What type of conic is represented by the polar equation?

r= 1 / 4-6cos theta

Ellipse
Hyperbola
Parabola
Circle

Help! Homework help needed

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?

To determine the type of conic represented by the given polar equation, we need to analyze the equation and simplify it.

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.

To determine the type of conic represented by the given polar equation, it's helpful to rewrite the equation in a standard form.

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.