Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Conics and polar coordinates. The standard polar equation for lines if the point P naughty, R naughty, Theta naughty is the foot of the perpendicular from the origin to the line L. And if R naughty is greater than or equal to 0, then an equation for L is cosine of Theta minus Theta naughty equal R naughty divided by R Or we could think of multiplying both sides by R and we would get R naughty equaling R cosine parenthesis Theta minus Theta naughty. Now this is one equation for L There are going to be many because it's a conic and we can come up with lots of different equations for an equation. If we look at an example, if we have two, π thirds, our polar equation, if we have R naughty equal R cosine Theta minus Theta naughty, we get 2 equal R cosine Theta minus π thirds. If we wanted to expand this out, we know that cosine Theta minus π thirds, so we get 2 equal R cosine Theta cosine π thirds plus sine Theta sine π thirds 2 equal R cosine 1/2 + r sine Theta sqrt 3 / 2. And we know that R cosine Theta is really just X&R sine Theta is really just Y. So our polar equation for this line would be two equal R cosine Theta minus π thirds. And our equivalent Cartesian equation would be two equal 1/2 X plus root 3 / 2 X. If we look at this graph, we can see that we're going to have a positive Y intercept if we kept going and a negative slope. And if we solved this out, if we multiplied by two, we'd get 4 equaling X plus root 3 Y. And then if we divided through by root 3 and subtracted X, we'd get negative root 3 over three X + 4 root 3 / 3 equaling Y, which is what we would expect a positive Y intercept in a negative slope circles. If we have a circle somewhere out here with a point on the circle being R Theta and the center of the circle P naughty being R naughty, Theta naughty, then by the law of cosines of triangle P naughty PO this triangle, here we can see that a ^2 the radius of the circle would equal b ^2 + C ^2. These two sides b ^2 + C ^2 -. 2 BC cosine angle A, where this angle is our angle A. So we could think of our angle A really as Theta minus Theta naughty. So this angle A in here is really Theta minus Theta naughty. So now our b ^2, it's just going to be our R naughty squared all the way out from the origin. And our C ^2 is going to be our R. So replacing our B and our C in our our angle a, we're going to get a squared equaling R naughty, squared plus r ^2 -, 2 R naughty, R cosine Theta minus Theta naughty. So that's a generic formula for a circle where whose origin is not at the pole. If we look at the circle passing through the origin of the pool, then we get the fact that R naughty would equal A. So now we'd have a squared equaling a ^2 + r ^2 -, 2 AR cosine Theta cosine of Theta minus Theta naughty. So if we subtract our A squareds and we take the R-squared to one side, we'd get 2 AR cosine Theta minus Theta not equaling R-squared. If we divide by R, we get R equal to a cosine Theta minus Theta naughty. Now if Theta naughty equals 0, thus the point on the pole axis or the X axis, so we get 2A cosine Theta -0. Well, Theta -0 is just Theta if Theta not equal π halves. Thus the point lies on Theta equaling Pi halves or the Y axis. So we'd get 2A cosine Theta minus π halves equaling R. Or we could also think of that as 2A cosine Pi halves minus Theta equaling R because cosine is an even identity. So then if we expanded that out, we get 2A cosine Theta cosine, Pi halves plus sine, Theta sine Pi halves cosine π halves is 0, sine Pi halves is 1. So we would get 2A sine Theta equaling R. That's if our Theta was Pi halves if Theta equaled Pi the left or the negative X axis. SO2A equal cosine Theta minus π equaling R2A cosine Theta cosine π plus sine Theta sine π equaling R So the cosine π is -1 sine π is 0, so we get -2 a cosine Theta equaling R. Our last possibility is a Theta equal Theta not equal 3 Pi halves. This would be the bottom or the negative Y axis. So we get 2A cosine Theta -3 Pi halves equaling R. Expanding it out again, cosine 3 Pi halves is 0 sine 3 Pi halves is -1 so R equal -2 a sine Theta equaling R. So these four equations are equations about where a circle would be. In this case the circle would be Theta naughty being three Pi halves or the negative Y axis. So the circle would be down here because our Theta naughty would have to be on the bottom or the negative Y axis direction. We're going to talk about eccentricity and eccentricity is really just a ratio and it's the ratio of PF divided by PD. So eccentricity is PF divided by PD, PF being point to focus divided by .2 directrix. So if our eccentricity is 0, it's an ellipse or it's a circle. Sorry. If eccentricity equals zero, it's a circle. If the eccentricity is in between zero and one, it's an ellipse. If the eccentricity is one, it's a parabola, and if the eccentricity is greater than one, it's a hyperbola. So to find equations for... parabolas, hyperbolas, we're going to place one focus at the origin and the corresponding directrix. In this case, we're going to do it to the right. There'll be four different equations, but this time we're going to corresponding directrix to the right of the origin along the vertical line X equal K So our eccentricity is just a point to the directrix times E equaling our point to our focus because eccentricity was PF divided by PDPF divided by PD. So when we look at this PD is really just going to be this K distance here minus the R cosine Theta in this triangle. If we look at this triangle, we can see that cosine Theta is just FB over R or FB is R cosine Theta. So the PD, this distance here, PD is going to be the whole line distance minus FB. So PD here turns into K -, r cosine Theta times our eccentricity. And our PF is just our distance R. So now if we distribute out our E, we get Ek minus ER cosine Theta equaling R We're