Problem 62
Question
Find a polar equation for the parabola with focus \((0,0)\) and direc\(\operatorname{trix} r \cos (\theta-\pi / 2)=2\)
Step-by-Step Solution
Verified Answer
The polar equation is \( r = \frac{2}{1+\sin(\theta)} \).
1Step 1: Identify the Standard Formula
A parabola in polar coordinates with focus at the pole is described by the formula: \( r = \frac{ed}{1 + e\cos(\theta)} \), where \( e \) is the eccentricity and \( d \) is the perpendicular distance from the focus to the directrix.
2Step 2: Understand Given Parameters
We are given that the directrix is represented as \(r \cos(\theta - \frac{\pi}{2}) = 2\). This simplifies to \( r\sin(\theta) = 2 \). The focus is at \((0,0)\), and we know that for a parabola, eccentricity \( e = 1 \). Hence, \( d = 2 \).
3Step 3: Plug Into Formula
Substitute \( e = 1 \) and \( d = 2 \) into the standard formula. Since the directrix is \( r\sin(\theta) = 2 \), the polar equation becomes: \[ r = \frac{2}{1 + \sin(\theta)} \].
4Step 4: Final Equation
Verify that substituting \( e = 1 \) and \( d = 2 \) into the formula results in the equation \( r = \frac{2}{1 + \sin(\theta)} \), which correctly represents the parabola with the given focus and directrix.
Key Concepts
Polar CoordinatesParabolaEccentricityDirectrix
Polar Coordinates
Polar coordinates provide a unique system for representing points in a plane. Unlike the Cartesian coordinate system, which uses
When working with polar coordinates, it's important to remember how they differentiate from Cartesian coordinates, and how equations in polar form can help simplify certain geometric problems.
- x and y values to locate points,
- polar coordinates use a radius and an angle.
- (r, \( \theta \)), where r is the distance from the origin,
- and \( \theta \) is the angle made with the positive x-axis.
When working with polar coordinates, it's important to remember how they differentiate from Cartesian coordinates, and how equations in polar form can help simplify certain geometric problems.
Parabola
A parabola is a curve where any point is equidistant from a:
Polar equations for parabolas often involve measuring distances using the radius, r, and angles, \( \theta \). In this exercise, the focus is at the point (0,0) or the pole. This simplifies the parabolic equation, allowing it to be expressed neatly in polar form.
- fixed point, known as the focus,
- and a fixed line, called the directrix.
Polar equations for parabolas often involve measuring distances using the radius, r, and angles, \( \theta \). In this exercise, the focus is at the point (0,0) or the pole. This simplifies the parabolic equation, allowing it to be expressed neatly in polar form.
Eccentricity
Eccentricity, denoted by e, measures the deviation of a curve from a perfect circle.
It is a crucial concept in defining conic sections like ellipses, parabolas, and hyperbolas.
It tells us that the parabola is equidistant in relation to its directrix and focus, which is why the equation can have such a simplified form.
- For a parabola, the eccentricity is always 1.
- This value indicates that the shape of the parabola will be open, unlike an ellipse which closes in on itself.
It tells us that the parabola is equidistant in relation to its directrix and focus, which is why the equation can have such a simplified form.
Directrix
The directrix is a crucial component in the definition of a parabola. It is a fixed line that works together with the focus
For this exercise, the directrix was given as r sin(\( \theta \)) = 2. This expression helps to set the perpendicular distance d from the focus.Directrix and the Equation:
- to help shape the curve,
- ensuring that the distance from any point on the parabola to the focus equals the distance to the directrix.
For this exercise, the directrix was given as r sin(\( \theta \)) = 2. This expression helps to set the perpendicular distance d from the focus.Directrix and the Equation:
- The simplified equation provides \( d = 2 \),
- which directly informs our calculations in the final equation of the parabola.
Other exercises in this chapter
Problem 61
Replace the Cartesian equations in Exercises \(49-62\) by equivalent polar equations. $$ (x-3)^{2}+(y+1)^{2}=4 $$
View solution Problem 61
Graph the lines and conic sections in Exercises \(47-56\) Find a polar equation for the parabola with focus \((0,0)\) and direc- \(\operatorname{trix} r \cos \t
View solution Problem 62
Replace the Cartesian equations in Exercises \(49-62\) by equivalent polar equations. $$ (x+2)^{2}+(y-5)^{2}=16 $$
View solution Problem 63
Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}+2 y^{2}-2 x-4 y=-1 $$
View solution