 Hey, welcome to episode 28. We've just finished our fourth exam, and now we're starting the new material for exam number 5. And the good news is that's the last regular exam of the course. The first three episodes of this next sequence will be about three curves called the conic sections. Let's go to the green screen so I can introduce these. Suppose I take what I would call a double cone. Suppose that I have two cones that are sitting end to end like this. So I refer to this as a double cone. And we're going to be studying three curves called the conic sections that are derived from this figure. But first of all, before we go to those, I want to show you that when you're studying, say, plane geometry, you study basically points, lines, and circles. And you notice that if I take a plane, let's say this is a plane, and if I cut through the cone right here at that vertex, then we get a point. And if I raise that plane up and I cut it, say, right here, I get a circle. So we also get a circle. And if I were to take the plane and cut it right along this edge, right down the side here, if you can see this overlapped, I would get, if you imagine this cone is extended, I would get a line right along here. So a point, a circle, and a line, these are three curves, you might say, if you think of a point as a curve. These are three curves that you study in plane geometry. Now, today's episode is going to be about the first of three new cross-sections of a double cone, and these are called the conic sections. Let's go to the first graphic. OK, now here we see this double cone, but you notice I'm cutting it at a new angle. Imagine that the plane cuts it parallel to the edge of the cone that runs up and down. Remember, when I cut it right along that edge, I got a line. Now, if I move it in just a little bit and cut only the lower cone, then I get a curve that we call a parabola. Now, you might say, well, I thought we already studied parabolas. Weren't those the quadratic functions? As a matter of fact, this turns out to be the very same curve, but this was originally where parabolas came from, was the cross-section of a cone parallel to the edge. Now, there are two other conic sections. If I cut the lower cone at an angle, not horizontal, because if I cut it horizontally, what did we say we got? Circle. We got a circle. If I tilt it just a little bit, then I get an ellipse. And if I tilt it even more so that it's parallel to the edge, then I get a parabola. And then finally, if I tilt it even more than that, in this case, we have it drawn straight up and down, then this is the only time that I actually cut both cones into their interior. And this two-branch curve we refer to as a hyperbola. Now, episode 28, today's episode, we'll be talking about a parabola. And the next two episodes, we'll talk about ellipses and hyperbolas. OK, let's go to our objectives for this episode. First of all, we want to introduce, well, we have introduced the conic sections. Then we want to look at fundamental equations of parabolas. And then we'll look at transformations of those parabolas. And finally, we'll conclude by looking at the reflective property of parabolas and some other applications for parabolas. OK, now, we have a definition for the parabola on the next conic. Let's take a look at this. OK, this says a parabola can be defined as the set of points in the plane equidistant from a fixed point, which is called a focus, and a fixed line, which is called the directorics. The point nearest the focus and the directorics is the vertex. OK, now, at this point, you may say, wait a minute, Dennis, I'm a little bit confused. We had parabolas as quadratic functions. We had parabolas as cross sections of a double cone. And now you're telling us parabolas have this definition, which is really quite different from any of the other things that we've just mentioned. Well, we're going to tie some of this together in just a moment. But let's take a look at this definition and the parabola that's drawn just below. Now, we have picked a point on the y-axis, 0p. So this must be p units above the x-axis. And then there's a horizontal line drawn across the bottom, which is p units below the x-axis. So we have the point 0p above the focus. And we have the line below y equals negative p. That's the directorics. Now, if you look at all the points up here in the first and second quadrant that are the same distance from the focus and the directorics, those points lie in a parabola. In other words, the distance from xy to 0p and the distance from xy to the directorics, y equals negative p, those two distances should be the same. And if I pick a point, let's say a point right here, another point on the parabola, the distance from that point to the focus should equal the distance from that point to the directorics. If I pick a point, let's say over here on the parabola, then the distance from that point to the focus should equal the distance from that point to the directorics. And finally, let's go to the vertex. This point that's nearest the focus and directorics is referred to as the vertex. How far is this point from the focus? Who can tell me? It's p. p units away, whatever the value of p is. And how far is that point from the directorics? Well, the directorics is p units below, so that would also be p. So we have p going up and p going down, so this point's equidistant. OK, well, let's look at why these three ideas for a parabola are equivalent to each other. And I'm going to draw that last illustration right over here. I had a parabola that comes in to the origin. I