Exploring the Polars

By: Russell Lawless

For this exploration we will be exploring the following equations:

(let theta = a)

First let's look at the red graph.

Cartesian Polar

First let's look at what we see first with the Cartesian graph. The graph appears to be an ellipse with a major axis length of 10 (along the x-axis) and a minor axis length of 8 (along x=3). The center appears to be located at (3, 0) [cartesian] or (0, 3) [polar]. When we look at the Polar graph we see that it is rotated 0o. We know that cosine varies between 1 and -1 so we know that our r value will vary between 16 / (5-3) = 8 and 16 / (5+3) = 2 [involves the polar graph]. This proves to us that the major axis is 10. To figure out if the maximum and minimum values are 4 and -4 we can look at when the output of cosine is 3/5. So r = 16 / (5 - 3 * 3/5) = 5. Since r is 5 we can use the 4/5 as the output of our sine function and get y = 5 * 4/5 = 4. Therefore 4 and -4 are in fact the maximum and minimum values. It also appears that the focal points for our graph are located at the origin and (6, 0) [cartesian] or (0, 6) [polar].

Now let's look at the purple graph.

Cartesian Polar

From looking at this graph in the Cartesian plane it appears to have a major axis length of 10 (along the polar graph of π/4) and a minor axis length of 8. It appears to be a 45o rotation of the previous graph about the origin. Because we know that distances and lengths are preserved in rotations, we can also say that the center of the red ellipse was rotated 45o. Let X = a - π/4. So then the equation 16 / 5 - 3 * cos(X) is the exact same equation as the previous equation given the angle X. It also appears that the focal points lie on the graph of the equation y = x.

Finally let's look at the orange graph.

Cartesian Polar

This graph seems to be a hyperbola. It appears the the location of the two asymptotes are at x = -2 and y = -2. To see if this is correct that means that the denominator would be equal to 0.

1 - sin(a) - cos(a) = 0 (Set equal to 1)

sin(a) + cos(a) = 1 (Square both sides)

sin2(a) + 2cos(a)sin(a) + cos2(a) = 1 (Use trig rules)

sin(2a) = 0

So a can either be 0 or π/2. Now we have found what angle gives us an asymptote.

It also appears that the focal point lies on the graph of the equation y = x. To see if this is correct check it out for yourself here.