Problem 29 - Thomas Calculus

Question

In Exercises \(23-30\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$ 8 y^{2}-2 x^{2}=16 $$

Step-by-Step Solution

Answer

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

Step 1: Simplify the equation

The given equation is \( 8y^2 - 2x^2 = 16 \). Divide every term by 16 to simplify it into standard form. The equation becomes \( \frac{y^2}{2} - \frac{x^2}{8} = 1 \). This represents a hyperbola centered at the origin with the transverse axis along the y-axis.

Step 2: Identify hyperbola parameters

From the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), identify \( a^2 = 2 \) and \( b^2 = 8 \). Thus, \( a = \sqrt{2} \) and \( b = \sqrt{8} = 2\sqrt{2} \).

Step 3: Calculate the eccentricity

For hyperbolas, the eccentricity \( e \) is calculated as \( e = \sqrt{1 + \frac{b^2}{a^2}} \). Substitute the values to get \( e = \sqrt{1 + \frac{8}{2}} = \sqrt{5} \). The eccentricity is \( e = \sqrt{5} \).

Step 4: Find the foci

The coordinates for the foci of a hyperbola are \( (0, \pm c) \) for a hyperbola opening vertically. Find \( c \) using \( c = ae = \sqrt{2} \times \sqrt{5} = \sqrt{10} \). Thus, the foci are at \( (0, \pm \sqrt{10}) \).

Step 5: Determine and graph the directrices

The equations for the directrices are \( y = \pm \frac{a^2}{c} \). Calculate \( \frac{a^2}{c} = \frac{2}{\sqrt{10}} = \frac{\sqrt{10}}{5} \) after rationalizing the denominator. Thus, the directrices are \( y = \pm \frac{\sqrt{10}}{5} \). To graph, plot the foci and draw the lines for the directrices parallel to the x-axis at these y-values.

Key Concepts

Eccentricity

In hyperbolas, eccentricity measures how "stretched" the conic section is. It is a crucial parameter that helps in distinguishing hyperbolas from other conic sections like ellipses or circles. Eccentricity, denoted as \( e \), of a hyperbola is always greater than 1. In this problem, it is calculated using the formula:

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

Here, \( a \) and \( b \) are derived from the equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
For our specific hyperbola with \( a^2 = 2 \) and \( b^2 = 8 \), substituting these values gives \( e = \sqrt{5} \).
This shows that the hyperbola is significantly elongated along its transverse axis.

Foci

The foci of a hyperbola are points that lie on the transverse axis and are symmetrical about the center of the hyperbola. For a hyperbola oriented along the y-axis, like the one given, the foci have coordinates \( (0, \pm c) \). The value of \( c \) defines how far apart these foci are from the origin.

• Calculate \( c \) as \( c = ae = \sqrt{2} \times \sqrt{5} = \sqrt{10} \)
• The foci are \( (0, \pm \sqrt{10}) \)

These foci are crucial in defining the hyperbola's geometric properties and are not located on the curve itself but help form it.
Each point on the hyperbola maintains a constant difference in distance from each focus.

Directrices

The directrices of a hyperbola are lines parallel to the conjugate axis that help define its shape. For the equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the directrices are given by the equations:

• \( y = \pm \frac{a^2}{c} \)

In this scenario, after computing \( c = \sqrt{10} \) and \( a^2 = 2 \), the directrices are found by:

• \( y = \pm \frac{2}{\sqrt{10}} = \pm \frac{\sqrt{10}}{5} \)

The directrices help in constructing the hyperbola's asymptotes and are an extension of its geometric property.

Graphing Hyperbolas

Graphing a hyperbola requires plotting several components: the center, the foci, the vertices, and the directrices. Given the equation \( \frac{y^2}{2} - \frac{x^2}{8} = 1 \), the graph centers at the origin. Here, the transverse axis is along the y-axis.
To accurately portray the hyperbola:

• Identify the vertices at \( (0, \pm \sqrt{2}) \).
• Plot the foci points \( (0, \pm \sqrt{10}) \).
• Draw the directrices at \( y = \pm \frac{\sqrt{10}}{5} \).

By connecting these points and considering the asymptotic behavior (lines through opposite vertices of the rectangle formed by \( a \) and \( b \)), a complete and accurate graph of the hyperbola is created.
This visual representation helps in better understanding the structure and properties of the hyperbola.