had a point that I call the focus that was at 0p. And I have a point just below, rather a line just below. That was y equals negative p. And this point, let me just put its coordinates up here, was 0p. Now, we said that if we were to pick a random point on the parabola, I'll call this point xy, that the distance to the focus should equal the distance to the directorics. I'm going to call this distance, distance number 1. And I'll call this distance, distance number 2. And distance number 1 should equal distance number 2. OK, now, if these two distances are equal, then I should be able to find a mathematical expression to represent the distance here and the distance over here. How can I find the distance from the point xy to the point 0p? Well, we have a distance formula. What were you going to say, Jeff? I'm going to say it's the distance formula. OK, and what is the distance formula saying? How would I write that? It's the square distance is equal to the square root of x1 minus x2 squared plus y1 minus y2 squared. Right, it's the difference in the x squared plus the difference in the y squared. Now, the x's in this case are x and 0. 0 is the x-coordinate of the focus point. So that would be x minus 0 squared. And the difference in the y's would be what? Who can tell us? y minus p. y minus p, yeah, y minus p, and then we square that. So that's the distance formula for distance d1. Now, for distance d2, I think we can take a shortcut. The distance from the point xy down to the directorics would be the distance down to the x-axis plus the distance from the x-axis down to the directorics. Now, the distance from the point to the x-axis is y. Because you remember when we located that point, we went over x and we went up y. So this vertical portion is y. And this distance, we said earlier, was p. So altogether, that has to be y plus p. y plus p. So in a sense, I could say this is the equation of the parabola. But we'd like to simplify that. So what would you do to simplify this expression a little bit? Square both sides. Square both sides. Sure, let's get rid of the square root. We have x minus 0 squared plus y minus p squared equals y plus p. And over here, I'm going to have to show the square. And if I multiply this out, well, the first term is just x squared. Let's see, what's the square of y minus p? David, what's the square of y minus p? Of y minus p? What do you get when you square that? y squared minus 2yp. OK, I'll say p y and put the thing in front. And then plus p squared. Plus p squared, OK. And the square of y plus p is y squared plus 2py plus p squared. Now back here, David said negative 2yp. I'm putting the p in front because I'm thinking of p as being a constant. And I'm thinking of y as being the variable. So that's just the reason why I chose that particular ordering. OK, now at this point, I could say, well, then Dennis, this must be the equation of the parabola because we got it by setting these two distances equal. But you know, this can still be simplified further. For example, what would you do to reduce this? Susan, what would you do to simplify that? I like terms on one side. OK, cancel off like terms. Which like terms could you cancel here? A y squared. The y squared's cancel. Yeah, anything else? 2py. OK, actually, those are going to add together. This one's negative. That one's positive. Why don't I move the negative 2py, add 2py to both sides, and make that 4py. OK, but there's still something else that'll cancel. David, what do you see? Is what, Susan? P squared. P squared. P squared's cancel off. OK, so the only thing left is the x squared. Now this is the way we're going to write the equation of the parabola when the focus is chosen on the positive y-axis and the directorics is chosen below the x-axis the same distance away. x squared equals 4py. OK, let's put a little box around that. You notice that if I were to solve for y, y would equal 1 over 4p times x squared. Now look how that compares with what we saw about parabolas earlier. You remember early on in this course, we said that the fundamental parabola was y equals x squared. And in fact, this is in that form. If you pick the right value for p, if you put the focus in the right position, this equation is the same thing as our initial quadratic function. So parabolas, based on this definition, do become quadratic functions like we saw earlier in the course. Now something that we can establish in this course is that parabolas are formed by taking a cross section of the cone parallel to the edge of the cone. That's actually more of a geometry problem than a college algebra problem. So let's see, keeping this equation in mind, let me just summarize now what we said about parabolas. From earlier in the course, we said, number one, parabolas are quadratic functions. They are quadratic functions. And what I mean by that is they're functions that can be written in the form f of x equals ax squared plus bx plus c for some choices of a, b, and c. And if we complete the square, we can find out where the vertex is located and we can shift it vertically, up or down, left or right. We can stretch the curve, et cetera. Number two, we said that parabolas are formed by cross sections of a cone taken parallel to an edge. Now, this is the statement that we have not been able to verify because this is more of a geometry problem. And number three, we said that a parabola is a set of points in the plane that are equidistant from a focus point and a directrix line. Now, we've been able to make a connection between number one and number three by establishing that