Problem 38 - Precalculus Mathematics for Calculus

Question

Find an equation for the hyperbola that satisfies the given conditions. Vertices: \((0, \pm 6),\) hyperbola passes through \((-5,9)\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{y^2}{36} - \frac{x^2}{20} = 1\).

Step 1: Identify the Standard Form

Given that the vertices are at \((0, \pm 6)\), the hyperbola is vertical, which has the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), where \(a\) is the distance from the center to a vertex along the y-axis.

Step 2: Determine 'a'

The distance from the center \((0,0)\) to a vertex \((0,6)\) is 6. Thus, \(a = 6\) and \(a^2 = 36\).

Step 3: Use Given Point to Find 'b^2'

The hyperbola passes through \((-5,9)\). Substitute \(x = -5\) and \(y = 9\) into the equation:\[\frac{9^2}{36} - \frac{(-5)^2}{b^2} = 1\]This simplifies to:\[\frac{81}{36} - \frac{25}{b^2} = 1\]

Step 4: Solve for 'b^2'

Solve the equation:\[\frac{81}{36} - \frac{25}{b^2} = 1\] which simplifies to \[\frac{9}{4} = 1 + \frac{25}{b^2}\]. Subtract 1 from \(\frac{9}{4}\) to get \(\frac{5}{4} = \frac{25}{b^2}\). Thus, \(b^2 = 20\).

Step 5: Write the Equation

Now that we have \(a^2 = 36\) and \(b^2 = 20\), the equation of the hyperbola is:\[\frac{y^2}{36} - \frac{x^2}{20} = 1\].

Key Concepts

Standard Form of a Hyperbola

The concept of a hyperbola is crucial in mathematics as it represents a type of conic section or curve. An important aspect of hyperbolas to understand is their standard form equation.
A hyperbola can be either horizontal or vertical, depending on the orientation given by its vertices. For a vertical hyperbola, like in our example where the vertices are

• \( (0, \pm 6) \),

the standard form is represented as

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

Let's break down this equation:

• The term \( y^2/a^2 \) is always positive in this configuration as that identifies this particular form as vertical.
• The term \( x^2/b^2 \) has a negative sign.
• The number 1 on the right-hand side represents the relationship between the squared terms for all values that lie on the hyperbola.

Understanding the components of this equation is important as it helps us determine the distance between the center and the vertices and also understand how to adjust the equation for any given hyperbola.

Vertices of a Hyperbola

Vertices are pivotal when working with hyperbolas because they help to define the curve itself. For hyperbolas, the vertices are the points where the branches of the hyperbola are closest together.
In the exercise, the vertices are located at coordinates

• \((0, \pm 6)\).

This implies that the center is at

• \((0, 0)\),

positioning the hyperbola along the y-axis.
The distance from the center to each vertex is denoted as

• \(a\).

This distance is crucial, as it equals 6 in this case, making

• \(a = 6\).

Moreover,

• \(a^2 = 36\).

Clearly identifying the vertices and their distance from the center allows you to establish the basic parameters of a hyperbola. This information is essential for later calculations and modifying the equation as needed.

Hyperbola Through a Point

Apart from determining the structure of a hyperbola by its vertices, knowing another point through which the hyperbola passes can be vital. This point provides additional necessary information to fully define the parameters of the hyperbola’s equation.
In the given exercise, there’s a specific point

• \((-5, 9)\)

that lies on our hyperbola.
This gives us a way to solve for an unknown part of the equation. After substituting this point into the standard form equation

• \( \frac{9^2}{36} - \frac{(-5)^2}{b^2} = 1 \),

we can calculate the unknown

• \(b^2\).

Through simple algebra, solving

• \( \frac{25}{b^2} = \frac{5}{4} \)

allows us to determine that

• \(b^2 = 20\).

Adding this step helps ensure our hyperbola fits not only the primary structural criteria but also that it aligns with other known points, refining our model and confirming solutions accurately.