Problem 37 - Thomas Calculus

Question

Exercises \(35-38\) give foci and corresponding directrices of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's eccentricity. Then find the hyperbola's standard-form equation. $$ \begin{array}{l}{\text { Focus: }(-2,0)} \\ {\text { Directrix: } \quad x=-\frac{1}{2}}\end{array} $$

Step-by-Step Solution

Answer

The hyperbola's eccentricity is 4. Its equation is \(4x^2 - \frac{4}{7}y^2 = 1\).

Step 1: Understand the Components

A hyperbola's foci are located at points that are equidistant from the center. Given the focus (-2, 0), we note that the center is at the origin (0, 0). The directrix is given as \( x = -\frac{1}{2} \). Hyperbolas have an eccentricity formula \( e = \frac{c}{a} \), where \( c \) is the distance from the center to a focus, and \( a \) is the distance from the center to the directrix.

Step 2: Calculate the Distance to the Focus

The distance from the center of the hyperbola (0, 0) to the focus (-2, 0) is \( c = 2 \). This is because the focus is 2 units to the left of the origin.

Step 3: Calculate the Distance to the Directrix

The distance from the center (0, 0) to the directrix \( x = -\frac{1}{2} \) is \( a = \frac{1}{2} \), because the directrix is \( \frac{1}{2} \) units to the left of the origin.

Step 4: Determine the Eccentricity

Using the formula for eccentricity \( e = \frac{c}{a} \), substitute the values \( c = 2 \) and \( a = \frac{1}{2} \). This gives \( e = \frac{2}{\frac{1}{2}} = 4 \).

Step 5: Write the Equation of the Hyperbola

The eccentricity \( e = 4 \) confirms the shape of the hyperbola. A hyperbola with an equation of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) has its transverse axis along the x-axis and is centered at the origin. Since \( c^2 = a^2 + b^2 \) and \( c = 2 \), \( a = \frac{1}{2} \), we can find \( b^2 \) from \( 2^2 = (\frac{1}{2})^2 + b^2 \). Calculating gives \( b^2 = \frac{7}{4} \), and the standard form becomes \( \frac{x^2}{\left( \frac{1}{2} \right)^2} - \frac{y^2}{\frac{7}{4}} = 1 \). This simplifies to \( 4x^2 - \frac{4}{7}y^2 = 1 \).

Key Concepts

Eccentricity

The eccentricity of a hyperbola is a measure of how "stretched out" it is. It's a unique value that describes the shape and can tell us a lot about the hyperbola. It is denoted by the symbol \( e \). To find it, we use the formula for eccentricity:

• \( e = \frac{c}{a} \)

where \( c \) is the distance from the center to the focus, and \( a \) is the distance from the center to the directrix.

In our example, we have a focus at (-2, 0), making \( c = 2 \), and a directrix at \( x = -\frac{1}{2} \), giving us \( a = \frac{1}{2} \).

Substituting these values into the formula, we calculate:

• \( e = \frac{2}{\frac{1}{2}} = 4 \)

This eccentricity of 4 indicates a hyperbola that is highly elongated along its transverse axis.

Standard-form equation

The standard-form equation of a hyperbola helps define its shape and orientation on the coordinate plane. It's typically expressed as:

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

for hyperbolas with the transverse axis along the \( x \)-axis. In these equations, \( a^2 \) and \( b^2 \) are determined by the distances related to the hyperbola's foci and vertices.

For the given exercise, we use \( a = \frac{1}{2} \) and substitute it into the equation. To find \( b^2 \), we use the relationship \( c^2 = a^2 + b^2 \) based on our existing calculations, where \( c = 2 \) and \( a = \frac{1}{2} \). We simplify this as follows:

• \( 2^2 = \left(\frac{1}{2}\right)^2 + b^2 \)
• \( 4 = \frac{1}{4} + b^2 \)
• \( b^2 = \frac{7}{4} \)

Finally, replacing \( a^2 \) and \( b^2 \) into our hyperbola equation gives:

• \( \frac{x^2}{\left( \frac{1}{2} \right)^2} - \frac{y^2}{\frac{7}{4}} = 1 \)

Simplifying the equation, we arrive at:

• \( 4x^2 - \frac{4}{7}y^2 = 1 \)

Foci and directrices

In the context of hyperbolas, the foci and directrices play crucial roles in shaping the curve.

Each hyperbola has two foci that are equidistant from its center. These points affect the "open" shape of the hyperbola. For our example, one focus is located at (-2, 0).

A directrix is a line that, together with the focus, helps to define the hyperbola. It is used alongside the distance from the focus to determine eccentricity, with the hyperbola opening away from the directrix. In the exercise, the directrix is \( x = -\frac{1}{2} \).

Understanding how foci and directrices interact allows students to appreciate how eccentricity impacts the overall geometry of the hyperbola. It also shines light on why particular forms and orientations are used in mathematics to describe these fascinating curves.