Problem 47
Question
For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=1 ; \quad e=1\)
Step-by-Step Solution
Verified Answer
The polar equation is \(r = \frac{1}{1 - \cos\theta}\).
1Step 1: Identifying the Type of Conic
Since the eccentricity \(e=1\), the conic is a parabola. The polar equation of a conic with focus at the origin is given by \(r = \frac{ed}{1 + e\cos\theta}\) or \(r = \frac{ed}{1 + e\sin\theta}\) depending on the position of the directrix.
2Step 2: Analyzing the Directrix
The given directrix is \(x=1\). Since this is a vertical line, the cosine component is used for conversion. In polar coordinates, the directrix \(x=\frac{d}{\cos\theta}\), so \(d=1\).
3Step 3: Formulating the Polar Equation
Substitute the values \(e=1\) and \(d=1\) into the polar equation formula. Thus:\[r = \frac{1 \cdot 1}{1 - 1 \cdot \cos\theta} = \frac{1}{1 - \cos\theta}\]
4Step 4: Verifying the Equation
This equation \(r = \frac{1}{1 - \cos\theta}\) implies a parabola with a directrix that coincides with the line \(x=1\). The directrix being negative shows symmetry to the line through the pole.
Key Concepts
Conic SectionsEccentricityParabolaFocusDirectrix
Conic Sections
Conic sections are an essential part of geometry, representing the intersection of a plane with a cone. They include:
Using polar equations, we can capture these conics mathematically, where the position of the focus and the eccentricity define the shape.
- Circles
- Ellipses
- Parabolas
- Hyperbolas
Using polar equations, we can capture these conics mathematically, where the position of the focus and the eccentricity define the shape.
Eccentricity
Eccentricity (\(e\)) is a measure that determines the shape of a conic section. It quantifies how much a conic section deviates from being circular:
- For circles, \(e=0\)
- For ellipses, \(0
- For parabolas, \(e=1\)
- For hyperbolas, \(e>1\)
Parabola
A parabola is one of the four types of conic sections, and it is defined as the set of all points that are equidistant to a point called the focus and a line called the directrix.
This \(e=1\) conic section forms a symmetric curve with the following characteristics:
This \(e=1\) conic section forms a symmetric curve with the following characteristics:
- It appears U-shaped in cartesian coordinates.
- Its polar equation can be expressed as \(r = \frac{ed}{1 + e\cos\theta}\)
- It is frequently seen in the path of projectiles and reflective properties.
Focus
The focus is an essential point in the construction of conic sections, especially in parabolas. It is a point from which distances are measured along with the directrix to define the conic. In the context of the given exercise, the focus is positioned at the origin of the polar coordinate system.
For parabolas:
For parabolas:
- The focus is located at one end of the axis of symmetry.
- This focal point is critical in determining the set of points that qualify as lying on the parabola.
Directrix
The directrix is a straight line used in tandem with the focus to define a conic section. For parabolas, this line exists such that it helps maintain the constant distance rule for all points on the parabola. In the exercise under examination, the directrix is given by the equation \(x=1\), signifying a vertical line.http://web.dev
In polar coordinates:
In polar coordinates:
- The directrix helps position the conic section relative to the origin.
- It influences the general form of the polar equation.
Other exercises in this chapter
Problem 46
For the following exercises, given information about the graph of the hyperbola, find its equation. Vertices at (0,6) and (0,-6) and one focus at (0,-8) .
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For the following exercises, use the given information about the graph of each ellipse to determine its equation. Center at the origin, symmetric with respect t
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For the following exercises, find the equation of the parabola given information about its graph. Vertex is (2,2)\(;\) directrix is \(x=2-\sqrt{2},\) focus is \
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For the following exercises, given information about the graph of the hyperbola, find its equation. Vertices at (1,1) and (11,1) and one focus at (12,1) .
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