 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# Families of Polar Curves: Conic Sections - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Share

We're looking at the family of polar curves, r equals a over 1 plus epsilon cosine theta. Now I want to graph r equals 6 over 1 plus 2 cosine theta. Now notice in this example, epsilon is 2.

In the previous examples, we've graphed epsilon equal to 1, we got a parabola. Epsilon equals to 1/2, and we got an ellipse. So let's see what this gives us. Let's also observe that, because there's a cosine here, because cosine of negative theta equals cosine theta, then when I plug negative theta I'm going to get r. Which means that this graph is going to be symmetric about the x axis and that will save me plotting some points. So the graph is symmetric about the x axis.

Let me plot points starting with 0. The cosine of 0 is 1, so I get 6 over 1 plus 2 is 6 over 3, 2. Let me try pi over 3. Cosine of pi over 3 is 1/2, 2 times 1/2 is 1, plus 1 is 2, 6 over 2 is 3. Pi over 2, cosine of pi over 2 is 0, so I get 6 over 1, 6. Then in 2 pi over 3 something interesting happens. The cosine of 2 pi over 3 is -1/2, times 2 is -1, -1 plus 1 is 0, 6 over 0 is undefined.

So let's just make a mental note. As theta gets closer and closer to 2 pi over 3, this value in the denominator is going to get closer and closer to 0. So this r value is going to get closer, and closer to infinity. So it goes up to infinity before it actually becomes undefined. Then on the other side, it's back again, 5 pi over 6. The cosine of 5 pi over 6 is negative root 3 over 2. So I get 6 over 1 plus 2 times negative root 3 over 2. This is equal to 6 over 1 minus root 3. If I use the approximation 7/8 for root 3 over 2, this gives me 1 minus 7/4, which is -3/4. 6 divided by -3/4 is the same as 6 times -4/3. -8, so this is an approximation. So approximately -8, that will give me a point that I can plot.

Then at pi, what's the cosine of pi? -1. So it's 1 minus 2, 6 over 1, -6. Let me plot these points. Actually I already plotted these points. I have 2,0, 3,pi over 3, 6,pi over 2, and so far I've got these here. 2, 0, where's my pi over 3? 3, that should go right here. Then this is 6 pi over 2 and remember it goes off to infinity. So it'll be just something like that. Then where does it come back? At -8,5 pi over 6. 5 pi over 6 is this direction, -8 is down here. So it's coming back from this direction here. So it comes in like that.

Remember, it's symmetric about the x axis, so this point will be reflected up here. So it's going to go right back out again, and it's going to come back in from the bottom here. We have reflections here and right about there. So it will come in like this, so that's what the graph looks like.

This is a hyperbola. Notice once again, remember that these conic sections all have a focus at the origin. This value is 6, the same value we have in the numerator here. So it looks like whenever epsilon is bigger than 1, we get a hyperbola. When epsilon is between 0 and 1, we get an ellipse. And when epsilon equals 1, we get a parabola.

An epsilon can also be negative. If our epsilon were -1, we get a parabola. If it were less than -1, like -2 we'd get a hyperbola. If it were between -1, and 0 we'd get an ellipse.