since the general form of a conic is
r = ep/(1-e cosθ), you have
r= 1/(4-6cosθ)
r = 1/(4(1-3/2 cosθ))
r = (1/4)/(1-3/2 cosθ)
so e = 3/2
which conic has e>1?
r= 1 / 4-6cos theta
Ellipse
Hyperbola
Parabola
Circle
Help! Homework help needed
r = ep/(1-e cosθ), you have
r= 1/(4-6cosθ)
r = 1/(4(1-3/2 cosθ))
r = (1/4)/(1-3/2 cosθ)
so e = 3/2
which conic has e>1?
The polar equation provided is: r = 1 / (4 - 6cos(theta))
To identify the type of conic, we can start by manipulating the equation:
Multiply both sides of the equation by the denominator (4 - 6cos(theta)) to get rid of the fraction:
r * (4 - 6cos(theta)) = 1
Expanding the left side of the equation:
4r - 6r * cos(theta) = 1
Next, we need to rearrange the equation to isolate 'r':
6r * cos(theta) = 4r - 1
Now, we can divide both sides of the equation by 'r':
6cos(theta) = 4 - (1/r)
Rearranging further:
4 - (1/r) = 6cos(theta)
Subtracting 6cos(theta) from both sides:
4 - 6cos(theta) = 1/r
Now, we can invert both sides to obtain:
1 / (4 - 6cos(theta)) = r
Comparing this form to the initial polar equation provided, we can observe that they are equal. Therefore, the original polar equation can be simplified to:
r = 1 / (4 - 6cos(theta))
From this simplified form, we can conclude that the type of conic represented by the polar equation is a Circle.
Starting with the given equation: r = 1/(4 - 6cosθ)
First, multiply both sides by (4 - 6cosθ) to eliminate the fraction:
r(4 - 6cosθ) = 1
Next, distribute r to both terms:
4r - 6r*cosθ = 1
Rearranging terms to isolate r:
6r*cosθ = 4r - 1
Divide both sides by r:
6cosθ = 4 - 1/r
Now, since r = 1/√(x^2 + y^2), we can substitute this into the equation:
6cosθ = 4 - √(x^2 + y^2)
Simplifying further, we get:
4 - √(x^2 + y^2) - 6cosθ = 0
So, we have the equation of a conic in Cartesian coordinates. From this equation, we can see that both x and y are involved in the equation, and there is a square root term. This indicates that the conic is an ellipse. Therefore, the answer is:
The type of conic represented by the given polar equation is an Ellipse.