sketch the conic r=2/1+cos theta
since e=1, it is a parabola, with vertex at (0,1) opening to the left.
Perhaps you feel more comfortable with rectangular graphing
r=2/1+cos theta
r = 2/(1+ x/r)
r + x = 2
√(x^2 + y^2) + x = 2
√(x^2 + y^2) = 2-x
square both sides
x^2 + y^2 = 4 - 4x + x^2
y^2 = 4 - 4x
Steve meant to say, vertex is (1,0)
in (r,θ) it is (0,1)
in (x,y) it is (1,0)
Dang - did I really say that?
(1,0) in either case!
To sketch the conic r = 2/(1 + cosθ), we can follow these steps:
1. Understand the equation: The given equation is in polar form, where r is the radial distance from the origin to a point on the curve and θ is the angle formed by the ray from the origin to the point.
2. Analyze the denominator: The denominator in the equation is 1 + cosθ. Since the cosine function oscillates between -1 and 1, the denominator will be greater than zero for all values of θ except when cosθ = -1. Thus, we can conclude that the curve is defined for all θ except θ = π.
3. Identify the points: To sketch the conic, we need to find some points that lie on it. We can start by choosing a few values of θ, such as 0, π/4, π/2, π, and 3π/4.
- For θ = 0, we have r = 2/(1 + cos(0)) = 1. Therefore, the point is (1, 0).
- For θ = π/4, we have r = 2/(1 + cos(π/4)) = 2/(1 + √2/2) = 4/ (2 + √2) ≈ 1.314. The point is approximately (1.314, π/4).
- For θ = π/2, we have r = 2/(1 + cos(π/2)) = 2/(1 + 0) = 2. The point is (2, π/2).
- For θ = π, the denominator becomes 1 + cos(π) = 1 + (-1) = 0, which is undefined. Therefore, there is no point at θ = π.
- For θ = 3π/4, we have r = 2/(1 + cos(3π/4)) = 4/(1 + √2/2) = 8/(2 + √2) ≈ 2.628. The point is approximately (2.628, 3π/4).
4. Plot the points: Now that we have a few points on the curve, we can plot them on a polar coordinate system. We place the origin at (0, 0) and measure the radial distance r from the origin for each point and the angle θ formed by the ray from the origin to the point.
5. Sketch the curve: Once the points are plotted, we can draw a smooth curve that passes through these points. The shape of the curve will resemble an oval or an ellipse with its center at the origin.
However, please note that the above steps provide a general approach to sketch the conic. To accurately plot the entire curve, it may be helpful to use software that can generate a polar plot or use a spreadsheet program to calculate additional points based on various values of θ.