equation that led to a quadratic function. OK, let's go to the next graphic and we'll summarize the information that we've just computed here about a parabola that's focus is on the positive y-axis. It says, for the focus, 0p. So here we have the point 0p, p units above the origin. And directrix y equals negative p. So this is p units below the x-axis. The equation of the parabola is x squared equals 4p y. And we just derived that. I wouldn't ask you to derive this equation, but now we know what the fundamental equation of a parabola is. You notice that this graph looks like it's symmetric about the y-axis. If you fold it over, it's exactly the same on both sides. So we refer to the y-axis as the axis of symmetry. Every parabola has an axis of symmetry that passes through the focus and is perpendicular to the directrix right here. OK, now you know if p had been a negative number instead of a positive number, then the point 0p would have been below the x-axis. And the directrix, y equals the negative of p. Now if p is a negative number already, y equals negative p is a positive, gives us a positive value. The directrix will be above the x-axis. And what that will do is cause the parabola to open down rather than up. So if we find a parabola that opens either up or down, the equation of the parabola is x squared equals 4p y. And the choice of p can be positive or negative based on whether the parabola opens up or down. And the axis of symmetry will still be the y-axis. OK, now on the other hand, if we go to the next graphic. OK, now here we have a parabola that opens to the side. In this case, to the right-hand side, we have the y-axis that's solid right here. You notice this time the focus is just to the right on the x-axis. So this would be the point p0 rather than 0p. So the focus is at p0. And the directrix is a vertical line at negative p. So the directrix is x equals negative p. And we could go through the very same argument that we did a moment ago where we reduced those square roots and find out that the equation of this parabola is y squared equals 4p x. Now if p is negative instead of positive, I'll have a negative coefficient right here in that case. If p is negative, that will put the focus over on the left-hand side and the directrix on the right-hand side. And I'll have a parabola that opens out to the left. So if we see a parabola that opens to the left or right, we'll know that its equation should be basically in this form and that p will be negative if it opens to the left. And this time you can see the axis of symmetry is the x-axis because this graph is symmetric about the x-axis. If I were to fold the upper half over to the lower half, they should match up exactly. OK, let's go to our first example. OK, this problem says find the equations of the parabolas with vertices at the origin, foci on the y-axis, and have here five different choices for p. p equals 2, p equals 1, a half, a fourth, and an eighth. OK, now if we come to the green screen, let me just write those down. We have p equals 2, we have p equals 1, p equals 1 half, p equals 1 fourth, and p equals 1 eighth. Now when I say find their equations, what I mean is if the vertex is going to be on the y-axis, excuse me, if the focus is on the y-axis, and if the directrix is parallel to the x-axis but below, then in the first case, this point would be the point 0, 2. That would be the focus. I'll just put an f beside it for focus. And the directrix would be y equals negative 2. And I know that these equations are in the form x squared equals 4 p y. That was the fundamental equation that we just derived. So when p is 2, this is going to be x squared equals 8 y. So x squared equals 8 y. I'm just substituting in a 2 where the p is. In the case that p is equal to 1, what would the equation be? x squared equals 4 y. Yeah, if I put in a 1 for p, x squared equals 4 y. And if I put in a 1 half, this will say x squared equals 2 y. And if I put in 1 fourth, this will say x squared equals y. Hey, now that one sounds familiar, because this is the equation y equals x squared. And finally, if I put in 1 eighth, this is x squared equals 1 half y. OK, now what I want to do is go to a graphing calculator and graph each one of these so we can compare what the graphs look like. And on the graphing calculator, I have to solve for y if I'm going to graph it. In the first case, this is y equals 1 eighth x squared. In the second case, y equals 1 fourth x squared. And then y equals 1 half x squared. And then y equals x squared. That's our fundamental parabola. And y equals 2x squared. Now, you see all that's changing is the coefficient in front of the x squared. And what happens to our fundamental graph if I put a 1 eighth in it? Do you remember what that does to the graph? Sure. It's going to compress it. Yeah, it's going to compress it vertically. So we're going to get a really wide parabola. That one's going to be really wide. And as I put in smaller fractions, it's not going to be compressed as much. And this one's going to be a stretch right here. OK, so let's zoom in on my graphing calculator. And let's graph all of these, one on top of the other. The first one we wanted to graph was 1 eighth x squared. So I'm going to enter 1 divided by 8 times x squared. Whoops, times. I must have hit division there. Times x squared. OK, then right below it, I'm going to graph 1 divided by 4 times x squared. And then 1 divided by 2. In other words, 1 half times x squared. And then x squared. And then the last one was 2x squared. OK, and let's pick a window for these. And let's choose our minimum x to be, let's say, negative 4. And our maximum x to be positive 4. And the scale, well, let's just make it one unit for the scale. And for our minimum y, I think we could pick 0 for our minimum y, because these curves aren't going to be negative. And for our maximum y, let's say we pick 50. And for our scale, let's say we pick 10 units. OK, so we're going to graph all four of these. And I think we're going to see these graphed one after the other consecutively. So first of all, we see y equals 1 eighth x squared. And you notice it hardly gets off the ground right there. OK, next comes 1 fourth x squared. 