Problem 56
Question
Identify the curve that is parameterized by the equations. In each case, use a trigonometric identity to eliminate the parameter. (It may help to consider the special case \(a=b=1\) first.) \(x=a \sec (\theta), y=b \tan (\theta), \theta \in[0, \pi / 2)\)
Step-by-Step Solution
Verified Answer
The curve is a hyperbola, specifically \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
1Step 1: Understand the Given Parameterization
The parameterization gives us two equations: \( x = a \sec(\theta) \) and \( y = b \tan(\theta) \). The variable \( \theta \) is in the range \([0, \pi/2)\), which means \( \theta \) can take values from 0 to just under \( \pi/2 \). We need to eliminate the parameter \( \theta \) to find the relation between \( x \) and \( y \).
2Step 2: Recall Trigonometric Identity
Remember that one of the trigonometric identities is \( \sec^2(\theta) = 1 + \tan^2(\theta) \). This identity will help us to eliminate \( \theta \) by expressing both \( \sec(\theta) \) and \( \tan(\theta) \) in terms of each other.
3Step 3: Express \( \tan(\theta) \) in Terms of \( x \)
From the equation \( x = a \sec(\theta) \), we can express \( \sec(\theta) \) as \( \sec(\theta) = \frac{x}{a} \). Substitute this into the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \) to find \( \tan(\theta) \).
4Step 4: Substitute and Simplify
Substitute \( \sec(\theta) = \frac{x}{a} \) into \( \sec^2(\theta) = 1 + \tan^2(\theta) \), leading to \( \left(\frac{x}{a}\right)^2 = 1 + \tan^2(\theta) \). Simplify to find \( \tan^2(\theta) = \left(\frac{x}{a}\right)^2 - 1 \).
5Step 5: Express \( y \) in Terms of \( x \)
Since \( y = b\tan(\theta) \), substitute \( \tan^2(\theta) = \frac{y^2}{b^2} \), equating it with the expression from the previous step: \( \frac{y^2}{b^2} = \left(\frac{x}{a}\right)^2 - 1 \).
6Step 6: Final Simplification to Standard Form
Multiply through by \( b^2 \) to clear the fraction: \( y^2 = b^2 \left(\frac{x^2}{a^2} - 1\right) \). Simplify to get the standard form of the hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
7Step 7: Identify the Curve Type
The equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is the equation of a hyperbola with a horizontal transverse axis and semi-axes of length \( a \) and \( b \).
Key Concepts
Trigonometric IdentitiesHyperbolaElimination of Parameter
Trigonometric Identities
Trigonometric identities are mathematical equations that express one trigonometric function in terms of others. They are essential tools for simplifying expressions and solving equations where trigonometric functions appear. In this exercise, we use the identity
The secant function, \( \sec(\theta) \), and the tangent function, \( \tan(\theta) \), are related through this identity.
Specifically, the secant function is expressed as the reciprocal of the cosine function, while the tangent function is the ratio of the sine function to the cosine function.
By substituting \( \sec(\theta) = \frac{x}{a} \) and using the identity, you can express \( \tan(\theta) \) in terms of \( x \), ultimately allowing you to find the relationship between \( x \) and \( y \) without \( \theta \).
This simplification through a trigonometric identity helps transform parameterized equations into recognizable standard forms, such as hyperbolas.
- \( \sec^2(\theta) = 1 + \tan^2(\theta) \)
The secant function, \( \sec(\theta) \), and the tangent function, \( \tan(\theta) \), are related through this identity.
Specifically, the secant function is expressed as the reciprocal of the cosine function, while the tangent function is the ratio of the sine function to the cosine function.
By substituting \( \sec(\theta) = \frac{x}{a} \) and using the identity, you can express \( \tan(\theta) \) in terms of \( x \), ultimately allowing you to find the relationship between \( x \) and \( y \) without \( \theta \).
This simplification through a trigonometric identity helps transform parameterized equations into recognizable standard forms, such as hyperbolas.
Hyperbola
A hyperbola is a type of conic section formed when a plane intersects both halves of a double cone.
It is defined by its distinct set of points where the difference in distances from any point on the hyperbola to two fixed points, called the foci, is constant.
The standard form of a hyperbola's equation is expressed as:
The transverse axis is the line segment that passes through the center and focuses parallel to the x-axis, and its length is related to \( a \).
The values \( a \) and \( b \) are the lengths of the semi-axes, which determine the hyperbola's shape and spread.
In our parameterization, by eliminating the parameter \( \theta \), you arrive at this standard hyperbola equation, clearly identifying the curve as such. Understanding this helps recognize how x and y relate in the context of conic sections.
It is defined by its distinct set of points where the difference in distances from any point on the hyperbola to two fixed points, called the foci, is constant.
The standard form of a hyperbola's equation is expressed as:
- \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
The transverse axis is the line segment that passes through the center and focuses parallel to the x-axis, and its length is related to \( a \).
The values \( a \) and \( b \) are the lengths of the semi-axes, which determine the hyperbola's shape and spread.
In our parameterization, by eliminating the parameter \( \theta \), you arrive at this standard hyperbola equation, clearly identifying the curve as such. Understanding this helps recognize how x and y relate in the context of conic sections.
Elimination of Parameter
Eliminating a parameter from a set of parametric equations is a process that simplifies the equations into a singular expression relating selected variables directly.
For example, given the equations \( x = a \sec(\theta) \) and \( y = b \tan(\theta) \), the objective is to remove \( \theta \) so we can derive a clear relationship between \( x \) and \( y \).
The elimination is crucial in moving from parametric to non-parametric forms, which are often easier to graph and analyze.
In this solution, our step-by-step elimination made use of trigonometric identities to rewrite and substitute forms of \( \tan(\theta) \) and \( \sec(\theta) \).
Replacing with \( \sec(\theta) = \frac{x}{a} \), the equation \( \sec^2(\theta) = 1 + \tan^2(\theta) \) becomes \( \left(\frac{x}{a}\right)^2 = 1 + \tan^2(\theta) \), simplifying to solve for \( \tan^2(\theta) \).
Once substituted into the equation for \( y \), this results in a hyperbola's standard equation, making it clear how the relationship evolves.
For example, given the equations \( x = a \sec(\theta) \) and \( y = b \tan(\theta) \), the objective is to remove \( \theta \) so we can derive a clear relationship between \( x \) and \( y \).
The elimination is crucial in moving from parametric to non-parametric forms, which are often easier to graph and analyze.
In this solution, our step-by-step elimination made use of trigonometric identities to rewrite and substitute forms of \( \tan(\theta) \) and \( \sec(\theta) \).
Replacing with \( \sec(\theta) = \frac{x}{a} \), the equation \( \sec^2(\theta) = 1 + \tan^2(\theta) \) becomes \( \left(\frac{x}{a}\right)^2 = 1 + \tan^2(\theta) \), simplifying to solve for \( \tan^2(\theta) \).
Once substituted into the equation for \( y \), this results in a hyperbola's standard equation, making it clear how the relationship evolves.
Other exercises in this chapter
Problem 56
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