Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Polar coordinates are the points PR, Theta. So the point is in the form the radius, the angle. So when we look at a polar coordinate, we have a pole, IE where the two lines come together, we have the polar axis, which is always going to be in the X direction positive. We have our angle Theta and we have some radius, some distance out on that ray. So if we look at some examples, A2, π force, we're going to go to our π force angle and we're going to count out to 1-2 and put in our point B3, 4 Pi thirds. So we're going to go to four Pi thirds here. We're going to count out three 123. Well, what if the radius was negative? We're going to go to the angle, in this case negative π halves this direction, and then we're going to go -2 So instead of coming out on negative π halves, we're going to go through the pull backwards 1-2. That's where our Point C would be D30. So our angle is 0 and we're going to come out three 123 and we get our point D What about -3 Pi? We're going to go to the angle π and instead of coming out on the ray in the positive direction, we're going to go through the pole backwards 123. So right there, D&E are actually on the same location. Now we're going to use that to figure out about conics. By definition, we're going to have PF, which is the focus to any point on the curve divided by PD, which is the point to the directrix. That ratio is going to equal E, where E is the eccentricity and E is greater than 0. So by definition, PF divided by PD equal E. Now that definition tells us what type of conic in a polar. So if E is equal to 1, the conic's a parabola, if E is less than one, it's an ellipse, and if E is greater than one, it's a hyperbola. So if we know that PF divided by PD equals E, we're going to figure out some other things like a formula for this. When we look here, we're going to define the distance between F&D to be P And we know that if F is our pull, that our R or our point could be thought of as Theta, rural route Theta. Sorry, we went over Theta and up our R Theta. So PF is really, really just a distance R. So we're going to take away PF and we're going to replace it with R, the radius. So then here PD, this distance here is really the same distance as down here. So this distance is P plus FQ for my PD. But FQ is really part of this right triangle. So FQ we could think of as R cosine Theta. If we look at this triangle here, hopefully we see that cosine Theta would be FQ divided by R. So R cosine Theta equal FQ. So take out FQ put in our cosine Theta. So now we're going to take this equation and we're going to solve for R So we're going to cross multiply R equal EP plus ER cosine Theta, take all the Rs to one side and then factor out the R. So R equal EP divided by 1 -, e cosine Theta. So when it's a -, e cosine Theta, what that's telling us is the directrix is to the left of the pole. The reason the directrix is to the left of the pole, the negative tells me whether it's to the right or left or positive or negative. No right or left or up or down. But the cosine or we might have a sine here depending on our direction, is going to tell me left or right for cosine, up and down for sine. So the negative here tells me the directrix is on the left. The cosine told me that the curve was opening right and left. Let's look at some examples. Oh, let's look at some given standard equations first. So let the pole be a focus on a conic section of eccentricity E with the directrix P units from the focus. The equation of the conic is given by one of the four equations listed. So R equal EP divided by 1 + e cosine Theta. So here we have the focus at the pole. Because it's a positive, the directrix is to the right. Because it's also because it's a cosine Theta. The cosine Theta tells me we're opening right or left. So we have our focus and we have our directrix here, our equal EP divided by 1 -, e cosine Theta. So the fact that it's a cosine Theta tells me it's opening right or left. The fact that it's a minus tells me my directrix is on the left side. If it was R equal EP divided by 1 + e sine Theta, the sine Theta tells me it's opening up or down and the positive tells me the directrix is above it. If it was R equal EP 1 -, e sine Theta, the sine Theta tells me up or down and the negative tells me the directrix is below. So if we have r = 3 / 1 plus cosine Theta, first of all EP is the top and the E is the coefficient on the trig function as long as we're starting with A1 here. So the fact that the coefficient was a one equal 1 means we know it's a parabola and we know EP is 3, so P is 3. So the distance