1 fourth x squared. It's a little bit taller. OK, here is 1 half x squared. And then we have y equals x squared. And oh, that last thing we have there was 2x squared. So you see what happens is, as the p changes, these parabolas tend to get taller. Now, let's see, the p's were actually getting smaller. The p's went from 2 to 1 to 1 half. So as the p's got smaller, that forced the focus to get closer to the vertex. And as the p gets smaller, the parabola tends to get steeper in there. So let's go to the next example. OK, this problem says, find the equation of the parabola with focus at minus 1 half 0 and the directories as at x equals 1 half. Now, we can sketch a graph of this problem over here. Here's the x-axis and the y-axis. And let's locate that focus and directories. The focus is at negative 1 half 0, so I'll put it right about there. This supposed to be negative 1 at that mark. And at plus 1 half, I have a vertical line coming through here that's x equals 1 half. OK, so this tells me that the parabola is going to come out of the origin, and it's going to open which direction? To the left. To the left, OK. So the parabola is going to generally be coming out in this direction. Now, parabolas that open to the left have the equation y squared equals 4p times x. And the value of p comes from the focus right here. The focus, we said, was at minus 1 half 0, so p is negative 1 half. So if I substitute that value in, that says that y squared is equal to negative 2x. So this is the equation in its standard form that we've just derived. That's the equation of that parabola. OK, let's go to the next example. OK, this problem says, find the focus and directories of the parabola that has this equation, and then sketch its graph. OK, now let's see. The first thing I notice is this isn't in standard form. We have two standard forms that we've just arrived. One of them was x squared equals 4p y. And you remember, these are parabolas that open up or down. And then the other equation was y squared equals 4px. These are parabolas that open either to the right or to the left. OK, so what I need to do is to get this equation in one of these forms. Which form do you think we can get it into? The second one. Into the second form, because it has a y squared in it, but not an x squared. And this has a y squared. So let's divide both sides by 3, and I'll have y squared equals 2 thirds times x. OK, now if I compare this equation and this one, what must be the value of 4p? 2 thirds. 4p must be 2 thirds, exactly. 4p is equal to 2 thirds. So p must be 1 fourth of that. 1 fourth of 2 thirds, which is 2 twelfths or 1 1 sixth. OK, so now I think we can find everything we need to know. The question says sketch the graph. Now we can't make a totally accurate sketch, but I think we can make a rough draft. Would you say that this parabola will be opening up or down or left or right? OK, it'll be right. Now first of all, Stephen, how do you know it's going to be the left or right choice? Well, because it's y squared. Because we have the y squared term in it, exactly. And why did you choose right? Because 1 sixth is positive. Yes, because the choice of p is positive. And you remember, the focus is the distance over to the, is at the point that p is 0. So if I go over to 1 sixth 0, 1 sixth 0, that'll be my focus point for this graph. So I'll just say f is 1 sixth and 0. So on your homework and on your exams, when you're abbreviating this, I would just say label the focus with an f. Now there's also a directrix line. And let's see now, parabolas that are opening to the right have a vertical directrix on the other side. And what will be the equation of that directrix? x equals negative 6. x equals negative 1 sixth, yeah. And so my parabola comes out of the origin where the vertex is and it turns and goes back to this in this direction. And the vertex at the moment is 0, 0. You know, all of the parabolas we've seen so far, the vertex has been at the origin. But we'll see some variations of that in just a moment. One more question that I didn't ask here is, what is the axis of symmetry? The x-axis. Is the x-axis, yeah. So this graph will be symmetric about the x-axis. All of the parabolas that have the vertex on the x-axis and have the directrix and focus on opposite sides of the origin are going to have the x-axis for the axis of symmetry. So I'll say the axis of symmetry is the x-axis right here. Let's just take one more example like that. If we just come to the green screen, let me just make up another problem like this. Suppose we had 8x equals, let's say, 9y squared. I'll tell you what, let's make it negative 8y equals 9x squared. OK, now the two fundamental equations are x squared equals 4py and y squared equals 4px. So in which form will this one be able to be placed? In the first form, because there's an x squared in it. So if I solve for x squared, I'll need to divide by 9. And so x squared is going to be negative 8 ninths times y. Now, as soon as we have the x squared term, we know that this one opens either up or down. Let me just make a note over here that this one's going to open either up or down. And I can see that 4p is a negative number. 