Problem 25 - Thomas Calculus

Question

Exercises \(25-28\) 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 in Cartesian coordinates. $$\begin{array}{l}{\text { Eccentricity: } 3} \\ {\text { Vertices: }(0, \pm 1)}\end{array}$$

Step-by-Step Solution

Answer

The hyperbola's equation is \(\frac{y^2}{1} - \frac{x^2}{8} = 1\).

Step 1: Identify Hyperbola Orientation

Given the vertices \((0, \pm 1)\), we see that the hyperbola is vertical. This is because the vertices are aligned along the \(y\)-axis. This means the equation will be of the form \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\).

Step 2: Find Value of \(b\)

The vertices \((0, \pm 1)\) indicate that the distance from the center to a vertex along the \(y\)-axis is \(1\). This gives us \(b = 1\). Thus, \(b^2 = 1\).

Step 3: Use Eccentricity to Find \(a\)

The eccentricity \(e\) is given as \(3\), and for a vertical hyperbola, \(e = \frac{c}{b}\). Given \(e = 3\) and \(b = 1\), we have \(c = 3\cdot 1 = 3\).

Step 4: Relate \(c\), \(a\), and \(b\)

For hyperbolas, the relation is \(c^2 = a^2 + b^2\). We have \(c^2 = 9\) and \(b^2 = 1\). Substituting these, we get \(9 = a^2 + 1\).

Step 5: Solve for \(a\)

The equation \(9 = a^2 + 1\) simplifies to \(a^2 = 8\). Thus, \(a = \sqrt{8}\).

Step 6: Write the Standard Form Equation

The standard form equation of the hyperbola is \(\frac{y^2}{1} - \frac{x^2}{8} = 1\).

Key Concepts

Cartesian coordinates

The Cartesian coordinate system is a two-dimensional plane defined by two perpendicular axes, usually labeled as the x-axis and the y-axis. Hyperbolas, a type of conic section, are often expressed using this coordinate system. They have equations in a standard form that can be either horizontal or vertical, depending on the orientation of the vertices.For example, the exercise relates to a vertical hyperbola. In such cases, the standard form equation is \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\).This form indicates that:

• The hyperbola is centered at the origin (0,0).
• The terms \(b^2\) and \(a^2\) represent the squares of the distances from the center to the vertices along the y-axis and x-axis respectively.
• The choice of positive and negative signs in the equation determines the hyperbola's orientation.

Understanding this format allows you to transform verbal descriptions or vertex and eccentricity information into the hyperbola's equation using Cartesian coordinates.

Eccentricity of hyperbolas

Eccentricity, denoted as \(e\), is a measure of how "stretched" a conic section such as a hyperbola is. This important parameter determines the hyperbola's shape and is calculated using the formula:\[ e = \frac{c}{b} \]where \(c\) is the distance from the center to each focus, and \(b\) is the distance from the center to each vertex.For hyperbolas:

• A higher eccentricity value indicates a more elongated hyperbola.
• The eccentricity value is always greater than 1.

In the featured exercise, the given eccentricity is \(e = 3\). Using this:

• We find \(c = 3 \times b\) since \(b = 1\), hence \(c = 3\). This gives the distance to the foci.

Knowing the eccentricity offers insight into the hyperbola's geometry and aids in forming its equation.

Vertices of hyperbolas

The vertices of a hyperbola are key points that define its main structure. They lie along the transverse axis, which correlates with the hyperbola's orientation.

• When the vertices are given, such as in this exercise with \((0, \pm 1)\), it indicates the hyperbola is vertical and these are located on the y-axis.
• The distance \(b\) from the center to a vertex helps set the hyperbola's scale and dimensions.

Using the vertices to derive the square of \(b\), we have \(b^2 = 1\) since each vertex is 1 unit from the center at the origin.In combination with the eccentricity, the vertices facilitate finding the complete standard form equation of the hyperbola, forming the basis for more complex mathematical analyses.