Problem 32
Question
For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. $$ r=\frac{2}{3+3 \sin \theta} $$
Step-by-Step Solution
Verified Answer
The conic is a parabola with eccentricity 1.
1Step 1: Identify the Conic Section
The given equation is in the polar form, \( r = \frac{2}{3 + 3 \sin \theta} \). Recognize that this is a conic section in polar coordinates. Compare it with the general form \( r = \frac{ed}{1 + e \sin \theta} \), where it represents different conics based on the eccentricity \( e \).
2Step 2: Determine the Parameters
In the polar equation \( r = \frac{2}{3 + 3 \sin \theta} \), the denominator \( 3 + 3 \sin \theta \) suggests the comparison constant term \( 3 \) is linked with \( ed \), and the \( 3 \sin \theta \) part is related to \( e \). Thus, the eccentricity \( e = \frac{3}{3} = 1 \).
3Step 3: Identify the Type of Conic
When the eccentricity \( e = 1 \), the conic is a parabola. Since \( e = 1 \), the equation \( r = \frac{2}{3 + 3 \sin \theta} \) represents a parabola.
4Step 4: Convert to a Recognizable Parabola Form
Multiply both sides by \( 3 + 3\sin\theta \) to eliminate the denominator: \( r(3 + 3\sin\theta) = 2 \). This simplifies to \( 3r + 3r\sin\theta = 2 \). In Cartesian coordinates, \( r\sin\theta = y \), so \( 3r = 2 - 3y \).
5Step 5: Simplify and Relate to Cartesian Coordinates
Express \( r \) in terms of \( x \) and \( y \): \( r = \sqrt{x^2 + y^2} \). Substitute into the equation: \( 3\sqrt{x^2 + y^2} + 3y = 2 \), leading to the Cartesian form after simplification.
6Step 6: Identify Vertex, Focus, and Directrix
After converting and simplifying, identify significant features of the parabola. Standard form provides the vertex, and properties of the parabola give the focus and directrix. With \( e = 1 \), relate calculations to define these features in polar and Cartesian coordinates.
Key Concepts
Polar CoordinatesParabolaEccentricityCartesian Coordinates
Polar Coordinates
Polar coordinates are used to locate points on a plane using a distance and angle. Instead of finding a point based on horizontal and vertical distances (as in Cartesian coordinates), polar coordinates use:
- Distance from a reference point, known as the origin.
- An angle from a reference direction, typically the positive x-axis.
Parabola
A parabola is a symmetrical curve that looks like a U or an inverted U. In the context of conic sections, a parabola is defined by its eccentricity. The given equation\[ r = \frac{2}{3 + 3 \sin \theta} \]is a parabolic form in polar coordinates when the eccentricity equals 1. Some essential features of a parabola include:
- Vertex: The highest or lowest point on the parabola.
- Focus: A point from which distances are measured to give the curve its shape.
- Directrix: A line opposite the parabola, used with the focus to define the curve.
Eccentricity
Eccentricity is a parameter associated with conic sections that helps determine their shape:
- If eccentricity \( e = 0 \), the conic is a circle.
- If \( 0 < e < 1 \), it is an ellipse.
- For \( e = 1 \), the conic section is a parabola.
- When \( e > 1 \), it becomes a hyperbola.
Cartesian Coordinates
Cartesian coordinates are used to locate points on a plane with horizontal and vertical axes, typically labeled as x and y. In many mathematical problems, converting between polar and Cartesian coordinates can simplify the process of graphing curves. This is done by transforming the polar equations into the Cartesian form using:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
Other exercises in this chapter
Problem 31
For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. $$ \frac{x^{2}}{49}-\frac{y^{2}}{16}=1 $$
View solution Problem 31
Sketch a graph of the hyperbola, labeling vertices and foci. \(\frac{x^{2}}{49}-\frac{y^{2}}{16}=1\)
View solution Problem 32
Graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbo
View solution Problem 32
For the following exercises, graph the given ellipses, noting center, vertices, and foci. $$ \frac{x^{2}}{25}+\frac{y^{2}}{36}=1 $$
View solution