Problem 17
Question
(a) Find the eccentricity and directrix of the conic \(r=1 /(1-2 \sin \theta)\) and graph the conic and its directrix. (b) If this conic is rotated counterclockwise about the origin through an angle 3\(\pi / 4\) . write the resulting equation and graph its curve.
Step-by-Step Solution
Verified Answer
The eccentricity is 2; directrix is \(r=1\). The rotated equation is \(r = \frac{1}{1 + \sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\).
1Step 1: Identify the type of conic section
The given polar equation is \(r = \frac{1}{1 - 2\sin\theta}\). Here, \(e = 2\) (from the coefficient of \(\sin\theta\)). Since \(e > 1\), this is a hyperbola.
2Step 2: Calculate the eccentricity
The eccentricity \(e\) of the conic is the coefficient of \(\sin\theta\) in the denominator, which is 2.
3Step 3: Determine the directrix
The standard form for a conic in polar coordinates is \(r = \frac{ed}{1-e\sin\theta}\), where \(d\) is the distance from the pole to the directrix. Since \(d = 1\) for the given equation, the directrix is \(r = 1\).
4Step 4: Graph the original conic and directrix
Graph the hyperbola in polar coordinates using \(r = \frac{1}{1 - 2\sin\theta}\). The directrix is a line at \(r = 1\). This graph shows a hyperbola with eccentricity 2 and a directrix parallel to the major axis.
5Step 5: Rotate the conic equation
The given conic needs to be rotated counterclockwise by \(\frac{3\pi}{4}\) (135 degrees). To rotate the equation, use substitution for \(\theta\). Let \(\theta' = \theta - \frac{3\pi}{4}\) to get the rotated equation. Substituting this into the original, you get: \( r = \frac{1}{1 - 2\sin(\theta - \frac{3\pi}{4})}\). This simplifies to: \(r = \frac{1}{1 + \sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\).
6Step 6: Graph the rotated conic
Graph the rotated hyperbola using the equation \(r = \frac{1}{1 + \sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\). This shows the hyperbola rotated by 135 degrees.
Key Concepts
EccentricityDirectrixPolar CoordinatesHyperbola
Eccentricity
The eccentricity of a conic section is a key parameter that dictates its shape. It's denoted by the letter \(e\). For different values of eccentricity, we get different types of conic sections:
- If \(e = 0\), it's a circle.
- If \(0 < e < 1\), it's an ellipse.
- If \(e = 1\), it's a parabola.
- If \(e > 1\), it's a hyperbola.
Directrix
The directrix is a straight line used to define and construct a conic section. Each type of conic has its own relationship with this line:
- For parabolas, it is the line from which every point on the parabola is equidistant to the focus.
- For ellipses and hyperbolas, it's part of the geometric property that defines them, based on the ratio of distances from any point on the conic to the focus and to the directrix.
Polar Coordinates
Polar coordinates provide a unique way to express points on a plane. Instead of using \(x\) and \(y\) coordinates, polar coordinates use \(r\) and \(\theta\), where:
- \(r\) is the distance from a point to the origin.
- \(\theta\) is the angle formed with the positive x-axis.
Hyperbola
A hyperbola is a type of conic section characterized by two distinct open curves. It appears when you slice through a double cone.
- Each point on the hyperbola maintains a constant difference in distance from two fixed points called foci.
- The hyperbola consists of two disconnected parts known as branches.
Other exercises in this chapter
Problem 16
\(15-20\) Identify the curve by finding a Cartesian equation for the curve. $$ r \cos \theta=1 $$
View solution Problem 16
(a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is trace
View solution Problem 17
Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. $$x=10-t^{2}, \quad
View solution Problem 17
\(15-20\) Identify the curve by finding a Cartesian equation for the curve. $$r=3 \sin \theta$$
View solution