Problem 38 - Thomas Calculus

Question

Exercises \(35-38\) give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation from the information given. $$ \begin{array}{l}{\text { Vertices: }(0, \pm 2)} \\ {\text { Asymptotes: } y=\pm \frac{1}{2} x}\end{array} $$

Step-by-Step Solution

Answer

The hyperbola's equation is \( \frac{y^2}{4} - \frac{x^2}{16} = 1 \).

Step 1: Understand Hyperbola Structure

A hyperbola centered at the origin with vertical transverse axis (vertices along the y-axis) has the standard form equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). For a horizontal transverse axis, the equation would be \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). In this problem, since the vertices are at \((0, \pm 2)\), the hyperbola has a vertical transverse axis.

Step 2: Identify Parameters from Vertices

The given vertices are \((0, \pm 2)\), meaning \( a = 2 \). Therefore, \( a^2 = 4 \). This is because the distance from the center to each vertex along the transverse axis is \( a \).

Step 3: Use Asymptote Information

The asymptotes for a hyperbola with vertical transverse axis have the form \( y = \pm \frac{a}{b}x \). Given asymptotes are \( y = \pm \frac{1}{2} x \), which tells us \( \frac{a}{b} = \frac{1}{2} \). With \( a = 2 \), solve for \( b \) by \( \frac{2}{b} = \frac{1}{2} \), hence \( b = 4 \) and \( b^2 = 16 \).

Step 4: Formulate the Hyperbola Equation

Now that we know \( a^2 = 4 \) and \( b^2 = 16 \), substitute these into the standard form of the hyperbola equation. For a vertical hyperbola, the equation is: \[ \frac{y^2}{4} - \frac{x^2}{16} = 1 \].

Key Concepts

Vertices of Hyperbola

In a hyperbola, vertices are key points that determine its shape and orientation. For hyperbolas centered at the origin, the vertices are aligned along the transverse axis. In the given exercise, the vertices are at \((0, \pm 2)\), indicating a vertical transverse axis since these points lie along the y-axis. The transverse axis is crucial because it defines the length and direction of the spread of the hyperbola. Hence, you can picture the hyperbola opening upward and downward.

The distance from the center \((0, 0)\) to each vertex is denoted by \( a \). In this case, the distance is 2, therefore \( a = 2 \). Consequently, you square this distance to get \( a^2 = 4 \), which will be used in the standard equation of the hyperbola. Recognizing the position of the vertices helps identify the direction the hyperbola opens, and it is a vital step when deriving the equation.

Asymptotes of Hyperbola

Asymptotes are lines that a hyperbola approaches but never actually touches. They provide a structural framework that helps sketch the hyperbola's arms as they extend infinitely. For hyperbolas with vertical transverse axes, their equations take the form \( y = \pm \frac{a}{b} x \). Understanding this relationship is key to finding the value of \( b \) when given the slope of the asymptotes.

In the exercise, the asymptotes are given as \( y = \pm \frac{1}{2} x \). Comparing these with the general form, it's clear that \( \frac{a}{b} = \frac{1}{2} \). Since \( a \) was previously determined to be 2, you can find \( b \) by solving \( \frac{2}{b} = \frac{1}{2} \). Thus, \( b = 4 \) and \( b^2 = 16 \). Asymptotes serve as a guide for the hyperbola's orientation in space, ensuring that the curve nestles between them but never converges fully.

Standard Form Equation

The standard form equation of a hyperbola is pivotal to understanding its structure and behavior. The equation lays out the core parameters that shape the hyperbola. Generally, for hyperbolas centered at the origin, the form is either \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), depending on which axis the transverse course takes.

In this problem, since the vertices were aligned vertically, the appropriate form was \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Plugging in the values calculated earlier, where \( a^2 = 4 \) and \( b^2 = 16 \), gives us the specific equation \( \frac{y^2}{4} - \frac{x^2}{16} = 1 \). This equation fully describes the hyperbola, allowing for visualization and further analysis. It defines the hyperbola's opening direction, the distance between its branches, and how it interacts with the asymptotes.