Problem 42
Question
Find a polar equation of the conic with its focus at the pole. $$\begin{array}{cc}\text{Conic} & \text{Eccentricity} & \text{Directrix} \\\ \text{Ellipse} & e=\frac{3}{4} & y=-2 \end{array}$$
Step-by-Step Solution
Verified Answer
The polar equation of the elliptical conic with a focus at the pole is \(r = \frac{8}{7 - 3* cos(\theta)}.\)
1Step 1: Understand the formula for the conic section with a focus at the pole
The general formula for a conic section with a focus at the pole is \(r = \frac{p}{1 - e \cdot cos(\theta)}\) where \(r\) is the radius, \(p\) is the distance from the focus to the directrix, \(e\) is the eccentricity, and \(\theta\) is the angular coordinate.
2Step 2: Calculate the value of p
The formula for \(p\) is \(p = \frac {d}{1+e}\), where \(d\) represents the distance from the pole to the directrix. The directrix is \(y=-2\), and the pole (focus) sits at the origin \((0, 0)\). The distance \(d\) is 2 and using the given eccentricity, \(p\) evaluates to \(p=\frac{2}{1+\frac{3}{4}} = \frac{8}{7}\).
3Step 3: Substitute the values of e and p to get the polar equation
Substituting the values of \(p = \frac{8}{7}\) and \(e = \frac{3}{4}\) in the general formula, we get \(r = \frac{\frac{8}{7}}{1 - \frac{3}{4} \cdot cos(\theta)}.\) Simplifying, we get \(r = \frac{8}{7 - 3* cos(\theta)}.\)
Key Concepts
Conic SectionsEccentricityDirectrixEllipse
Conic Sections
Conic sections are curves obtained by slicing a cone at various angles. These shapes include circles, ellipses, parabolas, and hyperbolas. Each type arises from changing the angle at which you slice through.
They can be expressed in Cartesian coordinates, but in polar coordinates -- like the exercise -- they are particularly useful when dealing with problems where symmetry around a point (the pole) is apparent.
- A circle forms when the slice is perfectly perpendicular to the cone's axis.
- An ellipse results from slicing at a shallow angle, while a parabola forms when the slice is parallel to one of the cone’s slopes.
- A hyperbola comes from intersecting the cone with a steeper angle, cutting through both halves of the cone.
They can be expressed in Cartesian coordinates, but in polar coordinates -- like the exercise -- they are particularly useful when dealing with problems where symmetry around a point (the pole) is apparent.
Eccentricity
Eccentricity is a measure that describes how stretched out a conic section is. It is denoted by the symbol \( e \).
In the exercise you are considering, the ellipse has an eccentricity of \( e = \frac{3}{4} \). This tells us it is an elongated circle, but does not have an open end like a hyperbola.
- For circles, \( e = 0 \). This means they have no stretch.
- Ellipses have eccentricity values between 0 and 1, \( 0 < e < 1 \), meaning they are slightly elongated.
- For parabolas, \( e = 1 \), indicating a perfect open curve.
- Hyperbolas have \( e > 1 \), showing more pronounced stretching.
In the exercise you are considering, the ellipse has an eccentricity of \( e = \frac{3}{4} \). This tells us it is an elongated circle, but does not have an open end like a hyperbola.
Directrix
A directrix is a fixed line used to help define and construct conic sections. This line maintains a constant relationship with all points on the conic.
In polar coordinates, a directrix serves as a reference line. Each point on the conic section is maintained at a specific location relative to this line.
The distance \( d = 2 \) from the pole to this directrix was used in solving the exercise.
In polar coordinates, a directrix serves as a reference line. Each point on the conic section is maintained at a specific location relative to this line.
- For an ellipse, like the exercise problem, the directrix helps define the elongation and orientation.
- Mathematically, it ties into the eccentricity and helps compute other parameters like \( p \), the semi-latus rectum.
The distance \( d = 2 \) from the pole to this directrix was used in solving the exercise.
Ellipse
An ellipse is a type of conic section characterized by its oval shape. Unlike a circle, its two focal points aren't equidistant from the center, causing it to elongate.
Ellipses can range from nearly circular to highly flattened shapes, depending on their eccentricity.
In the exercise, the polar equation derived describes such a shape:
\[ r = \frac{8}{7 - 3 \cdot \cos(\theta)} \]This equation reflects an ellipse with specific conditions:
Ellipses can range from nearly circular to highly flattened shapes, depending on their eccentricity.
In the exercise, the polar equation derived describes such a shape:
\[ r = \frac{8}{7 - 3 \cdot \cos(\theta)} \]This equation reflects an ellipse with specific conditions:
- An eccentricity of \( \frac{3}{4} \), indicating moderate elongation.
- A directrix located horizontally at \( y = -2 \).
Other exercises in this chapter
Problem 41
Find the standard form of the equation of the hyperbola with the given characteristics. Foci: (0,±8)\(;\) asymptotes: \(y=\pm 4 x\)
View solution Problem 41
Find the center, vertices, foci, and eccentricity of the ellipse. Then sketch the ellipse. $$9 x^{2}+4 y^{2}+36 x-24 y+36=0$$
View solution Problem 42
Use a graphing utility to find the rectangular coordinates of the point given in polar coordinates. Round your results to two decimal places. $$(8.2,-3.2)$$
View solution Problem 42
Sketch the graph of the polar equation using symmetry, zeros, maximum \(r\) -values, and any other additional points. $$r=3 \sin 3 \theta$$
View solution