Problem 40
Question
A polar equation of a conic is given. (a) Find the eccentricity and the directrix of the conic. (b) If this conic is rotated about the origin through the given angle , write the resulting equation. (c) Draw graphs of the original conic and the rotated conic on the same screen. $$ r=\frac{9}{2+2 \cos \theta} ; \quad \theta=-\frac{5 \pi}{6} $$
Step-by-Step Solution
Verified Answer
Eccentricity is 1 (parabola), directrix at \( r = \frac{9}{2} \); the new equation is rotated using angle \( \theta = -\frac{5\pi}{6} \).
1Step 1: Identify the Form of the Polar Equation
The given polar equation is \( r = \frac{9}{2 + 2 \cos \theta} \). This can be compared to the standard form of conic in polar coordinates: \( r = \frac{ed}{1 + e \cos \theta} \), where \( e \) is the eccentricity and \( d \) is the directrix. Our task is to rewrite the equation in standard form to determine these properties.
2Step 2: Calculate the Eccentricity
Compare \( r = \frac{9}{2 + 2 \cos \theta} \) to the standard form \( r = \frac{ed}{1 + e \cos \theta} \). Notice that the equation can be rewritten as \( r = \frac{9/2}{1 + (1)\cos \theta} \), showing that \( ed = 9/2 \) and \( e = 1 \). Hence, the eccentricity \( e \) of the conic is 1, indicating it is a parabola.
3Step 3: Determine the Directrix
Given \( ed = 9/2 \) and \( e = 1 \), substitute to find \( d \): \( d = \frac{9/2}{1} = \frac{9}{2} \). Thus, the directrix is at \( r = \frac{9}{2} \) or \( x = \frac{9/2} \) depending on the reference axis.
4Step 4: Apply Rotation Transformation
When rotating a polar equation by an angle \( \theta = -\frac{5\pi}{6} \), the transformed angle becomes \( \theta' = \theta + \frac{5\pi}{6} \). Substitute \( \theta = \theta' - \frac{5\pi}{6} \) into the original equation: \( r = \frac{9}{2 + 2 \cos(\theta' - \frac{5\pi}{6})} \). Apply the cosine angle addition formula: \( \cos(a - b) = \cos a \cos b + \sin a \sin b \) to find the resulting equation.
5Step 5: Calculate the Cosine Addition Components
Using \( \cos(-\frac{5\pi}{6}) = -\sqrt{3}/2 \) and \( \sin(-\frac{5\pi}{6}) = -1/2 \), plug these into the transformations. This yields the resulting equation: \( r = \frac{9}{2 + 2(-\sqrt{3}/2 \cos \theta' - 1/2 \sin \theta')} \). Simplify the cosine and sine components in the equation and finalize the new polar equation.
6Step 6: Draw the Graphs of Both Conics
Using graphing software or plotting on paper, graph the original equation \( r = \frac{9}{2 + 2 \cos \theta} \) and the transformed equation on the same axis system. The original is a parabola with its vertex at the pole, and the transformed parabola is the rotated version according to the angle given.
Key Concepts
Conic SectionsEccentricityDirectrixRotation of ConicsPolar Equations
Conic Sections
Conic sections refer to the curves obtained by intersecting a plane with a double right circular cone. These shapes are essential in mathematics as they appear in many real-world applications. The four primary types of conic sections are:
- Circle: All points are equidistant from a central point.
- Ellipse: Similar to an elongated circle, having two focal points.
- Parabola: A U-shaped curve, with each point being equidistant from a single focus and directrix.
- Hyperbola: Two mirrored curves, with two focal points.
Eccentricity
Eccentricity is a measure that defines the shape of a conic section. It indicates how much the conic deviates from being circular. The eccentricity, represented by the symbol \( e \), varies for different conic types:
- Circles have \( e = 0 \).
- Ellipses have \( 0 < e < 1 \).
- Parabolas have \( e = 1 \).
- Hyperbolas have \( e > 1 \).
Directrix
In conic sections, the directrix is a reference line that, together with the focus, helps define a conic. The position of the directrix can affect the shape and orientation of the conic section. To find the directrix in polar coordinates, we use the relationship:Given the polar equation \( r = \frac{ed}{1 + e \cos \theta} \), where \( ed = \) constant.For our equation, \( r = \frac{9}{2 + 2 \cos \theta} \), with \( e = 1 \), we have \( ed = \frac{9}{2} \). Thus, the directrix is at \( r = \frac{9}{2} \). This line helps shape the parabola by specifying how every point relates to both the focus and the line itself.
Rotation of Conics
Rotating a conic section means changing its orientation in the plane without altering its shape. This concept is particularly useful when studying polar equations, as it involves modifying the angle component, \( \theta \). To rotate a conic given by a polar equation, a transformation is applied:- For a rotation, the angle \( \theta \) is replaced by \( \theta - \alpha \) or \( \theta + \alpha \) depending on the direction of rotation.For our exercise, a rotation by \( \theta = -\frac{5\pi}{6} \) is performed. The cosine addition formula, \( \cos(a - b) \), is utilized to adjust the equation, resulting in a new orientation for the conic. This transformation shows how a simple geometric operation can alter the spatial relationship of the conic in the coordinate plane.
Polar Equations
Polar equations express conic sections in a coordinate system based on angle and radius, rather than the usual x-y coordinates. This approach is suitable for problems involving circular symmetries and rotations.For conics, a common polar form is:\[ r = \frac{ed}{1 + e \cos \theta} \]Where \( r \) is the radius, \( e \) is the eccentricity, \( d \) is the directrix, and \( \theta \) is the angle.The given equation, \( r = \frac{9}{2 + 2 \cos \theta} \), can be interpreted using this structure. Converting conics into polar equations can simplify understanding their geometric properties, and assist in graphing and transformations, ensuring a comprehensive exploration of these fascinating shapes.
Other exercises in this chapter
Problem 39
\(29-40\) Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Opens upward with focus 5 units from the ver
View solution Problem 40
Find an equation for the hyperbola that satisfies the given conditions. Foci: \(( \pm 3,0),\) hyperbola passes through \((4,1)\)
View solution Problem 40
Find an equation for the ellipse that satisfies the given conditions. Endpoints of minor axis: \((0, \pm 3),\) distance between foc: 8
View solution Problem 40
\(29-40\) Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Focal diameter 8 and focus on the negative \
View solution