Problem 25 - College Algebra with Corequisite Support

Question

For the following exercises, convert the polar equation of a conic section to a rectangular equation. \(r=\frac{5}{5-11} \sin \theta\)

Step-by-Step Solution

Answer

The rectangular equation is \(25x^2 = 121y^2 + 110y + 25\).

Step 1: Identify the Polar Equation

The given polar equation is: \[ r = \frac{5}{5 - 11\sin \theta} \]This is the polar form of a conic section equation where \(e = 11\sin \theta\).

Step 2: Compare with Standard Form

Recall the standard form of a polar conic: \(r = \frac{ed}{1 - e\sin\theta}\) or \(r = \frac{ed}{1 - e\cos\theta}\). Recognizing that in our equation, \(5 - 11\sin\theta\) matches \(1 - e\sin\theta\), we can deduce that \(e=11\).

Step 3: Cross Multiply

Multiply both sides of the polar equation by \((5 - 11\sin\theta)\) to remove the fraction:\[ r(5 - 11\sin\theta) = 5 \]This simplifies to:\[ 5r - 11r\sin\theta = 5 \]

Step 4: Substitute Polar to Rectangular

Use the relationships \(r^2 = x^2 + y^2\), \(x = r\cos\theta\), and \(y = r\sin\theta\). Specifically, substituting for \(r\sin\theta\):\[ y = r\sin\theta \rightarrow r = \frac{y}{\sin\theta} \rightarrow y = \frac{11y}{5} \]

Step 5: Simplify the Equation

Express all in terms of x and y: \[ 5\sqrt{x^2 + y^2} = 5 + 11y \]Square both sides to get rid of the square root:\[ 25(x^2 + y^2) = (5 + 11y)^2 \]Expand it:\[ 25x^2 + 25y^2 = 25 + 110y + 121y^2 \]

Step 6: Rearrange to Form Rectangular Equation

Re-arrange the terms involving x and y:\[ 25x^2 + 25y^2 - 121y^2 - 110y - 25 = 0 \]Combine and simplify:\[ 25x^2 - 96y^2 - 110y - 25 = 0 \]\[ 25x^2 = 121y^2 + 110y + 25 \]

Key Concepts

Polar Equations

Polar equations represent curves on a plane using the polar coordinate system. They are written in terms of the radius, \( r \), and the angle, \( \theta \). In this context, conic sections such as ellipses, parabolas, or hyperbolas can be defined through a polar equation.
For conic sections, a standard polar equation can take the form of:

• \( r = \frac{ed}{1 - e\sin\theta} \)
• or \( r = \frac{ed}{1 - e\cos\theta} \)

Here, \( e \) is the eccentricity of the conic, which determines its shape:

• \( e = 1 \) indicates a parabola.
• \( e < 1 \) represents an ellipse.
• \( e > 1 \) denotes a hyperbola.

By recognizing how polar equations correspond to these geometric shapes, we can convert them into rectangular equations that incorporate \( x \) and \( y \), the Cartesian coordinates.

Rectangular Equations

Rectangular equations use the Cartesian coordinate system to describe curves. Unlike polar coordinates, which use a radial and angular relationship, rectangular equations express a relationship using x and y coordinates directly. This form is more commonly used for graphing and analyzing curves, especially when working with straight lines and parabolas.
To convert a polar equation into a rectangular form, we use the relationships:

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

By substituting these relationships into our polar equation, we can express everything in terms of \( x \) and \( y \) and rearrange the terms to isolate and simplify the equation.
This conversion allows for more straightforward calculus applications, like finding tangents and normals, evaluating limits, and more conventional graphing methods.

Coordinate Conversion

Converting between polar and rectangular coordinates involves understanding the relationship between the two systems. Polar coordinates specify a point based on its distance from the origin (r) and the angle \( \theta \) from the positive x-axis. Rectangular coordinates specify the same point using horizontal \( (x) \) and vertical \( (y) \) displacements.
Here are essential conversion steps:

• To find \( x \), use \( x = r \cos \theta \).
• To find \( y \), use \( y = r \sin \theta \).
• To find \( r \), use \( r = \sqrt{x^2 + y^2} \).
• To find \( \theta \), use \( \theta = \arctan \left( \frac{y}{x} \right) \), keeping in mind the quadrant of the point.

This conversion is particularly useful when the problem requires shifting from one coordinate system to the other to make computation easier or to offer a different perspective on the spatial relationships of existing points and curves.