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: Identify the Hyperbola Orientation

The vertices given are \((0, \pm 2)\). This indicates that the transverse axis is vertical. Therefore, the standard form of the hyperbola is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).

Step 2: Determine the Value of \(a\)

The vertices \((0, \pm 2)\) indicate that \(a = 2\), since the vertices are \((0, \pm a)\). Therefore, \(a^2 = 4\).

Step 3: Use Asymptotes to Find \(b\)

The equations of the asymptotes are \(y = \pm \frac{1}{2}x\). For the vertical hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the slope of the asymptotes is \(\pm \frac{a}{b}\). Thus, we have \(\frac{2}{b} = \frac{1}{2}\), leading to \(b = 4\). So, \(b^2 = 16\).

Step 4: Write the Equation of the Hyperbola

Now that we have \(a^2 = 4\) and \(b^2 = 16\), we can write the equation of the hyperbola as \(\frac{y^2}{4} - \frac{x^2}{16} = 1\).

Key Concepts

Equation of Hyperbola

The equation of a hyperbola is essential as it defines the shape and orientation of the curve on a plane. Hyperbolas can either open horizontally or vertically. The general forms depend on their orientation:

• For a horizontally oriented hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
• For a vertically oriented hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

In this exercise, the vertices provided are \((0, \pm 2)\), indicating a vertical orientation. Consequently, we will use the second format, where \(a\) represents the distance from the origin to each vertex along the vertical axis. Hence, the equation we need here is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Knowing \(a = 2\), we find \(a^2 = 4\).

Asymptotes of Hyperbola

Asymptotes are the straight lines that a hyperbola approaches as it extends to infinity. They help in understanding the geometry and extent of the hyperbola. For hyperbolas centered at the origin, the asymptotes can be defined using the formula

• \(y = \pm \frac{b}{a}x\) for horizontal hyperbolas
• \(y = \pm \frac{a}{b}x\) for vertical hyperbolas

In our case with a vertical hyperbola, checking the given asymptotes \(y = \pm \frac{1}{2} x\), the slope \(\pm\frac{1}{2}\) informs us about the relationship between \(a\) and \(b\). Comparing \(\frac{a}{b} = \frac{1}{2}\) with our values allows us to solve for \(b\) by setting \(\frac{2}{b} = \frac{1}{2}\), giving us \(b = 4\) and consequently \(b^2 = 16\).

Vertices of Hyperbola

Vertices are key points that help define the basic shape of a hyperbola. They lie on the transverse axis, which is the axis where the hyperbola 'opens'. In this exercise, we are given vertices at \((0, \pm 2)\). This means the hyperbola's transverse axis is vertical.
This placement on the axes is crucial as it confirms the vertical orientation of the hyperbola. The distance from the center, which is at the origin in this instance, to each vertex is denoted by \(a\). For our hyperbola, \(a\) is obtained as 2, directly from these given vertex coordinates. Thus, these vertices confirm the pattern and equation shape we've determined.

Transverse Axis of Hyperbola

The transverse axis is a significant line that passes through the vertices of the hyperbola and divides it symmetrically. It is part of what visually defines the direction of a hyperbola: vertically or horizontally.
In this specific hyperbola, with vertices located at \((0, \pm 2)\), the transverse axis is vertical (the line that runs up and down), confirming the orientation as vertical.

• The length of the transverse axis in this case is \(2a\), which equals 4.
• This occurs because \(a\) is the distance from the center to each vertex, and as there are two such distances in total, summing each way from the center.

The transverse axis plays a critical role, as it not only identifies the hyperbola's orientation but also provides insight into its equation's structure.