Problem 29 - College Algebra with Corequisite Support

Question

For the following exercises, convert the polar equation of a conic section to a rectangular equation. \(r=\frac{6 \sec \theta}{-2+3 \sec \theta}\)

Step-by-Step Solution

Answer

The rectangular equation is \( 5x^2 - 24x + 9y^2 = 36 \).

Step 1: Understand the Conic Section Equation

We begin with the polar equation \( r = \frac{6 \sec \theta}{-2 + 3 \sec \theta} \). Recall that \( \sec \theta = \frac{1}{\cos \theta} \), which can be substituted in the equation. Our goal is to convert this into a rectangular form that uses \( x \) and \( y \).

Step 2: Substitute Secant Function

Substitute \( \sec \theta = \frac{1}{\cos \theta} \) into the equation: \( r = \frac{6 \cdot \frac{1}{\cos \theta}}{-2 + 3 \cdot \frac{1}{\cos \theta}} \). Simplify this by multiplying the numerator and denominator by \( \cos \theta \) to eliminate fractions: \( r = \frac{6}{-2 \cos \theta + 3} \).

Step 3: Express in Terms of \( r \) and \( \theta \)

The cosine function can be expressed in terms of \( r \), \( x \), and \( y \) in the polar coordinate system: \( x = r \cos \theta \). Thus, \( \cos \theta = \frac{x}{r} \). Substitute this into the equation: \( r = \frac{6}{-2\left(\frac{x}{r}\right) + 3} \).

Step 4: Clear the Denominator

Multiply both sides of the equation by the denominator to clear the fraction: \( r \left(-2\frac{x}{r} + 3\right) = 6 \). Simplifying gives: \( -2x + 3r = 6 \). Rearrange to express \( r \): \( 3r = 2x + 6 \).

Step 5: Substitute \( r \) and Simplify

Recall that \( r = \sqrt{x^2 + y^2} \). Substitute \( r \) with \( \sqrt{x^2 + y^2} \) in the equation: \( 3\sqrt{x^2 + y^2} = 2x + 6 \). Isolate the square root: \( \sqrt{x^2 + y^2} = \frac{2x + 6}{3} \).

Step 6: Eliminate Square Root

Square both sides to eliminate the square root: \( x^2 + y^2 = \left(\frac{2x + 6}{3}\right)^2 \). Expand the right side: \( x^2 + y^2 = \frac{(2x + 6)^2}{9} \). Simplify the equation: \( 9(x^2 + y^2) = 4x^2 + 24x + 36 \).

Step 7: Simplify the Equation

Distribute and further simplify: \( 9x^2 + 9y^2 = 4x^2 + 24x + 36 \). Rearrange the terms to get \( 5x^2 - 24x + 9y^2 = 36 \). This is the rectangular form of the given polar equation.

Key Concepts

Conic Sections

Conic sections are a fascinating topic in mathematics, dealing with shapes that result from intersecting a plane with a cone. These shapes include circles, ellipses, parabolas, and hyperbolas. Each has distinctive properties and equations that define it.

Polar equations are often used to describe conic sections, as they can simplify expressions and calculations. Particularly, these are defined using parameters like the radius and angle. However, to better analyze or graph these conic sections on a Cartesian plane, we often need to express them as rectangular equations using the familiar variables \(x\) and \(y\).

Converting between these forms allows us to use the wealth of techniques available for manipulating and understanding algebraic equations. By understanding conic sections and their properties in both polar and rectangular forms, we can tackle a broader range of mathematical problems and applications.

Rectangular Equations

Rectangular equations use the Cartesian coordinate system, where the position of each point is specified using \(x\) and \(y\) coordinates. These equations make it easier to visualize geometric shapes and analyze their properties on a graph.

In the conversion process from polar to rectangular, the challenge lies in correctly substituting the polar expressions for \(r\) and \(\theta\) using \(x\) and \(y\). In our example, given the equation \(r = \frac{6 \text{ sec } \theta}{-2 + 3 \text{ sec } \theta}\), we can use three key substitutions:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( r = \sqrt{x^2 + y^2} \)

Through systematic substitution and simplification, the original polar equation becomes a rectangular equation: \(5x^2 - 24x + 9y^2 = 36\). This equation provides a clearer understanding of the shape and dimensions of the conic section it represents, making further analysis more straightforward.

Understanding rectangular equations is vital for working with geometric and algebraic problems, as they provide a foundation for calculus and further mathematical studies.

Polar Coordinates

Polar coordinates offer an alternative way to describe locations on a plane, using a distance from a reference point (radius \(r\)) and an angle \(\theta\) from a reference direction. These can be particularly useful for describing phenomena with rotational symmetry, such as circular or spiral patterns.

When working with polar coordinates, converting them to rectangular coordinates is often necessary. This is because rectangular coordinates, \(x\) and \(y\), are more intuitive for plotting on standard graphs and linking to physical space measurements. It is here that identities like \(\sec \theta = \frac{1}{\cos \theta}\) become pivotal, as seen in our problem, where they aid in transforming the polar description into a familiar rectangular form.

Key relationships for these conversions include:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( \tan \theta = \frac{y}{x} \)

Mastering polar coordinates and their conversion into rectangular form enables us to navigate smoothly between different mathematical frameworks, optimizing our ability to analyze and solve problems.