4p is a negative 8 ninths. So p is going to be a negative number. So will it open up or will it open down? It'll open down. So let's cross off the up choice. It's got to open down. 4p equals negative 8 over 9. And so p equals negative 2 over 9. And when I draw the graph, I'm going to be placing, let's say if this is negative 1 right here, I'm going to place the focus at negative 2 ninths, which is just below the origin. And I'll draw the directorics just above the origin. And let's say this is 1 up above. OK, so what would you say are the coordinates of the focus for this graph? 0 negative 2 ninths. Yeah, 0 negative 2 ninths. That's the focus right there, 0 and negative 2 ninths. And the directorics has what equation? Y equals 2 ninths. Yeah. Y equals 2 ninths. You remember the directorics always has the equation, if it's a horizontal line, y equals the negative of p. And p is already a negative number. So this would be y equals positive 2 ninths. The vertex is at the origin. And the axis of symmetry is what? Y axis. Is the y axis. OK, now you know, sometimes we are given information about the parabola, but we don't have an equation for the parabola. So let me make up an example where we have a parabola and we have some information about it and the question is to find the equation of it. So let me just write this problem out here on the screen. Find the equation of the parabola with vertex at the origin, which is what we've been expecting in the problem so far. And directorics, y equals negative 2. Y equals negative 2. OK, so we have some clues about the parabola, but we don't have an equation for it. And our goal is to find the equation. And furthermore, we can sketch the graph. We could sketch the graph from this information, or we could sketch the graph from the equation of it. Well, what can you tell me about the parabola based on this information? It's going to be a parabola that opens either up or down. OK, let's see. It's going to be a parabola opening either up or down. And if I graph the line y equals negative 2, here's y equals negative 2. And yeah, the vertex is right here. So it looks like it has to open around the focus and away from the directorics. So it's got to open up. Would you say this is going to be a fairly wide parabola or a fairly thin parabola, relatively speaking? Relative to what we've seen before. Well, I think to figure that out, we need to locate where is the focus. Because remember when I used the graphing calculator a while ago, and I graphed those parabolas for various choices of p? When the p is small, we tend to get rather thin, tall parabolas. And when the p is large, we tend to get flatter parabolas. Those initial ones that I graphed on the graphing calculator looked rather flat. So we need to figure out what is p. Can anyone tell me the choice? What is the value of p in this problem? 2. p is equal to 2. Because you see I'm thinking the directorics will have the equation y equals the negative of p, and we have y equals negative 2. So p must be 2. So p is rather large. So that tells me this is going to be a rather wide looking parabola coming in like this. Where's the focus of this parabola? It's 2 above. It's going to be 2 above. So it's going to be up here. And this will be the point 0, 2. And what's the fundamental equation of a parabola that looks like this? x squared equals 4py. x squared equals 4py. And now we know that p is 2. So this is the parabola x squared equals 8y. And that's the answer to the problem. Now along the way, we got a lot of other information about it. We located not only the directorics, which was given to us, but we located the focus up here. We knew that the parabola was going to be rather wide, that it opens up. And now we ultimately found its equation. And that's what we were after. OK, this information can be given in other ways. So let's look at the next example. And we'll see a different set of information that describes the parabola. This is find the equation of the parabola illustrated to the right. And we have a parabola that opens down. Its vertex is at the origin. And it passes through the point negative 2, negative 4. So it passes through the origin. And it opens down. That means we have a fundamental equation for this. And we're given one point on the parabola. Can anyone tell me what's the fundamental equation of this parabola? It's x squared. x squared equals? y, 4py. 4py, yeah. x squared equals 4py. And in order to completely write the equation of this parabola, the only thing I need to know is what is p. That's really what the problem boils down to. What is the choice of p? But we don't have the directorics given. The directorics should be somewhere up above the x-axis. We don't have the focus, which is somewhere on the negative y-axis. But we have this point. How is that going to help me calculate p? If we put negative 2 and negative 4 and for the x and the y, then we could calculate p. Right. Yeah. We could substitute x equals negative 2 and y equals negative 4. Because you see, if this is a point on the parabola, it should satisfy the equation of the parabola. Every point on the parabola should satisfy the equation of it. So if I substitute these numbers in, that's going to tell me that negative 2 squared equals 4p times negative 4. And now I have an equation with only p in it. So I can solve for p. And once I calculate p, I