Problem 20 - Thomas Calculus

Question

In Exercises \(17-24\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$y^{2}-x^{2}=4$$

Step-by-Step Solution

Answer

Eccentricity: \( \sqrt{2} \), Foci: (0, \( 2\sqrt{2} \)) and (0, \( -2\sqrt{2} \)), Directrices: \( y = \sqrt{2} \) and \( y = -\sqrt{2} \).

Step 1: Identify the Canonical Form

The given equation of the hyperbola is \( y^2 - x^2 = 4 \). This can be rewritten in the standard form of a hyperbola centered at the origin as \( \frac{y^2}{4} - \frac{x^2}{4} = 1 \), which resembles the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Here, \( a^2 = 4 \) and \( b^2 = 4 \), so \( a = 2 \) and \( b = 2 \).

Step 2: Calculate the Eccentricity

The eccentricity \( e \) of a hyperbola is given by the formula \( e = \sqrt{1 + \frac{b^2}{a^2}} \). Substituting the given values, we get \( e = \sqrt{1 + \frac{4}{4}} = \sqrt{2} \). Therefore, the eccentricity of the hyperbola is \( \sqrt{2} \).

Step 3: Find the Foci

For hyperbolas, the foci are located along the axis of symmetry. The distance from the center to each focus \( c \) is found using the formula \( c^2 = a^2 + b^2 \), which means \( c^2 = 4 + 4 = 8 \) and \( c = 2\sqrt{2} \). The foci for this hyperbola are at \( (0, \pm 2\sqrt{2}) \).

Step 4: Find the Directrices

The directrices of a hyperbola with a vertical transverse axis can be found using the formula \( y = \pm \frac{a}{e} \). Substituting the values, \( y = \pm \frac{2}{\sqrt{2}} = \pm \sqrt{2} \). Thus, the equations of the directrices are \( y = \sqrt{2} \) and \( y = -\sqrt{2} \).

Step 5: Sketch the Graph

Plot the center of the hyperbola at the origin (0,0). The foci are at points (0, \( 2\sqrt{2} \)) and (0, \( -2\sqrt{2} \)). The directrices are the horizontal lines \( y = \sqrt{2} \) and \( y = -\sqrt{2} \). Draw the hyperbola opening upwards and downwards with vertices at (0, 2) and (0, -2).

Key Concepts

Eccentricity of a Hyperbola

The eccentricity (\( e \)) of a hyperbola is a measure of how "stretched" or "flat" the hyperbola appears. It's a key feature that helps in distinguishing the shape of a hyperbola from other conic sections. For a hyperbola, the eccentricity is always greater than 1 because of the hyperbola's unique open shape.

To calculate the eccentricity of a hyperbola, we use the formula:

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

Here, \( a \) and \( b \) are the distances from the center to the vertices and the co-vertices of the hyperbola, respectively. In our example with the equation \( y^2 - x^2 = 4 \), after converting it to its canonical form, we found that \( a = 2 \) and \( b = 2 \).

Plug these values into the formula to find that the eccentricity is \( \sqrt{2} \), indicating that the curves are indeed hyperbolic.

Foci of a Hyperbola

The foci of a hyperbola are two fixed points that lie along the axis of symmetry of the hyperbola. Points on the hyperbola have the unique property that the difference in their distances to the foci is constant. This fixed difference is what gives the hyperbola its distinct shape.

To determine the position of the foci, we use the relationship:

• \( c^2 = a^2 + b^2 \)

Here, \( c \) represents the distance from the center to each focus. For our hyperbola \( y^2 - x^2 = 4 \), we compute \( c \) by finding \( c = \sqrt{8} = 2\sqrt{2} \).

Therefore, the foci are located at the points \((0, \pm 2\sqrt{2})\), along the vertical axis, since our hyperbola opens upward and downward. Understanding the location of the foci is essential for graphing the hyperbola correctly.

Directrices in Hyperbolas

Directrices are lines associated with hyperbolas that help in defining the locus of points that make up the hyperbola. For each directrix, there is a corresponding focus, and points on the hyperbola maintain a constant ratio of distances to these.

For a hyperbola with a vertical transverse axis, like the one in our equation \( y^2 - x^2 = 4 \), the directrices are parallel to the transverse axis and are found using the formula:

• \( y = \pm \frac{a}{e} \)

Inserting our values, where \( a = 2 \) and \( e = \sqrt{2} \), we find the directrices are \( y = \pm \sqrt{2} \).

These directrices help in understanding and visualizing the geometry of the hyperbola, as they guide its curving shape.

Canonical Form of a Hyperbola

The canonical, or standard form, of a hyperbola equation makes it easier to analyze and graph the hyperbola by clearly identifying all necessary parameters. For hyperbolas, the canonical form depends on the orientation of the transverse axis, which can be horizontal or vertical.

For the given hyperbola equation \( y^2 - x^2 = 4 \), it is rearranged into its canonical form:

• \( \frac{y^2}{4} - \frac{x^2}{4} = 1 \)

This corresponds to the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), where \( a^2 = 4 \) and \( b^2 = 4 \).

The variables \( a \) and \( b \) are critical because they determine the dimensions of the hyperbola. By knowing \( a \) and \( b \), you can subsequently locate the foci, draw the directrices, and understand the hyperbola's opening - central pieces in sketching an accurate representation.