Recall that the polar equations of conic sections can come in the following forms:

, where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

will give an ellipse.

will give a parabola.

will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by .

Now, for the given conic section, so it must be an ellipse.

Recall that the polar equations of conic sections can come in the following forms:

, where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

will give an ellipse.

will give a parabola.

will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by .

Now, for the given conic section, so it must be an ellipse.

Recall that the polar equations of conic sections can come in the following forms:

, where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

will give an ellipse.

will give a parabola.

will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by .

Now, for the given conic section, so it must be a parabola.

Recall that the polar equations of conic sections can come in the following forms:

, where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

will give an ellipse.

will give a parabola.

will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by .

Now, for the given conic section, so it must be a parabola.

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, 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

.

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:

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 down-opening parabola. The vertex is at . We can figure out the location of the focus by solving . This means that , so the focus is at .

Because the focus is at the origin, and the parabola opens down, the polar form of the equation is .

This equation is

.

This is the equation for a down-opening parabola. The vertex is at . We can figure out the location of the focus by solving . This means that , so the focus is at .

Because the focus is at the origin, and the parabola opens down, the polar form of the equation is .

This equation is

.