INSTRUCTOR: Welcome to a lesson on Polar equations of Conic Sections. The goals of the video are to determine the type of conic section in polar form, and also, to graph a conic section in polar form. The graph of a polar equation in either of these two forms is going to be a conic section where the eccentricity is greater than 0 and a focus is at the pole. 

Notice, in both cases, the numerator is e times d, and our denominator is 1 plus or minus e cosine theta, or 1 plus or minus e sine theta. Notice, if the equation contains cosine theta, we have a vertical directrix at x equals plus or minus d. And if the equation contains sine theta, we have a horizontal directrix at y equals plus or minus d. 

And then, just to review, remember, if the eccentricity is between 0 and 1, we have an ellipse. If it's equal to 1, we have a parabola. And if it's greater than 1, we have a hyperbola. Let's go ahead and take a look at an animation to reinforce this idea. 

We didn't mention it, but if eccentricity is equal to 0, we have a circle. Notice, when eccentricity is between 0 and 1, we have an ellipse with a focus at the pole, as we see here. And then when the eccentricity is equal to 1, we have a parabola with a focus at the pole. And then when the eccentricity is greater than 1, we have a hyperbola with one focus at the pole, as we see here. 

OK, let's go back to our presentation. Let's take a closer look at polar equations in both of these forms. Let's first take a look at the equation that has cosine theta in it. Here's r equals 2 divided by the quantity 1 plus cosine theta. And here is r equals 2 divided by the quantity 1 minus cosine theta. 

The first thing we should notice is that the coefficient of cosine theta in both cases is equal to 1. Therefore, eccentricity is equal to 1. The next thing is if eccentricity is equal to 1, then in both cases, d must equal 2. Because these equations contain cosine theta, that means we have a directrix at x equals plus or minus d. 

So if it's plus cosine theta, we have x equals 2 is a vertical directrix. If we have 1 minus cosine theta, we have a vertical directrix at x equals negative 2, as we see here. One more connection we should make is, remember, on the unit circle, cosine theta is equal to x. And I use this to remember that if the equation contains cosine theta and it's a parabola, it's going to open along the x-axis. 

And if it's an ellipse, the major axis will be along the x-axis. And if it's a hyperbola, it'll open in both directions along the x-axis if the equation contains cosine theta. Now let's take a look at the same equations with sine theta. Once again, the eccentricity is equal to 1, because the coefficient of sine theta is equal to 1, and then d is equal to 2 again. 

But now, because the equation contains sine theta, here, we have a horizontal directrix y equals positive 2. And here, we have a horizontal directrix y equals negative 2. Remember, on the unit circle, sine theta is equal to y. And I use that to help me remember that if it's a parabola, it will open up or down along the y-axis. If it's an ellipse, the major axis will be along the y-axis, and if it's a hyperbola, it'll open up and down along the y-axis if the equation contains sine theta. 

OK, let's go ahead and take a look at one of our own examples. A lot of times, the equation that we're given is not on the form that we need to gather this important information. Notice, we need our denominator to be 1 plus or minus e cosine theta. 

The key here is this value here has to be 1. It's given to us as 2. So what we'll have to do is divide everything on the right side by positive 2. So we'll have 8 divided by 2 all over 2 divided by 2 minus 2 divided by 2 cosine theta. So this will give us 4 divided by 1 minus 1 cosine theta. 

So from this form, we can tell that e is equal to 1. So we have another parabola. d, however, this time must equal 4. And then, since this equation contains 1 minus cosine theta in the denominator, we'll have a vertical directrix at x equals negative 4. 

So let's go ahead and take this information on to the next slide and start to graph our equation. So notice I have the simplified form of the equation here. Let's go ahead and sketch our vertical directrix at x equals negative 4. 

The next thing we should remember is the focus is at the pole. So this would be our focus. And then, next, since this equation does contain cosine theta. We know that the parabola will open along the x-axis. And since we have the directrix here and the focus here, we know the parabola is going to open in this direction here. So let's just go ahead and make a note of that. 

Our parabola, when we get done, should look something like this. Now, there's a lot to remember about a parabola, but remember that the vertex will be halfway between the focus and the directrix. So we know that our vertex will be right here. And then, from here, what we can do is find additional points by completing this T table to make a nice sketch of this graph. 

But there is one more thing that we can remember to help us. Notice this parabola also has symmetry across the polar axis. So for every one point that we plot, we can actually plot two by reflecting it across the polar axis. Let's go ahead and start to complete our T table by selecting theta and then determining what r would need to be. 

Let's start by selecting theta equals 0. Well, cosine 0 is equal to 1. So notice, when theta is equal to 0, we'd have 4 divided by 1 minus 1, or 4 divided by 0. So this would be undefined. 

So if it's undefined at 0 radians, let's go ahead and see what r would be at pi radians. Well, the cosine of pi is negative 1. So we'd have 4 divided by 1 minus negative 1. That'll be 4 divided by 2, which is equal to 2. And this is actually the vertex of our parabola. 

So since it's undefined at 0 radians, but it is defined at pi radians, let's go ahead and work our way back toward 0 radians. So let's pick 2 pi over 3 radians, which is equal to 120 degrees. So we'd be over here in the second quadrant with a 30, 60, 90 reference triangle. 

So the cosine of pi over 3 radians is equal to negative 1/2. So we'd have 4 divided by 1 minus negative 1/2, or 4 divided by 3/2. That's going to give us 8/3. So when theta is 2 pi over 3 radians, right along here, r is equal to 8/3. So there's 1, 2. Let's call this 8/3 right here. 

Now, remember, it's symmetrical about the polar axis. We have another point over here. Now, let's select theta equals pi over 2. Cosine of pi over 2 is equal to 0. So we'll have 4 divided by 1. That's equal to 4. So that point would be here. And we have a corresponding point in the opposite direction. 

Let's go ahead and select one more angle. Let's select pi over 3 radians. Well, the cosine of pi over 3 is going to be positive 1/2. So we'll have 4 divided by 1 minus 1/2, which is 4 divided by 1/2. That'll give us 8. 

So when theta is pi over 3, here, theta is equal to 8. And again, we have a corresponding point on the other side of the polar axis right here. And that should be plenty of information to make a nice sketch of this parabola. It would look something like this. 

That's going to do it for this video. We'll take a look at an ellipse and a hyperbola in the next video. Thank you for watching.