find the eccentricity to identify the conic, then sketch the conic and its directrix
r=1/1-cos theta
r=2/3+2 sin theta
r=4/1-2 cos theta
To find the eccentricity and identify the conic, we'll express each equation in terms of the Cartesian coordinates (x, y).
1. For r = 1/(1 - cos(theta)):
To convert to Cartesian coordinates, we'll use the following relationships:
x = r * cos(theta)
y = r * sin(theta)
Substituting r and simplifying:
x = (1/(1 - cos(theta))) * cos(theta)
y = (1/(1 - cos(theta))) * sin(theta)
Simplifying further:
x = cos(theta) / (1 - cos(theta))
y = sin(theta) / (1 - cos(theta))
If we eliminate the parameter theta, we get:
x - cos(theta) = 0
y - sin(theta) = 0
Squaring both equations and adding them:
(x - cos(theta))^2 + (y - sin(theta))^2 = 0
This equation represents a single point, which means the conic is a degenerate form of a conic section known as a point.
2. For r = 2/(3 + 2sin(theta)):
Using the same substitution for Cartesian coordinates:
x = r * cos(theta)
y = r * sin(theta)
Substituting r and simplifying:
x = (2/(3 + 2sin(theta))) * cos(theta)
y = (2/(3 + 2sin(theta))) * sin(theta)
Simplifying further:
x = 2cos(theta) / (3 + 2sin(theta))
y = 2sin(theta) / (3 + 2sin(theta))
If we eliminate the parameter theta, we get:
2x - 3y = -4
This equation represents a line. Hence, the conic is a degenerate form of a conic section known as a line.
3. For r = 4/(1 - 2cos(theta)):
Using the same substitution for Cartesian coordinates:
x = r * cos(theta)
y = r * sin(theta)
Substituting r and simplifying:
x = (4/(1 - 2cos(theta))) * cos(theta)
y = (4/(1 - 2cos(theta))) * sin(theta)
Simplifying further:
x = 4cos(theta) / (1 - 2cos(theta))
y = 4sin(theta) / (1 - 2cos(theta))
If we eliminate the parameter theta, we get:
x^2 + y^2 = 16/(1 - 2cos(theta))^2
This equation represents a circle centered at the origin with radius 4 units. Since it is a circle, the eccentricity is 0.
To find the equation of the directrix, we can rewrite the equation of the conic in its standard form.
For the line conic:
2x - 3y = -4
The directrix is a horizontal line given by the equation: y = (-4 + 2x) / 3
For the circle conic:
x^2 + y^2 = 16/(1 - 2cos(theta))^2
Since a circle does not have a directrix, it does not apply in this case.
Sketching the conic and directrix for each equation will require plotting the points using the Cartesian coordinates and drawing the corresponding lines or shapes.
To find the eccentricity and identify the conic for each equation, we will convert them into their standard forms by using the definitions of eccentricity for different conic sections.
1. Equation: r = 1 / (1 - cos(theta))
To find the eccentricity, we'll rearrange the equation by multiplying both sides by (1 - cos(theta)):
r(1 - cos(theta)) = 1
r - r cos(theta) = 1
r = 1 + r cos(theta)
From this form, we can identify that the equation represents a circle with a center at (1,0). The eccentricity of a circle is always 0 since it is a special case of an ellipse.
Sketching the circle and finding the directrix: Since the eccentricity is 0, the directrix is not applicable for circles. You can sketch a circle with a radius of 1 centered at (1,0) on a polar coordinate system.
2. Equation: r = 2 / (3 + 2sin(theta))
To find the eccentricity, we'll again rearrange the equation by multiplying both sides by (3 + 2sin(theta)):
r(3 + 2sin(theta)) = 2
3r + 2r sin(theta) = 2
From this form, we can identify that the equation represents an ellipse. The eccentricity of an ellipse is given by the formula e = √(1 - (b²/a²)), where a is the semi-major axis and b is the semi-minor axis. However, since the equation is not given in the usual form (r = (a(1 - e²))/(1 + ecos(theta))), it might be a bit complex to directly determine the eccentricity without further manipulation.
Sketching the ellipse and finding the directrix: To sketch the ellipse, it might be easier to convert the equation into Cartesian form by substituting r = √(x² + y²) and sin(theta) = y / √(x² + y²):
√(x² + y²) = 2 / (3 + 2y / √(x² + y²))
Square both sides:
x² + y² = 4 / [(3 + 2y / √(x² + y²))²]
Simplify the equation:
(x² + y²)(3 + 2y / √(x² + y²))² = 4
From this equation, it might be difficult to directly determine the directrix unless we further manipulate it. However, we can sketch the ellipse by plotting points or utilizing software that can graph polar equations.
3. Equation: r = 4 / (1 - 2cos(theta))
To find the eccentricity, we'll rearrange the equation by multiplying both sides by (1 - 2cos(theta)):
r(1 - 2cos(theta)) = 4
r - 2r cos(theta) = 4
r = 4 + 2r cos(theta)
From this form, we can identify that the equation represents a hyperbola. The eccentricity of a hyperbola is defined as e = √(1 + (b²/a²)), where a is the distance from the center to the vertex and b is the distance from the center to the foci. However, determining the exact eccentricity might be complicated without further manipulation.
Sketching the hyperbola and finding the directrix: Similarly, if we convert the equation into Cartesian form, it might enable us to sketch the hyperbola more easily.