between the focus and the directrix was three. The focus is always going to be at the pole. So this distance here is three. We knew it was going to open right and left because it's a cosine and it's a plus. So we need the direct was going to be on the right, a distance of the P away. So P equal 3 here. So to graph this, we're going to look at what happens when Theta zero Pi halves π three Pi halves. Well, when Theta is zero, we get 3 / 1 plus cosine 0. Cosine zero is 1, so we get three halves, 0. So if we come our zero direction, we get three halves out. And that's going to be our vertex in this case, because we know our vertex is halfway between our focus and our directrix. Here's our focus. Here's our directrix. The whole distance was three, so three halves is halfway between and it's R, Theta, so three halves, zero. When Theta is π halves, that cosine Pi halves is 0, so we get three Pi halves. So go to the Pi halves angle and out three 123. Now that's just a point on the curve when Theta is π it's not possible because the denominator becomes undefined. When Theta is 3 Pi halves, we get 33 Pi halves, so out 123. So that's a graph, a sketch of our graph and polar of 3 / 1 plus cosine Theta. It's going to be a parabola. It's opening to the left. We know where our focus is and our directrix and our vertex looking at another one, R equal 20 / 5 -, 9 cosine Theta. Well, first thing, this five has to always be a one. This location right here has to be a one. So we're going to divide everything through by 5. We're going to get 4 / 1 -, 9 fifths cosine Theta. So we know that our E is the coefficient. So E is 9 fifths on the trig function and our EP is 4. So if E is 9 fifths and EP is 4, the E is 9 fifths tells us we're expecting a hyperbola. So 9 fifths times P equal 4 P equal 20 ninths. So if we look at what happens, so that actually tells us a bunch of things. We know that our directrix is going to be to the right of the pole, to the left of the pole, to negative left of the pole, and we're going to be opening right or left. We know that the distance is going to be 20 ninths away. So we're going to have X equal -20 ninths. Now when Theta equals zero, we're going to literally stick 0IN and solve. So 20 / 5 -, 9 or -5 S when Theta zero, here's our zero, we're going to go -5. So if this is our zero, our -5 means we're going backwards to right here distance of -5. Now remember -20 ninths is -2 and two ninths. So we knew we had to go past that. So that's going to be one of my vertices. When Theta is π when we do our math here, we get Theta equal to π So we get Pi 10 sevenths. So from the focus, we're going π and we're going out 10 sevenths or one and three sevenths. So we get this is our second vertices. We could put other points in. We could say what happens when Theta equal π halves. Now we know that the cosine of π halves is 0, so we'd get 20 / 5 or 4, Pi halves. And when Theta is 3 Pi halves, we're going to get the same thing. So we're going to get 4, three Pi halves. So at Pi halves we'd be out at 4:00 and three Pi halves we'd be at 4:00, so 43 Pi halves. So those are just points on the curve. And then we would sketch in the curve if R equal 13 / 6 + 2 sine Theta. We need a one here, so we're going to divide everything through by 6 this time. Thirteen 6 / 1 + 1/3 sine Theta. So my EP is 13 six, my E is 1/3. Thus we know it's an ellipse. So we have 1/3 P equal 13 six or P equal 13 halves. This time, because it's a sign, we know we're going up and down and it's a plus. So we know the directrix is above. So here's our directrix, and it's a distance of that P, so 13 halves away from the focus. So the focus to the directrix is 13 halves. So because it's a sign, we're going to want to think about what happens at Pi halves and three Pi halves. Those are going to be our extreme values, how far away they are. So if I put in Pi halves 13 / 6 + 213 eights π halves, so we're going to go to the Pi halves angle, we're going to go out 13 eighths, and that's one of our vertices. When Theta is 3 Pi halves, we're going to get 13 / 6 -, 2 or 13 fourths. So at three Pi halves, we're going to go out 13 fourths for other vertex to get a few more points. Let's do one Theta zero because we know that sine of 0 just goes away. So we're going to get 13-6, zero, and at π we're going to get 13-6, π. So here's zero. We're going to go out 13/6. Here's Pi, we're going to go out thirteen, six again. Now, that's not the endpoints of the minor axis because it's going through the focus, not the center. They're just two points that give us a more accurate representation of the graph. Thank you and have a wonderful day.