Problem 31
Question
(a) Find the eccentricity and identify the conic. (b) Sketch the conic and label the vertices. $$r=\frac{2}{1-\cos \theta}$$
Step-by-Step Solution
Verified Answer
Eccentricity is 1; the conic is a parabola.
1Step 1: Recognize the Conic Type
Observe that the polar equation of the form \( r = \frac{ed}{1 - e \cos \theta} \) represents a conic section where \( e \) is the eccentricity. In this exercise, \( r = \frac{2}{1 - \cos \theta} \), which matches the form with \( ed = 2 \) and \( e = 1 \). Hence, the equation has eccentricity \( e = 1 \).
2Step 2: Calculate Eccentricity
From the equation \( r = \frac{2}{1 - \cos \theta} \), we see \( ed = 2 \) and \( e = 1 \). So, the eccentricity \( e \) is 1, which indicates that this conic is a parabola.
3Step 3: Identify and Define Conic Type
Since the eccentricity of the conic is \( e = 1 \), this defines a conic section known as a parabola. A parabola is a specific type of conic with an eccentricity of exactly 1.
4Step 4: Sketch the Parabola
In polar coordinates, a parabola described by \( r = \frac{2}{1 - \cos \theta} \) opens to the right. To sketch, note that its vertex is located at the pole (\( r = 0 \)), when \( \theta = 0\). The directrix is a vertical line to the left of the pole, and the focus is at the pole itself.
5Step 5: Label the Vertices on the Sketch
Since this is a polar form of a conic and specifically a parabola, the vertex for \( r = \frac{2}{1 - \cos \theta} \) is at the origin (pole), as it opens to the right. On the sketch, label the vertex at the pole, with the understanding it lies at the minimum radius point.
Key Concepts
Understanding EccentricityThe Role of Polar CoordinatesExploring the Parabola
Understanding Eccentricity
Eccentricity is a key concept in understanding different types of conic sections. It is denoted by the letter \( e \) and essentially measures how much a conic section deviates from being circular. The value of eccentricity dictates the type of conic:
Understanding eccentricity not only helps in identifying conic types but also plays a role in sketching them correctly.
- If \( e = 0 \), the conic is a circle.
- If \( 0 < e < 1 \), the conic is an ellipse.
- If \( e = 1 \), the conic is a parabola.
- If \( e > 1 \), the conic is a hyperbola.
Understanding eccentricity not only helps in identifying conic types but also plays a role in sketching them correctly.
The Role of Polar Coordinates
Polar coordinates are an alternative to the traditional Cartesian coordinate system. They present points in the plane using a distance from a reference point and an angle from a reference direction. This system is particularly useful when dealing with curves like circles and conics. Here's how it works:
- Instead of \( (x, y) \) in Cartesian coordinates, we use \( (r, \theta) \) where \( r \) is the radius or distance from the origin, and \( \theta \) is the angle measured from the positive x-axis.
- This approach simplifies equations for many types of curves, especially in capturing symmetry and rotations.
Exploring the Parabola
A parabola is a type of conic section with a distinctive curve. It is defined geometrically as the set of points equidistant from a point called the focus and a line called the directrix. Key characteristics of a parabola include:
- Its shape resembles a smooth, symmetric U-shaped curve.
- Each parabola is defined by its vertex, the point where it makes its sharpest turn.
- The distance from any point on the parabola to the focus is equal to the perpendicular distance from that point to the directrix.
Other exercises in this chapter
Problem 30
Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). $$\text { Focus: } F\left(0,-\frac{1}{2}\right)$$
View solution Problem 31
Find an equation for the hyperbola that satisfies the given conditions. Foci: \((\pm 5,0),\) vertices: \((\pm 3,0)\)
View solution Problem 31
(a) Use the discriminant to identify the conic. (b) Confirm your answer by graphing the conic using a graphing device. $$6 x^{2}+10 x y+3 y^{2}-6 y=36$$
View solution Problem 31
Use a graphing device to graph the ellipse. $$6 x^{2}+y^{2}=36$$
View solution