A Conic section is the intersection of a plane and a double-napped right circular cone. By changing the location and angle the plane intesects the cone we can get a parabola,ellipse, and hyperbola:

We can use polar coordinates to describe the conic sections above. The definition of a conic in terms of polar equations is:

Let **F** be a fixed point and ** l**
a fixed line in a plane. The set of all points

**k **is
called the eccentricity of the conic. Therefore the constant **k
**is sometimes labeled **e**.

Let the focus **F** be at the origin
of our polar coordinate system and** l **be a vertical
line at y = -p, where p>0 then we will have:

So we have the distance from p to the directrix is and the distance from the Focus to p is r. Therefore by the definition of a conic we have .

Solving for r we get .

Recall the definition of a **parabola:** The set of all points in a plane equidistant
from a fixed point F (the** focus**) and a fixed line *l*
(the **directrix)** in the plane.

If we look at the definition for the
conic above we see that when **k=1** the definition is the same as that of a **parabola with Focus F and directrix l**.

Next recall the definitions for an ellipse and a hyperbola:

An **ellipse** is the set of all points
in a plane, the **sum of** whose **distances** from two
fixed points in the plane (the foci) is **constant**.

A **hyperbola **is the set of all
points in a plane, the d**ifference **of whose **distances**
from two fixed points in the plane (the foci) is a **positive
constant.**

So with p > 0 for 0<k<1 the sum of the distances from the Foci is a constant , but for k>1 the differences of the distances from the point to the Foci are constant. So we have an ellipse if 0<k<1 and for k > 1.

Lets look at some examples: First let p=1 (the directrix is at x=-1) and k=1 we get:

What happens as we increase p? Let p = 2:

We see increasing p widens our parabola.

Now lets consider 0<k<1,p=1

Let p=2:

Now let k > 1,p=1:

For the same k, p=2:

So we see for all graphs as p increases our graph widens.

If we had considered our Focus to the right of the directrix:

then we will get

for our ploar equation.

If we choose our directrix parallel to the polar axis we have:

So we would have or . Similarly with the Focus above the directrix parallel to the polar axis, we would get:.

For where k=2, and p=1 we get :

As you can see the conic is rotated 90 degrees from that equation with the directrix perpendicular to the x-axis.

Another way to experiment with the graphs of these equations, that doesn't involve a computer is with a claculator such as the TI-83. I will go through the key strokes

Note: actually keys will be shown in green.

necessary to plot: for p,k=.5.

First to choose Polar Coordinate Mode:

Press **Mode
**to display Mode screen

**v v v > > Enter **to choose Pol graphing mode. v - the down arrow
and > the right arrow.

To enter the equation Press

**Y = .5 divide key (1-.5xcos x,T,,n) ENTER**

Press **WINDOW
**and set xmin=-2,xmax=2,ymin=-2,ymax=2.

Be sure to use the negative sign and not the minus operator for defining negative number.

Press **GRAPH.
**Now if you want to channge k
go back to Y = , adjust the equation and press GRAPH. You may
need to use the insert key when changing the equation (it is 2nd
del). You may also need to redo the window command to see the
whole graph. You might also try pressing ZOOM 6 to get the standard
window.

These graphs can also be created using a spreadsheet. To see the exploration of the Polar Form of Conics:

**To see an Example included with GSP,Click
Here**