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# Families of Polar Curves: Conic Sections - Problem 1

###### 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.

We're graphing members of the family r equals a over 1 plus epsilon cosine theta. These are actually conic sections with the focus at the origin or pole. Here is an example. R equals 12 over 2 plus cosine theta. Actually this is not in the exact form that we want it. We want a 1 here, not a 2, but I can get that by dividing the top and bottom by 2. I get r equals 6 over 1 plus .5 cosine theta. So for this graph, the epsilon value is point 5. It turns out the epsilon value determines what kind of conic section we get.

In the previous example we had epsilon equal to 1, and that gives us parabola. So let's first of all observe that because we have a cosine here, and cosine is even, we're going to get symmetry across the x axis again, so let's watch for that. Then we have to plot some points.

Let's start with theta equals 0. When theta equals 0, cosine is 1. So we get 12 over 2 plus 1, 12 over 3 which is 4. When theta equals pi over 3, cosine is 1/2. So we get 12 over 2 point 5 which is the same as 24 over 5. That's 4.8.

When theta is pi over 2, cosine is 0. 12 over 2 plus 0, we get 12 over 2 which is 6. When theta is 2 pi over 3, cosine is -1/2, and we get 12 over 1.5. So this is the same as 24 over 3 which is 8. I want to plot 5 pi over 6, because we'll need the extra points. 5 pi over 6 let me just skip the process of evaluating, and it's approximately 10 and 2/3.

For pi, pi is easy because cosine of pi is -1, we get 12 over 2 minus 1, 12. Now I've taken the liberty of plotting these points already. We just have the top half of some graph here. Let me just fill them out. Starting with theta equals 0, and going around pi over 3, pi over 2, 2 pi over 3, 5 pi over 6, and pi.

Remember this graph is symmetric about the x axis, so this point is reflected down here. This point is reflected to here. This point is reflected to here. This point, the 4.8, is reflected to about here, and then it comes back to 0. So it continues around. You can see that if this is a conic section, it's got to be an ellipse.

Remember, all of these conic sections, all the graphs of this family of conics have a focus at the origin here. And this length here is the value 6, the value that we had in our numerator. This length is called latus rectum, even an ellipse has a latus rectum. It's the width of the ellipse that passes through the focus. Anyway that's the graph of our ellipse.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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