Problem 9
Question
Identify the conic with a focus at the origin, and then give the directrix and eccentricity. $$ r=\frac{5}{1+2 \sin \theta} $$
Step-by-Step Solution
Verified Answer
The conic is a hyperbola with eccentricity 2 and a directrix at \( d = \frac{5}{2} \).
1Step 1: Identify the Conic Type
The given equation is of the form \( r = \frac{ed}{1 + e \sin \theta} \). This indicates that the conic section is represented in polar coordinates. In this standard polar equation, if the form is \( 1 + e \sin \theta \), then the conic section has its focus at the origin and the directrix is parallel to the polar axis.
2Step 2: Determine the Eccentricity
In the given equation \( r = \frac{5}{1 + 2 \sin \theta} \), compare it to the standard form \( r = \frac{ed}{1 + e \sin \theta} \) to identify the eccentricity \( e \). We conclude that \( e = 2 \). The value of the eccentricity \( e \) determines the type of conic: if \( e = 2 \), which is greater than 1, the conic is a hyperbola.
3Step 3: Find the Directrix
The directrix \( d \) can be found using the equation \( ed = 5 \). Substituting \( e = 2 \), we have \( 2d = 5 \). Solving for \( d \), we get \( d = \frac{5}{2} \). Therefore, the directrix is a line located at \( \frac{5}{2} \) units away from the pole, positioned such that the directrix is parallel to the line \( \theta = \frac{\pi}{2} \).
Key Concepts
Polar CoordinatesEccentricityHyperbola
Polar Coordinates
Polar coordinates provide an alternate way to describe the location of points in a plane. Instead of using the typical Cartesian system (x, y), we use
For conic sections, we often see equations that use these polar coordinates, as shown in the original exercise. The equation \( r = \frac{5}{1 + 2 \sin \theta} \) not only gives us the shape's relationship with its focus but also tells us about its alignment and orientation in polar coordinates. With the focus at the origin, this description becomes concise and meaningful, making it easier to analyze the conic's properties in this frame.
- the radial distance from the origin \( r \), and
- the angle \( \theta \) from the positive x-axis.
For conic sections, we often see equations that use these polar coordinates, as shown in the original exercise. The equation \( r = \frac{5}{1 + 2 \sin \theta} \) not only gives us the shape's relationship with its focus but also tells us about its alignment and orientation in polar coordinates. With the focus at the origin, this description becomes concise and meaningful, making it easier to analyze the conic's properties in this frame.
Eccentricity
Eccentricity is a key number that helps in classifying conic sections. It denotes how "stretched" a conic section is.
- If the eccentricity \( e \) is less than 1, the figure is an ellipse.
- If \( e \) equals 1, it forms a parabola.
- When \( e \) is greater than 1, as in our example where \( e = 2 \), the conic is a hyperbola.
Hyperbola
A hyperbola consists of two parts, known as branches, and these branches open either horizontally or vertically. When a hyperbola is defined in polar coordinates, as in our exercise, its equation particularly reveals its dynamic drawing.
Since the eccentricity \( e > 1 \), the given equation \( r = \frac{5}{1 + 2 \sin \theta} \) shows a hyperbola.
Since the eccentricity \( e > 1 \), the given equation \( r = \frac{5}{1 + 2 \sin \theta} \) shows a hyperbola.
- The directrix for this hyperbola is found parallel to one of the coordinate axes, maintaining a fixed distance from the focus.
- In this case, the directrix is positioned at \( \frac{5}{2} \) units away, parallel to the line \( \theta = \frac{\pi}{2} \).
Other exercises in this chapter
Problem 8
For the following exercises, determine which conic section is represented based on the given equation. $$2 x^{2}-2 y^{2}+4 x-6 y-2=0$$
View solution Problem 9
For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. $$ r=\frac{5}{1+2 \sin \theta} $$
View solution Problem 9
For the following exercises, determine whether the given equations represent ellipses. If yes, write in standard form. $$ 4 x^{2}+9 y^{2}=1 $$
View solution Problem 9
For the following exercises, determine which conic section is represented based on the given equation. $$4 x^{2}-y^{2}+8 x-1=0$$
View solution