going to solve for R So we're going to take the term that had an R in it all to one side, and we're going to factor the R out. So we get Ek equaling R times the quantity 1 + e cosine Theta. And then to get R by itself, we're just going to divide 1 + e cosine Theta on each side. So this is one equation for a polar graph where our directrix is to the right of the graph. Now the positive is going to tell us the directrix is to the right, and the cosine Theta is actually going to tell us that our polar graph is going to be going right and left. So when we look at some other examples, if we wanted to have our directrix above, we'd have the same concept, PF equaling EPD. This time our PF is R again, just like before E. Now we have K minus PB. So here's our distance K Here's PB. Well, PB this time is really our sine Theta. Sine Theta would be PB over R or R sine Theta equal PB. So we have R equaling E times the quantity K -, r sine Theta. Multiply it out. Get all the Rs to one side and then divide to get R by itself. We get R equal Ek divided by 1 + e sine Theta. So this is when the directrix is above the polar graph and we know it's above by that plus there and this time the polar graph is going to open up and down due to the fact that it's a sine. If we look at the directrix is the line X equal negative K and the origin is the focus. Now if we did the same exact steps, we'd get a negative here, which tells us it's to the left. Cosine Theta tells us it was opening left and right. Our last possibility, Oops, our last possibility. Where did it go? Is here. That's telling us the directrix is below and the signs telling us it's opening right and left. If we look at an example, if we have our equaling 3 halves and X equaling 2, the fact that E is greater than one tells us it's a hyperbola. So we're going to have R equaling Ek divided by 1 + e cosine Theta. Well, our X distance is really RK, so E times K3 halves times 2 / 1 + e cosine Theta. So the polar graph for this would be 6 / 2 + 3 cosine Theta. We don't use leave a fraction inside a fraction. We don't do complex fractions. Another example, if we have R equaling 25 / 10 + 10 cosine Theta, and we want to figure out what this is and what the directrix is, we first of all need this 10 to turn into a one. So we're going to divide everything through by 10 and we're going to get twenty 5/10 / 1 plus cosine Theta. Well, our E is the number that's right here, and our E this time is an understood one, so we can see that it's got to be a parabola. And if E is one, Ek is 25 tenths, so E is 1K is five halves. So our directrix is X equal 5 halves. So this is a parabola. If we wanted to make a rough sketch, this is a parabola. The directrix is going to be to the right because it's a positive. So our directrix is over here somewhere. X equal 5 halves are. Focus is at the pole, so we can find our vertex by plotting some points. If we wanted our vertex, we could say what happens when Theta is 0 Pi halves π and three Pi halves. So if we have Theta being zero, we have our R being cosine of zeros. 125 twentieth, 25 twentieths would be 5 fourths. So when our Theta is zero, we come out of radius of 5 fourths, which we would expect for a parabola because that's exactly halfway between the focus and the directrix. If we had Pi halves, cosine of π halves is 0, so we'd get 25 tenths, which would reduce to five halves. And we know also that three Pi halves has a cosine of 0. So at Pi halves we're out five halves for my radius and we're three Pi halves. We're at 5 halves. Also at π. We know it's -1, so this is actually undefined here. So if we just did a rough sketch, we'd have a parabola looking something like that. If we keep going for our next topic, we get... and our ellipse is eccentricity equals C / a. Or we could think of C equaling EA center to focus is our C, center to vertex is our A. So now if we look at this, we're going to have one of the vertices or one of the focus at the pole. The distance from the focus to the directrix is going to be K, So X equal K. So FV is really all of A -, C so FV equal A -, C Up here we figured out was E times A. So FV equal A -, e AFP. Some point on the curve equals E times the point, the distance of the point to the directrix. So FV equal EVB because V is a point on the ellipse. So if A equal EAFV is now just EVB, we can now solve for VB by dividing everything through by E. So we get AE minus A equal VB, the center to be CV plus VB. So A plus AE minus A or just A / E Because the A's cancel K is going to equal FV plus VB or A minus EA plus A / E -, A. The A's once again are going to cancel. So we get K equaling AE minus EA or K * E = A -, e ^2 A. So our generic formula, remember, was R equal Ek over 1 + e cosine Theta. Now this 1 + e cosine Theta in the bottom will do different things based on where our focuses and our directrix. It could have been 1 -, e cosine Theta or 1 + e sine Theta or 1 -, e sine Theta. But the KE, here's the important piece. So instead of KE, we're going to put in A -, e ^2 A. Or we could factor out an A here. So then if e = 0, then R equals A 1 - 0 ^2 / 1 + 0 cosine Theta. And that's really just a. So remember, if E is 0, it's a circle. So if E is 00, we really just get R equal in our A, which is what we would expect. So this is our important formula here. R equal a times the quantity 1 -, e ^2 / 1 + e cosine Theta. So an example if we have E equal 1/3 and y = 6, E is 1/3 tells us it's less than one, hence it's an ellipse. So if we have R equal Ek over 1 + e sine Theta y = 6, and we knew that our directrix was above, hence the plus. And because it's AY, it's a sine value. So now we're literally going to plug stuff in R equal 1/3 * 6 / 1 + 1/3 sine Theta. We don't leave complex fractions, so we're going to multiply everything through by three R equal 6 / 3 plus sine Theta. Thank you and have a wonderful day.