can plug it in up here and write the final answer. So let's do that. We have 4 equals, well, equals negative 16p. And therefore, p is equal to 4 over negative 16, which is negative 1 fourth. Let's see, that's p, not y. So p is negative 1 fourth. And if I substitute that number right there, I get x squared equals 4 times negative 1 fourth times y, which reduces to be negative y. So therefore, the equation of this parabola is x squared equals negative y. Now if you want to put the negative on the other side, if you multiply both sides by negative 1, you could say it's y equals negative x squared. If you write it in this form, you're probably thinking of parabolas in the sense that they're quadratic functions like we saw earlier. If you write it in this form, you're thinking of a parabola as being in the standard form, x squared equals 4p y. Either one of those is the equation of the parabola. So either one's fine to leave your answer in that condition. OK, so you will see in your homework problems. First of all, where you're given the equation of a parabola and you're asked to sketch it, find its focus, find its directories, then you'll be given problems where you're given information about the parabola. For example, you're given the directories, you're given the focus, you're told the vertexes at the origin, or maybe you're given a point on the parabola and you find the equation of the parabola. So you'll see quite a few problems like that. Now let's go to some new information about parabolas that we haven't considered yet. Let's go to the next graphic. You notice here we have a parabola drawn and I have that upper line that's sort of diagonal across the parabola. It just goes from one point on the parabola to another point on the parabola. That's referred to as a chord. A segment that joins two points on any of our conic cross sections is referred to as a chord. For example, in a circle, if you remember in plane geometry, if you draw a segment that goes from one side of the circle to another that's referred to as a chord, and there's a special chord on a circle that goes right through the center that we call a diameter. Well, in this case, there's a special chord that if it goes parallel to the directories down below, and if it passes right through the focus that's referred to as the lattice rectum. Lattice rectum comes from Latin. It actually means straight side. So it's the straight side, so to speak, of the parabola. And the lattice rectum has a particular length relative to the parabola. And I think we can figure it out right here. The distance from this point to the focus has to equal the distance in this point down to the directories, which isn't shown here. Now, I know that the distance from that point to the x-axis is the same as the distance from the focus to the x-axis. So what would this distance be from this end point to the x-axis? P. And the distance from there down to the directories is also P. So what's the total distance from that point down to the directories? 2P. It's 2P. Therefore, how far is it from this point over to the focus? 2P. It's also 2P. How did you get that, Jeff? Well, it has to be equal distance from the focus and from the directories. Yeah, this goes back to the definition of a parabola. You remember, every point on a parabola is the same distance from the focus and the directories. And we figure this point is 2P away from the directories. So it must be 2P away from the focus. Therefore, what is the total length of the lattice rectum? 4P. It's always 4P. Where have you seen 4P already in these equations? x squared equals 4P. Why? Exactly. You see, in the equation of the parabolas, we have x squared equals 4Py, or y squared equals 4Px. That coefficient in front of the x or the y, the 4P, is actually the length of the lattice rectum. So we've been using P as merely the distance from the focus to the directories, or the coordinate of the focus. But now we find out it has other meanings as well. OK, now before we introduce the next example, let me just point this out if we come to the green screen. If I write any equation in this form, x squared, or ax squared plus by squared plus cx plus dy plus e equals 0, any equation in this form will be a line. Now, for example, if a and b were both 0, if I just cover up a and b, this is the equation of a diagonal line on the x-y plane. Or it could be a circle. If a and b are both 1, for example, if a and b are both 1, this is an equation of a circle. We talked about these problems earlier in the course where we completed the square and found the center of a circle. This could be the equation of a parabola. It would be a parabola if, for example, b were 0. If b were 0, I'd have only the x squared term plus an x term, a y term, and a constant term right there. That would be the equation of a parabola. We'll talk about this in a moment. If a and b are different but both positive, this is an ellipse. We'll talk about ellipses next time. And if a and b have opposite signs, this will be that other conic section that we'll talk about later. This will be a hyperbola. But my point is, anything that can be written in this form will be one of the cross sections of a conic that we talked about. Remember, we've got a line. We've got a circle. And then the three that are listed right here refer to as conic sections. And we're talking about this one today, the parabola. Now I want to look at some equations that would ultimately be in that form. Let's go to the next example and we'll see a case of this. OK, it says sketch each of the following, showing the vertex, the