Problem 7 - Precalculus

Question

Write the polar equation of the conic for \(e=1, e=0.5,\) and \(e=1.5\) Identify the conic for each equation. Verify your answers with a graphing utility. $$r=\frac{2 e}{1-e \sin \theta}$$

Step-by-Step Solution

Answer

For \(e = 1\), the conic section is a parabola with equation \(r = \frac{2}{1 - \sin\theta}\). For \(e = 0.5\), the conic section is an ellipse with equation \(r = \frac{1}{1 - 0.5\sin\theta}\). For \(e = 1.5\), the conic section is a hyperbola with equation \(r = \frac{3}{1 - 1.5\sin\theta}\).

Step 1: Substitute \(e = 1\) into the given equation

Substituting \(e = 1\) into \(r = \frac{2e}{1 - e\sin\theta}\) gives \(r = \frac{2}{1 - \sin\theta}\). We now have the equation of the curve for \(e = 1\), which corresponds to the conic section of a parabola.

Step 2: Substitute \(e = 0.5\) into the given equation

Substituting \(e = 0.5\) into \(r = \frac{2e}{1 - e\sin\theta}\) gives \(r = \frac{1}{1 - 0.5\sin\theta}\). We have now found the equation of the curve for \(e = 0.5\), which corresponds to the conic section of an ellipse.

Step 3: Substitute \(e = 1.5\) into the given equation

Substituting \(e = 1.5\) into \(r = \frac{2e}{1 - e\sin\theta}\) gives \(r = \frac{3}{1 - 1.5\sin\theta}\). We now have the equation of the curve for \(e = 1.5\), which corresponds to the conic section of a hyperbola.

Key Concepts

Conic Sections

Conic sections are the curves obtained by slicing a cone with a plane. Depending on the angle of the plane with respect to the cone, we get different shapes: a circle, an ellipse, a parabola, or a hyperbola. In polar coordinates, these shapes can be represented using equations that involve eccentricity.

• A circle is a special type of ellipse where the eccentricity, \( e = 0 \).
• An ellipse, including a circle, has an eccentricity \( e \) such that \( 0 \leq e < 1 \).
• A parabola has exactly \( e = 1 \).
• A hyperbola has an eccentricity \( e > 1 \).

In the provided equation, \( r = \frac{2e}{1 - e \sin \theta} \), each value of \( e \) signifies a different conic section:
- \( e = 1 \) results in a parabola.
- \( e = 0.5 \) results in an ellipse.
- \( e = 1.5 \) results in a hyperbola.
Understanding the role of eccentricity aids in identifying the type of conic section represented by a particular polar equation.

Eccentricity

Eccentricity in mathematics determines how much a conic section deviates from being circular. It's a descriptor of the shape of conic sections, helping classify them.
For conic sections, eccentricity \( e \) defines the following:

• \( e = 0 \): The conic is a circle. Here, the shape is perfectly round.
• \( 0 < e < 1 \): The conic is an ellipse. This results in an elongated circle or oval.
• \( e = 1 \): The conic is a parabola, indicating a curve where the distance from a point to a fixed line and a fixed point is equal.
• \( e > 1 \): The conic is a hyperbola, showcasing two separate curves.

By substituting each value of \( e \) into the polar equation \( r = \frac{2e}{1 - e \sin \theta} \), we find different shapes based on the given eccentricity.
Eccentricity provides a simple numeric value that helps in understanding and classifying the geometry of conic sections.

Graphing Utilities

Graphing utilities are invaluable tools for visualizing mathematical equations, particularly with conic sections. They help in converting complex polar equations into understandable visuals, allowing us to verify results effectively.
Using graphing utilities, such as graphing calculators or software like Desmos and Geogebra, one can:

• Plot the polar equation \( r = \frac{2e}{1 - e \sin \theta} \) for different \( e \) values.
• Identify and confirm the type of conic section—ellipse, parabola, or hyperbola—by analyzing the graph.
• Gain insights into symmetry, orientation, and size of the conic through visual representation.

Verifying solutions through graphing utilities enhances one's understanding by linking abstract equations with tangible graphs, improving comprehension and problem-solving skills.