Problem 20 - Calculus Volume 3

Question

Sketch the parametric equations by eliminating the parameter. Indicate any asymptotes of the graph. $$ x=4 \sec \theta, \quad y=3 \tan \theta $$

Step-by-Step Solution

Answer

The graph is a hyperbola with asymptotes at \( y = \pm \frac{3}{4}x \).

Step 1: Understand the Parametric Equations

We are given the parametric equations: \[ x = 4 \sec \theta \] \[ y = 3 \tan \theta \] Our task is to eliminate the parameter \( \theta \) to express \( y \) as a function of \( x \).

Step 2: Rewrite Secant in Terms of Tangent

Note that \( \sec^2 \theta = 1 + \tan^2 \theta \). Using \( x = 4 \sec \theta \), we have \( \sec \theta = \frac{x}{4} \). Therefore, \( \sec^2 \theta = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \).

Step 3: Express Tan in Terms of X

Substitute \( \sec^2 \theta \) into the identity \( \sec^2 \theta = 1 + \tan^2 \theta \): \[ \frac{x^2}{16} = 1 + \tan^2 \theta \] Simplifying gives \( \tan^2 \theta = \frac{x^2}{16} - 1 \).

Step 4: Substitute Tan into Y Equation

Since \( y = 3 \tan \theta \), and \( \tan^2 \theta = \frac{x^2}{16} - 1 \), we have: \[ y^2 = 9 \tan^2 \theta = 9 \left( \frac{x^2}{16} - 1 \right) \] This simplifies to \( y^2 = \frac{9x^2}{16} - 9 \).

Step 5: Final Equation and Graph Characteristics

Rearrange the equation \( y^2 = \frac{9x^2}{16} - 9 \) to find: \[ \frac{y^2}{9} - \frac{x^2}{16} = -1 \] This is the equation of a hyperbola centered at the origin with asymptotes \( y = \pm \frac{3}{4}x \).

Step 6: Identify Asymptotes

The hyperbola implies the asymptotes are lines it approaches infinity: \( y = \frac{3}{4}x \) and \( y = -\frac{3}{4}x \). These represent the behavior of \( \tan \theta \) and are confirmed by the calculation.

Key Concepts

Understanding Asymptotes in Parametric Equations

Asymptotes are lines that a graph approaches but never actually touches. When dealing with parametric equations, identifying asymptotes helps understand the behavior of the graph as it trends towards infinity.
For the given parametric equations \( x=4 \, \sec \theta \) and \( y=3 \, \tan \theta \), the asymptotes can be found by expressing the equation in its simplified form, and noticing the paths it takes at extreme values of \( \theta \).
In the step-by-step solution, after eliminating the parameter \( \theta \), we derived the standard form of a hyperbola: \( \frac{y^2}{9} - \frac{x^2}{16} = -1 \).
Hyperbolas naturally have asymptotes formed by the equation \( y= \pm \frac{b}{a}x \). Thus, for this hyperbola, the asymptotes are \( y= \frac{3}{4}x \) and \( y= -\frac{3}{4}x \). These represent the directions the hyperbola will never cross. Understanding asymptotes ensures proper sketching and comprehension of how the hyperbola behaves.

Understanding Hyperbolas in Relation to Graphs

A hyperbola is a type of conic section that appears as an open curve with two branches. The standard form of a hyperbola equation is either \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \). For these parametric equations, eliminating the parameter led us to \( \frac{y^2}{9} - \frac{x^2}{16} = -1 \).
This represents a hyperbola centered at the origin with its transverse axis along the y-axis.
Hyperbolas consist of two separate branches that mirror each other across the center. The curves approach the asymptotes but never intersect them.
To sketch a hyperbola, plot the center, the vertices, and the asymptotes. The asymptotes guide the drawing of the hyperbola’s branches, showing how they gradually extend away from the origin.
Recognizing the relationship of hyperbolas to their equations helps in understanding the broader topic of conic sections and their graphs.

Eliminating Parameters to Simplify Equations

Eliminating parameters is a process used to convert parametric equations into a single, simplified equation. This is especially useful for sketching and identifying all key features of the graph.
In our example, we had \( x = 4 \sec \theta \) and \( y = 3 \tan \theta \). By eliminating the parameter \( \theta \), we combined these into a single equation \( \frac{y^2}{9} - \frac{x^2}{16} = -1 \).
The parameter elimination involved rewriting \( \sec \theta \) and \( \tan \theta \) in terms of \( x \) and \( y \), making use of the trigonometric identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
This process simplified the parametric form into the recognized hyperbola equation, which clarifies the graph's shape and its asymptotes.
Understanding how to eliminate parameters is crucial in mathematics, providing a streamlined path from complex parametric equations to their simplified forms. This makes it easier to analyze the behavior and properties of various graphs.