Problem 26 - Thomas Calculus

Question

In Exercises \(23-30\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$ y^{2}-x^{2}=4 $$

Step-by-Step Solution

Answer

Eccentricity is \( \sqrt{2} \). Foci at \( (0, \pm 2\sqrt{2}) \); Directrices are \( y = \pm \sqrt{2} \).

Step 1: Identify the Equation of the Hyperbola

The given equation is \( y^2 - x^2 = 4 \). This is the standard form for a vertical hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). We can rewrite the equation as \( \frac{y^2}{4} - \frac{x^2}{4} = 1 \). Thus, \( a^2 = 4 \) and \( b^2 = 4 \).

Step 2: Calculate Eccentricity

The formula for the eccentricity \( e \) of a hyperbola is given by \( e = \frac{\sqrt{a^2 + b^2}}{a} \). Substituting the values \( a^2 = 4 \) and \( b^2 = 4 \), we find \( e = \frac{\sqrt{4 + 4}}{2} = \frac{\sqrt{8}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2} \).

Step 3: Find the Foci

For a hyperbola in the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the foci are located at \( (0, \pm c) \), where \( c = \sqrt{a^2 + b^2} \). Using \( a^2 = 4 \) and \( b^2 = 4 \), we find \( c = \sqrt{8} = 2\sqrt{2} \). Therefore, the foci are at \( (0, \pm 2\sqrt{2}) \).

Step 4: Find the Directrices

The equations for the directrices of a vertical hyperbola are given by \( y = \pm \frac{a^2}{\sqrt{a^2 + b^2}} \). With \( a^2 = 4 \), we calculate \( y = \pm \frac{4}{\sqrt{8}} = \pm \frac{4}{2\sqrt{2}} = \pm \sqrt{2} \). Thus, the equations of the directrices are \( y = \sqrt{2} \) and \( y = -\sqrt{2} \).

Step 5: Graph the Hyperbola, Foci, and Directrices

To graph the hyperbola, draw the equivalent vertical hyperbola \( \frac{y^2}{4} - \frac{x^2}{4} = 1 \) with its center at the origin \( (0,0) \). Mark the foci at \( (0, 2\sqrt{2}) \) and \( (0, -2\sqrt{2}) \). Draw the directrices \( y = \sqrt{2} \) and \( y = -\sqrt{2} \) as horizontal dashed lines. This completes the graphical representation.

Key Concepts

Eccentricity

Eccentricity is a measure that defines the "curviness" or elongation of a hyperbola. It helps distinguish hyperbolas from other conic sections, like ellipses or parabolas. The eccentricity, denoted as \( e \), is always greater than 1 for hyperbolas. The formula to calculate eccentricity for hyperbolas is \( e = \frac{\sqrt{a^2 + b^2}}{a} \). Here, \( a^2 \) and \( b^2 \) are terms from the hyperbola equation.
Given the hyperbola \( y^2 - x^2 = 4 \), we rearrange it to \( \frac{y^2}{4} - \frac{x^2}{4} = 1 \). This reveals that both \( a^2 \) and \( b^2 \) equal 4. By substituting into the formula, we find \( e = \sqrt{2} \). This eccentricity value confirms that the hyperbola is indeed stretched along its central axis.

Foci

The foci of a hyperbola are two fixed points located symmetrically outside the center. These points play a crucial role in defining the hyperbola’s shape and properties. The relationship with the foci is that, for any point on the hyperbola, the absolute difference of the distances to the foci is constant.
For the standard vertical hyperbola equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the foci are positioned at \( (0, \pm c) \), where \( c = \sqrt{a^2 + b^2} \).
In our exercise, with \( a^2 = 4 \) and \( b^2 = 4 \), we calculate \( c = \sqrt{8} = 2\sqrt{2} \). Consequently, the foci for this hyperbola are located at \( (0, 2\sqrt{2}) \) and \( (0, -2\sqrt{2}) \). Including these points in your graph helps in visualizing the hyperbola’s extended shape.

Directrices

Directrices are lines that work alongside the foci to describe a hyperbola’s precise shape. These lines are positioned symmetrically about the center, and their equations correspond to the orientation of the hyperbola.
The equations for the directrices of a vertical hyperbola are given by \( y = \pm \frac{a^2}{\sqrt{a^2 + b^2}} \). These provide limits that guide the curvature of the hyperbola closer to its vertices.

• Firstly, identify \( a^2 = 4 \) from the hyperbola equation.
• Next, calculate the directrices as \( y = \pm \frac{4}{\sqrt{8}} = \pm \sqrt{2} \).

Hence, the hyperbola includes horizontal directrices at \( y = \sqrt{2} \) and \( y = -\sqrt{2} \). On a graph, these lines help in outlining the boundary within which the hyperbola stretches.

Hyperbola Equation

The given equation \( y^2 - x^2 = 4 \) represents a hyperbola. To work with this equation, we reformulate it into the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). This process involves recognizing the coefficients and organizing them into fractions.
By rewriting as \( \frac{y^2}{4} - \frac{x^2}{4} = 1 \), we derive that \( a^2 \) and \( b^2 \) are each equal to 4. This indicates both axes are of equivalent magnitude in the hyperbola's graph which presents symmetrical features.
Such transformation into standard form allows for easier identification and computation of essential characteristics, such as eccentricity, foci, and directrices. This form thus serves as the groundwork from which to analyze the hyperbola’s geometric properties.