Problem 60 - Precalculus: Functions and Graphs

Question

Find an equation for the indicated part of the hyperbola. Right branch of \(\frac{x^{2}}{16}-\frac{y^{2}}{4}=1\)

Step-by-Step Solution

Answer

The equation for the right branch is \(x = \sqrt{16 + 4y^2}\).

Step 1: Identify the Form of the Hyperbola

Recognize the given equation \(\frac{x^{2}}{16} - \frac{y^{2}}{4} = 1\) as representing a hyperbola in the standard form \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\). Here, \(a^2=16\) and \(b^2=4\), which implies \(a=4\) and \(b=2\). This hyperbola opens left and right along the x-axis.

Step 2: Write the Parametric Equations

For hyperbolas of the form \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\), the parametric equations are given by \(x = a \sec(\theta)\) and \(y = b \tan(\theta)\), where \(\theta\) is the parameter. For the given hyperbola, these become \(x = 4 \sec(\theta)\) and \(y = 2 \tan(\theta)\).

Step 3: Identify the Right Branch Segment

The right branch of this hyperbola corresponds to values of \(\theta\) where \(\sec(\theta) > 0\), which occurs when \(\theta\) is between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) not including \(\frac{\pi}{2}\) or \(-\frac{\pi}{2}\). In this range, \(x = 4 \sec(\theta)\) will be positive.

Step 4: Express as a Function, Not Parameters

To express the right branch for \(x\) as a function of \(y\), or vice versa, use the relation \(x^2 = 16 + 4y^2\) derived from eliminating \(\theta\). Since we're dealing with the right branch, choose the positive root, resulting in \(x = \sqrt{16 + 4y^2}\).

Key Concepts

Parametric Equations

To better understand hyperbolas, let's dive into parametric equations. For hyperbolas, parametric equations give a way to express the coordinates

• of a hyperbola in terms of a variable known as a parameter.
• This is helpful for plotting and analyzing the curve.
• In our hyperbola equation, \(\frac{x^{2}}{16} - \frac{y^{2}}{4} = 1\), the parametric equations are useful to describe its shape more clearly.

Here, the parameters are given by \(x = 4 \sec(\theta)\) and \(y = 2 \tan(\theta)\). Remember:- \(\sec(\theta)\) is the reciprocal of \(\cos(\theta)\), - while \(\tan(\theta)\) stands for \(\sin(\theta) / \cos(\theta)\). This means we use trigonometric identities to represent each point on the hyperbola in terms of \(\theta\). The use of parameters makes it easy to explore the hyperbola's properties and graph it for different values of \(\theta\). It also helps in identifying specific parts of the hyperbola, such as the right branch.

Standard Form of Hyperbola

Hyperbolas have a specific equation format that indicates their orientation and structural characteristics. The standard form of the hyperbola is:\[\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\]This equation tells us:- \(x^2/a^2\) being positive means the hyperbola opens along the x-axis.- \(a^2\) and \(b^2\) are constants that define the hyperbola's inner geometry and the distance of vertices from the center.For the given equation, \(\frac{x^{2}}{16} - \frac{y^{2}}{4} = 1\), we recognize

• \(a^2 = 16\) (thus \(a = 4\)) and
• \(b^2 = 4\) (thus \(b = 2\)).

The deduction allows us to determine the shape and direction of the branches of the hyperbola. Since the \(x\) term is positive, the hyperbola opens to the left and right, parallel to the x-axis. This standard form helps you instantly know where the hyperbola's branches will be, even before graphing it.

Right Branch of Hyperbola

When focused specifically on the right branch of the hyperbola, we consider only positive x-coordinate values.In the geometric context:- The right branch comprises the section of the hyperbola that extends to the right side of the y-axis.To clarify: - The given hyperbola \(\frac{x^{2}}{16} - \frac{y^{2}}{4} = 1\) has two branches symmetric about the y-axis. - For hyperbolas opening along the x-axis, the right branch focuses on where \(x > 0\).Using parametric equations, \(x = 4 \sec(\theta)\), the hyperbola's right branch is mapped when \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\). This ensures that \(\sec(\theta) > 0\), resulting in positive x-values. Additionally, when transforming the equation parametrically to yield a Cartesian relation:\[x = \sqrt{16 + 4y^2}\]This restriction confirms that only the positive expression represents the right branch, detailing dependencies in terms of \(y\). Visualization hinges on understanding these computational transformations to identify and segregate the branches during analysis.