Problem 25 - Thomas Calculus

Question

In Exercises \(23-30\) , 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 the hyperbola's standard form

The given equation is \( y^2 - x^2 = 8 \). This needs to be compared with the standard form of a hyperbola, which is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Here, we rewrite the equation as \( \frac{y^2}{8} - \frac{x^2}{8} = 1 \). Thus, \( a^2 = 8 \) and \( b^2 = 8 \).

Step 2: Calculate the eccentricity

The formula for the eccentricity \( e \) of a hyperbola is \( e = \sqrt{1 + \frac{b^2}{a^2}} \). Substitute \( a^2 = 8 \) and \( b^2 = 8 \) to get \( e = \sqrt{1 + \frac{8}{8}} = \sqrt{2} \). So, the eccentricity is \( \sqrt{2} \).

Step 3: Find the coordinates of the foci

For a hyperbola centered at the origin with the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the foci are located at \( (0, \pm c) \), where \( c = \sqrt{a^2 + b^2} \). Here, \( c = \sqrt{8 + 8} = \sqrt{16} = 4 \). Therefore, the foci are at \( (0, 4) \) and \( (0, -4) \).

Step 4: Determine the equations of the directrices

For a hyperbola, the vertical directrices are given by \( y = \pm \frac{a^2}{c} \). With \( a^2 = 8 \) and \( c = 4 \), substitute to find \( y = \pm \frac{8}{4} = \pm 2 \). Therefore, the equations of the directrices are \( y = 2 \) and \( y = -2 \).

Step 5: Graph the elements of the hyperbola

To graph, plot the hyperbola given by \( \frac{y^2}{8} - \frac{x^2}{8} = 1 \), noting that it opens vertically. The foci at \( (0, 4) \) and \( (0, -4) \) are points inside the hyperbola. The directrices \( y = 2 \) and \( y = -2 \) are horizontal lines to be shown outside the hyperbola, indicating where the hyperbola points more steeply away from the center.

Key Concepts

Eccentricity

Eccentricity is a crucial value used in understanding the shape and nature of a hyperbola. It is denoted by the symbol \(e\). In general terms, eccentricity measures how much a conic section deviates from being circular. For a hyperbola, the eccentricity is always greater than 1.

To find the eccentricity of a hyperbola, you can use the formula \( e = \sqrt{1 + \frac{b^2}{a^2}} \), where \(a\) and \(b\) are the semi-major and semi-minor axes of the hyperbola, respectively. These are derived from the standard form equation of the hyperbola.

In the original problem, we determined the values of \(a^2 = 8\) and \(b^2 = 8\). Substituting these into the formula gives \( e = \sqrt{1 + \frac{8}{8}} = \sqrt{2} \), indicating that this hyperbola's shape is extended and thin, characteristic of hyperbolas with higher eccentricity values. Understanding the eccentricity helps grasp how broad or narrow the hyperbola appears.

Foci of a Hyperbola

The foci (plural of focus) are specific points related to any conic section, including hyperbolas. These are the points inside the hyperbola that help define its shape. For a hyperbola centered at the origin, the foci are located at \((0, \pm c)\), where \(c = \sqrt{a^2 + b^2}\).

The values of \(a^2 = 8\) and \(b^2 = 8\) as given in the exercise allow us to calculate \(c\). Specifically, \(c = \sqrt{8 + 8} = \sqrt{16} = 4\). Hence, the foci are at \((0, 4)\) and \((0, -4)\).

The significance of the foci is profound in the context of the hyperbola's geometric properties. Any point on the hyperbola is such that the absolute difference of its distances to the two foci is constant. This differential property elucidates why hyperbolas have their unique shape, distinguishing them from other conic sections.

Directrices of a Hyperbola

Directrices are lines associated with the hyperbola, used to define its geometric nature along with the foci. They are lines that help those studying conic sections to recognize how the shape flares outward.

For vertical hyperbolas, like the one in our problem, the directrices are horizontal lines given by \( y = \pm \frac{a^2}{c} \). With \(a^2 = 8\) and \(c = 4\), we find the directrices to be \( y = \pm 2 \).

These directrices provide a boundary reference for the slope of the asymptotic directions of the hyperbola. They are positioned such that the hyperbola approaches these lines as you move further away from the center along the transverse axis. Directrices thus help understand the hyperbola's 'opening' configuration, acting in tandem with the foci to elucidate the full picture of the hyperbola's construct.