focus, and the directories. As a matter of fact, both of these will be parabolas. Let's take this first example. x plus 2 squared equals 4 times 3 minus y. Now this doesn't look like a parabola. It doesn't look like one of our fundamental equations. But what I'd like for us to do is to compare this with x squared equals 4p times y. Now I notice there's a change on the x. I have 2 added to the x. x plus 2 squared. And what have we seen in the past happens when you add 2 to the x term on a graph? Horizontal shift? It's a horizontal shift. And if you add 2, it moves it to the left two units. Now on the right-hand side, I'm thinking that this is a bit more complicated because I have a negative on the y. So what I'm going to do is factor out the negative and make this a negative 4. And write this as y minus 3. And I see that 3 has been subtracted from the y. Now the behavior or the effect of subtracting 3 on the y is very similar to the effect of adding the 2 to the x. If I subtract 3 from the y, this actually moves the graph up 3. So this is a parabola that's been shifted two units to the left, and it's been shifted three units up. And the 4p coefficient is a negative 4. And I want to graph this parabola that's been translated off of the origin. So to do that, I'll just put the graph over here on the side. We'll get our axes set up here. Here's the x-axis. Here's the y-axis. We said we were going to shift it 2 to the left, and we were going to shift it 3 up. So my origin is shifted over 2 to the left and up 3. And this is my new vertex right there. So the vertex will be, at the point, negative 2, 3. OK, now what will be the choice of p in this problem? Well, 4p is negative 4. And so p is negative 1. By the way, because the x term is squared, do you think this parabola opens up or down or left or right? It opens up or down. And the fact that p is negative means it opens down. The focus will be one unit below the vertex. So the focus, one unit below, will be right about here. And the coordinates of that point should be, let's see, if I go down one, that should be negative 2, 2. Negative 2, 2. And the directrix should be horizontal and one unit above. So if I draw my directrix in right here, that should be one unit above, that would be the line y equals 4. So the directrix will be y equals 4. And when I draw my parabola, what I've done is to shift it off the origin, and it now has the vertex up here. And this will be a relatively wide parabola, because I have a relatively large value for p, relative to some of the other problems we've been looking at here. And so I'm going to draw my parabola relatively wide. By the way, this parabola is going to cross the y-axis. How could I find out the y-intercept of this parabola? I said x equal to 0. I said x equal to 0. OK, now let's just do that back up here. If I go back up the original equation and let x equals 0, that's going to say 2 squared. And this will be 4 times 3 minus y. OK, right above it, let me just write that down. That's going to be 4 equals 4p, or rather 4, times 3 minus y. That's a multiplication there. So that says that 3 minus y is equal to 1. 3 minus y is equal to 1. See, I kind of wrote that on top of the problem there. But that says that 3 minus y is equal to 1. And I think that in this case, y would have to be 2. I don't have much space to write that out, so I'll just say y is 2. So the y-intercept is 2, and it's going to go right through this point right here. So it is rather wide, this parabola. And I could find the x-intercepts if I let y be 0, but I don't think I'll do that. So we've been able to graph this parabola, and we've been able to find its vertex, its focus, and its directories. Now let's go to the other equation that's given there. And this says 3y squared equals x minus 5. This is also a parabola, but it's been shifted off the origin. And if I were to write this in an expanded form, I'd say that 3 times y minus 0 squared equals the quantity x minus 5. And if I isolate the y-squared term, y minus 0 squared equals 1 third times x minus 5. Now just compare that with the equation y-squared equals 4p times x. It looks like there's only been one translation or one shift made here. What would you say is the shift made in this graph from the standard equation? Horizontal? There's a horizontal shift because this is x minus 5. And which way has it moved? Five units. To the left? Five units to the right. Because if you subtract 5, you move it to the right. So we'll move this to the right-hand side. But there's no vertical shift because 0 was subtracted from y. It's just a y-squared. The other piece of information we have here is that 4p is equal to 1 third. And so p is equal to 1 12th. Well, now there's a small value for p compared to what we've seen before. So that's going to make this into a rather thin looking parabola. Let's graph it over here in the space that's available to us. Would you say this parabola opens up or down or left or right? So we have a y-squared term. See, p is positive. And p is positive. Yeah. It's going to be to the right. Yeah, you can't see David, but he's going to the right. OK, so let's see now. To find the vertex, though, I'll need to shift it over five units. One, two, three, four, five. So here's five. I'll put a solid dot right there and just put a v above it. And the vertex is at 5, 0. Now, to find the focus, I'm going to have to go over to the right 1 12th, because if this parabola is opening to the right-hand side, the focus is to the right-hand side. Now, it's going to be difficult to show much distinction there, but the