This would be an ellipse.

The polar form of any conic is [or cosine], where e is the eccentricity. If the eccentricity is between 0 and 1, then the conic is an ellipse, if it is 1 then it is a parabola, and if it is greater than 1 then it is a hyperbola. Circles have eccentricity 0.

To figure out what the eccentricity is, we need to get our equation so that the denominator is in the form . Right now it is , so multiply top and bottom by :

.

Now we can identify our eccentricity as which is between 0 and 1.

All polar forms of conic equations are in the form [or cosine] where e is the eccentricity.

If the eccentricity is between 0 and 1 the conic is an ellipse, if it is 1 then it is a parabola, if it is greater than 1 it is a hyperbola. Circles have an eccentricity of 0.

We want the denominator to be in the form of , so we can multiply top and bottom by one half:

The eccentricity is 1, so this is a parabola.

Although it's not immediately obvious, this is a circle. One way we can see this is by converting from polar form to cartesian:

multiply both sides by r

we can now replace with and with :

We can already mostly tell this is a circle, but just to be safe we can put it all the way into standard form:

complete the square by adding to both sides

condense the left side

Now this is clearly a circle.

To determine the polar equation, first we need to interpret the original cartesian graph. This is an ellipse with a vertical major axis with half its length . The minor axis has half its length . To find the foci, use the relationship

so

Since the center is , this means that the foci are at

Having a focus at the origin means we can use the formula  [or sine] where e is the eccentricity, and for an ellipse .

In this case, the focus at the origin is above the directrix, so we be subtracting. The major axis is vertical, so we are using sine.

Solving for p gives us:

The eccentricity is , in this case

This gives us an equation of:

We can simplify by multiplying top and bottom by 2:

To convert this cartesian equation to polar form, we will use the substitutions and .

First, we should expand the expression:

square x - 3

subtract 9 from both sides

group the squared variables next to each other - this will help see how to re-write this in polar form:

now make the substitutions:

This is a quadratic in r with

Solve using the quadratic formula:

Subtracting which would give us a circle with no radius, but adding so our answer is:

The directrix is to the left of the focus at the origin, and the major axis is horizontal, so our equation is going to take the form

where e is the eccentricity. For a hyperbola specifically,

First we need to solve for c so we can find the eccentricity. For a hyperbola, we use the relationship where is half the length of the minor axis and is half the length of the major axis, in this case the horizontal one.

In this case, and

add 9 to both sides

take the square root

To find the eccentricity: so in this case

gives us

Plugging in e and p:

To simplify we can multiply top and bottom by 3:

This is the equation for a parabola, so the eccentricity is 1. It opens up, so the focus is above the directrix.

This means that our equation will be in the form

where a is the distance from the focus to the vertex.

In this case, we have , so .

Because the vertex is , the focus is so we can use the formula without adjusting anything.

The equation is

.

This is the equation for a parabola with a vertex at . This parabola opens left because it is negative. The distance from the focus to the vertex can be found by solving , so . This places the focus at the origin.

Because of this and the fact that the focus is to the left of the directrix, we know that the polar equation is in the form

.

That means that this equation is

.

To put this in polar form, we need to understand its structure. We can find the foci by using the relationship where a is half the length of the major axis, and b is half the length of the minor axis.

In this problem,

.

Plugging these values into the above realationship we can solve for .

 divide by -1

Since the center is , the foci are at and .

The major axis is vertical, and the lower focus is at the origin, so the polar equation will be in the form where and for an ellipse, .

Here,  and 

Simplify the top by cancelling out 

 Simplify by multiplying top and bottom by 

Before writing this in polar form, we need to know the structure of this hyperbola. We can determine the foci by using the relationship

In this case, subtract 27

The center is at and the major axis is horizontal, so the foci are and , adding or subtracting 6 from the x-coordinate. Because the left focus is at the origin and the major axis is horizontal, our polar equation will be in the form where and for a hyperbola

Here,

Plugging this in gives:

simplify the numerator by cancelling the 2's

simplify once more by multiplying top and bottom by