Question

Graph the following conic sections, labeling the vertices, foci, directrices, and asymptotes (if they exist). Use a graphing utility to check your work. $$r=\frac{4}{2+\cos \theta}$$

Short answer

A: The given equation represents a parabola.

Step-by-step solution

Identify the conic section

Inspecting the given polar equation, \(r=\frac{4}{2+\cos \theta}\), we can see that it is in the form of the polar equation for a conic section: $$r =\frac{ed}{1±e\cos\theta},$$ where \(e\) is the eccentricity and \(d\) is the distance from the focus to the directrix. In this case, we have \(ed = 4\) and \(1±e\cos\theta = 2 + \cos\theta\). Thus, we can determine the eccentricity \(e\) and the distance \(d\) for this conic section. First, let's find the eccentricity \(e\). For this, we need to determine the denominator within the equation: $$2 + \cos\theta = 1 + e\cos\theta$$ Subtracting \(\cos\theta\) on both sides, we get $$1 = e$$ Since the eccentricity is 1, we conclude that this is a parabola.

Change the polar coordinates to rectangular coordinates

To find the parabola equation in rectangular form, we will use the conversions: $$x = r\cos\theta$$ $$y = r\sin\theta$$ Substitute \(r=4/(2+\cos\theta)\) in these equations: $$x = (\frac{4}{2+\cos\theta})\cos\theta$$ $$y = (\frac{4}{2+\cos\theta})\sin\theta$$

Simplify for y and x

Now we will solve for \(y\) and \(x\) by multiplying both sides by \((2 + \cos\theta)\): $$x(2+\cos\theta)=4\cos\theta$$ $$y(2+\cos\theta)=4\sin\theta$$ Now, divide both sides by \(\cos\theta\) for the \(x(2+\cos\theta)=4\cos\theta\): $$x = 2\cfrac{\cos\theta}{\cos\theta} + \cfrac{\cos^2\theta}{\cos\theta}$$ $$x = 2 + \cos\theta$$ Divide both sides by \(\sin\theta\) for the \(y(2+\cos\theta)=4\sin\theta\): $$y = 2\cfrac{\sin\theta}{\sin\theta} + \cfrac{\sin\theta\cos\theta}{\sin\theta}$$ $$y = 2 + \cfrac{\cos\theta}{\sin\theta}$$

Solve for r

Now, we'll substitute \(x = 2 + \cos\theta\) and \(y = 2 + \cfrac{\cos\theta}{\sin\theta}\) back into the polar equation of the parabola: $$4 = \cfrac{4}{2+\cos\theta}r$$ Solve for \(r\): $$r = \frac{4}{2 + \cos\theta}$$

Plot the parabola

From the equation \(x = 2 + \cos\theta\), we see that the parabola has a vertex at (2, 0). Since the vertex is horizontally shifted, we know this must be a parabolic curve either opening left or opening right. And since \(x>2\), it will open to the right side. Also, we have already found out that the eccentricity \(e=1\) and \(ed=4\). Since \(e=1\), this means that \(d=4\). Thus, the distance from the vertex to the focus is 4. Now, let's plot the parabola and its components: 1. Draw the parabola's axis of symmetry, which is a vertical line through the vertex. 2. Mark the vertex on the graph as (2, 0). 3. Move 4 units to the right of the vertex and mark the focus at (6, 0). 4. Move 4 units to the left of the vertex and draw the directrix as a vertical line through x=-2. Now, check the graph with a graphing utility to ensure that the parabola and its components are correct.

Key concepts

Polar Coordinates

Polar coordinates offer a unique way of representing points in a plane. Instead of the traditional rectangular grid layout, polar coordinates use a distance from the origin and an angle from the positive x-axis to denote a point's position.
These coordinates are typically expressed as \((r, \theta)\), where:

• \(r\) indicates the radial distance from the origin (also known as the pole).
• \(\theta\) is the angle measured in radians (or degrees) from the positive x-axis.

Polar coordinates are especially handy when working with curves that are naturally circular or spiral. For instance, conic sections like parabolas can be represented using specific polar equations, such as \(r=\frac{ed}{1\pm e\cos\theta}\). In this exercise, the given polar equation is \(r=\frac{4}{2+\cos\theta}\), showing how polar coordinates help in efficiently capturing and working with the shape of a curve.

Eccentricity

Eccentricity is a measure of how an orbit deviates from being circular, but in the context of conic sections, it serves as a crucial factor distinguishing between different types of conics. For a parabola, the eccentricity is always equal to 1. This unique characteristic sets parabolas apart from ellipses, which have an eccentricity less than 1, and hyperbolas, which have an eccentricity greater than 1.
In the given exercise, the formula \(r=\frac{ed}{1\pm e\cos\theta}\) helps identify the eccentricity by the equation in the polar coordinates format. After analysis, we found that the eccentricity \(e\) equals 1, confirming that the conic section is a parabola. Understanding eccentricity is essential because it not only identifies the type of conic but also explains the curve's extent of deviation from being circular.

Parabola

A parabola is a special curve known as a conic section, resulting from the intersection of a right circular cone and a plane parallel to one of the cone's generating lines. Parabolas have several defining features:

• The vertex is the highest or lowest point on the curve.
• The axis of symmetry is a vertical line that passes through the vertex.
• Each parabola has a focus and a directrix, equidistant from the curve at any point along the parabola.

In the exercise, the equation derived from the polar coordinates leads to a parabola that opens to the right. The vertex is identified using the rectangular coordinates representation of the parabola. The focus of our parabola lies at (6, 0), and the directrix is situated as a vertical line at \(x=-2\). Understanding these components is crucial for graphing and analyzing parabolas' properties, making them essential in fields such as physics, engineering, and architecture.

Rectangular Coordinates

Rectangular coordinates are perhaps the most familiar system for plotting points on a graph. In this coordinate system, the horizontal axis (x-axis) and the vertical axis (y-axis) intersect at the origin, denoted as (0, 0). Any point in this plane is expressed as \((x, y)\), which represents the horizontal and vertical distances from the origin.
To convert from polar to rectangular coordinates, the relationships \(x = r\cos\theta\) and \(y = r\sin\theta\) are used. In the context of the exercise, these conversions facilitated plotting the parabola's equation into the familiar \((x, y)\) plane. Using the polar equation \(r=\frac{4}{2+\cos\theta}\), the transformation into rectangular coordinates helped to determine the position and shape of the parabola. This process is invaluable in scenarios requiring visualization or further analytical work using familiar grid-line systems.