Problem 32 - Thomas Calculus

Question

Exercises \(31-34\) give the eccentricities and the vertices or foci of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation. $$ \begin{array}{l}{\text { Eccentricity: } 2} \\ {\text { Vertices: }( \pm 2,0)}\end{array} $$

Step-by-Step Solution

Answer

The hyperbola's equation is \( \frac{x^2}{4} - \frac{y^2}{12} = 1 \).

Step 1: Understand the Hyperbola's Features

The eccentricity \( e \) is given as 2, and the vertices are \((\pm 2, 0)\). The hyperbola is centered at the origin, meaning its equation will be in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Since the vertices are on the x-axis, use the first form.

Step 2: Identify \(a\)

The distance from the center to each vertex \( c \) along the x-axis is the absolute value of \( \pm 2 \), meaning \( a = 2 \).

Step 3: Calculate \(c\)

Using the formula for eccentricity of a hyperbola, \( e = \frac{c}{a} \), substitute \( e = 2 \) and \( a = 2 \) to find \( c = 4 \).

Step 4: Solve for \(b\)

Recall the hyperbola relationship \( c^2 = a^2 + b^2 \). Substitute \( a = 2 \) and \( c = 4 \), to solve \[ c^2 = a^2 + b^2 \ 16 = 4 + b^2 \ b^2 = 12.\]

Step 5: Write the Standard-Form Equation

Now that we know \( a^2 = 4 \) and \( b^2 = 12 \), the standard-form equation of the hyperbola is \[ \frac{x^2}{4} - \frac{y^2}{12} = 1.\]

Key Concepts

Eccentricity

Eccentricity is a key concept when dealing with hyperbolas. It is represented by the letter \(e\) and tells us how "stretched" the hyperbola is. For a hyperbola, the eccentricity is greater than 1. This is what separates it from ellipses, which have an eccentricity between 0 and 1.

Eccentricity is calculated using the formula:

• For hyperbolas, \(e = \frac{c}{a}\)

Where:

• \(c\) is the distance from the center to a focus.
• \(a\) is the distance from the center to a vertex.

In the given exercise, the eccentricity is 2. This means the distance from the center to a focus is twice the distance from the center to a vertex. Understanding \(e\) helps us see how wide or narrow the hyperbola is when compared to its foci and vertices.

Vertices

Vertices are another vital part of understanding hyperbolas. The vertices are the points where the hyperbola intersects its transverse axis. They are easy to spot and important for graphing the hyperbola correctly.

In this exercise, the vertices are given as \((\pm 2, 0)\). These points tell us the hyperbola stretches horizontally, as the \(x\)-coordinates change while the \(y\)-coordinate stays zero.

To summarize, for a hyperbola centered at the origin and with its transverse axis along the \(x\)-axis:

• \(a\) represents the distance from the center to each vertex, which is 2 in this exercise.
• The vertices show how far the hyperbola opens along the transverse axis.

Knowing the vertices helps you identify the dimensions of the hyperbola and select the appropriate standard form.

Standard Form

The standard form of a hyperbola's equation is essential, as it allows us to easily identify key features and graph it. Based on the orientation of the hyperbola, the standard form changes slightly:

• If the transverse axis is horizontal: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
• If the transverse axis is vertical: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)

In this exercise, the hyperbola's transverse axis is along the \(x\)-axis, so we use the equation with \(x^2/a^2\) first.

With \(a^2 = 4\) (from the vertex distance), and having calculated \(b^2 = 12\) using \(c^2 = a^2 + b^2\), the final equation becomes:

• \(\frac{x^2}{4} - \frac{y^2}{12} = 1\)

This equation is complete and reflects all the important characteristics of the hyperbola, including its width, orientation, and how it opens along its transverse axis.