = \pm \frac{a^2}{c} \). Substituting \( a^2 = 8 \) and \( c = 4 \), the equations are \( y = \pm \frac{8}{4} = \pm 2 \). So, the directrices are \( y = 2 \) and \( y = -2 \).

Step 5: Graph the Hyperbola

To graph the hyperbola, plot the foci at \( (0, 4) \) and \( (0, -4) \). Draw the vertical transverse axis through these points and sketch the hyperbola branches opening upwards and downwards. Draw the directrices at \( y = 2 \) and \( y = -2 \) as horizontal lines indicating their position relative to the hyperbola.

Key Concepts

Eccentricity

The concept of eccentricity is crucial in understanding hyperbolas. Eccentricity, denoted by \( e \), measures how "stretched" a conic section is. For hyperbolas, eccentricity is always greater than 1.
When given the hyperbola's equation in standard form, \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \),eccentricity is calculated using the formula:

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

In our example, \( a^2 = 8 \) and \( b^2 = 8 \).
Substituting these values, we find:

• \( e = \sqrt{1 + \frac{8}{8}} = \sqrt{2} \)

This depicts a hyperbola that is moderately stretched along its axis.
Eccentricity helps visualize the hyperbola's shape and focus arrangement.

Foci

The foci of a hyperbola are points located along the transverse axis. These points are essential for the curve's definition.
The foci are associated with the eccentricity and further define the hyperbola's properties. In a hyperbola with the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the foci are found using:

• \( c = e \cdot a \)

Thus, substituting \( a = \sqrt{8} \) and \( e = \sqrt{2} \), we calculate:

• \( c = \sqrt{2} \cdot \sqrt{8} = 4 \)

This gives us the foci at locations \( (0, 4) \) and \( (0, -4) \).
The distance between the center and each focus marks the hyperbola's spread along the primary axis.

Directrices

The directrices of a hyperbola are lines parallel to the transverse axis, playing a key role in its definition.
They are related to the hyperbola's eccentricity and help in characterizing the curve's width and shape. For vertically oriented hyperbolas like \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the equation for directrices is:

• \( y = \pm \frac{a^2}{c} \)

For our problem, \( a^2 = 8 \) and \( c = 4 \), leading to:

• \( y = \pm \frac{8}{4} = \pm 2 \)

Thus, the directrices are the lines \( y = 2 \) and \( y = -2 \). They help in structuring the hyperbola's graph, showing where the curve approaches but never touches.

Standard Form

Understanding the standard form of a hyperbola is foundational to studying its features.
A hyperbola's equation, similar to \( y^2 - x^2 = 8 \), can often be arranged into a standard form which aids in identifying its properties.

• For hyperbolas, it generally takes the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

This standard form indicates that the transverse axis is vertical, as the \( y^2 \) term appears first.
By rewriting \( y^2 - x^2 = 8 \) as \( \frac{y^2}{8} - \frac{x^2}{8} = 1 \), both terms become denominated with their respective squared lengths \( a^2 \) and \( b^2 \).
In this way, recognizing and manipulating the standard form equips students with the tools to analyze the hyperbola's structure and key elements like eccentricity, foci, and directrices.