focus is going to be 1 12th further over. Now, if I add a 12th to 5, I get 61 12ths, 0. That's 5 plus a 12th. And the directrix is a vertical line that comes in just to the left of the vertex, 1 12th to the left. And the directrix is going to be the line x equals 59 12ths. That's 5 minus 1 12th. And my parabola coming out of the vertex is going to be rather thin. So I'm going to have this thing turn rather quickly and go out. And so I get a parabola that looks like that. OK, let's go to the next graph. You can look at the next problem. It says sketch each of the following showing the vertex, the focus, and the directrix. No, I'm sorry, I'm looking at the wrong graphic. Sketch the graph of this equation y squared equals negative 12x plus 4y plus 16 equals 0. Now, you remember a moment ago, I said that the conic sections could all be written in this form. A x squared plus B y squared plus C x plus D y plus E is 0. And here we have something exactly like that. We have y squared minus 12x plus 4y plus 16 equals 0. Now, you see this problem is really like the two that I just did a moment ago, except those squares have been multiplied out. So for me to get it back into the form of the previous example, I'm going to need to complete the square. So what I'll do is to group the y's together. And I'm going to put the other terms on the other side. The 12x and the 16 becomes a minus 16. So what I want to do is to isolate the y terms because they have the square in them. Everything else goes on the other side. Now, what number should I add on right here that would complete the square? Plus 4. Yeah, remember the rule says if the coefficient of the square term is 1, you take half the middle term coefficient and square it. So if half of 4 is 2, square it is 4. And I have to add 4 over here on the other side as well. So that tells me this is y plus 2 squared equals 12x minus 12. Well, we're almost there. Now, if I just factor out a 12, I have y plus 2 squared equals 12 times x minus 1. And now, this problem is in the form of a translated quadratic. This tells me that the vertex of the parabola is going to be at the point 1, negative 2. 1 and negative 2. And it tells me that 4p is 12, so I think I can just see from that that p is equal to 3. And this parabola opens to the right because there's a y squared term and because p is positive. So when I graph it, I'm going to go to the point 1, negative 2. I'll just put a dot right here to say this is the vertex, 1, negative 2. The value of p is 3, so I'm going to have to move 3 units to the right to get the focus. So that says the focus would be at the point, if I move 3 to the right, 4, negative 2. And the directrix is 3 units to the left. What line would that be if I move 3 units to the left? If this is 1, negative 2. If I go 3 units to the left, that's at 1. It'd be x, negative 2. x equals negative 2. x equals negative 2. That's the directrix. And this parabola would be really wide because the choice of p is so big. So I get a parabola that's fairly wide like that. So we've been able to sketch the graph of a conic in which the terms have been multiplied out by simply completing the squares. And then rearranging terms in the proper order. OK, we have an example that we can close with. In this next example, it says a search light has a parabolic reflector 12 inches wide, rim to rim at 8 inches deep. How far from the vertex should the light source be located? Now, this involves a reflective property about parabolas that I'll just mention here briefly. It says that if you have light that comes in toward this reflector and bounces off a parabolic mirror, it will be reflected directly to the focus point. And the same thing for light coming in over here on this side, it'll be reflected to this focus point right below here. OK, so if this is a parabolic mirror, what I mean is this is a parabola that's been rotated so that it scoops out a sort of a parabolic dish behind it. And we're given a bit of information about the mirror. It says that it's 12 inches from side to side. So it's 12 inches across here, and it's 8 inches deep in the middle. So that tells me this point must be, let's see, if that's 12 inches, this must be 6 and 8. So on my parabola, I have the point 6 and 8 up here. 6 and 8. And I have a parabola that passes through that dish. Through that point. Now, I'd want to put the light source at the focus, we said here, so that light going out will be reflected parallel to the central axis. And so what I need to do is figure out where is the focus of this parabola. Well, the parabola opening up has the equation x squared equals 4py. And it passes through the point 6 and 8. So if I plug in 6 here and 8 for y, then we have that p is equal to 36 over 32. 36 over 32. And so that makes that 9 over 8. p is 9 over 8. So that tells me the focus is 9 eighths of a unit above the origin. Well, 9 eighths of a unit, that means it's 1 and 1 eighth of an inch above. So the light source should be placed 1 and 1 eighth inches above the vertex. I'll just put a v on that. It should be placed 1 and 1 eighth inches above. OK, you know, this reflective property is useful in telescopes and in satellite dishes and a number of other apparatuses, where if you have a parabola and if you have your receiver right here, then you can receive signals that bounce in, come off the parabola, and bounce directly to the focus. And same thing over here bounces right back to the focus. And for a headlight, if you put the light source right here, the light goes out, bounces off, and it's reflected straight down in front of the headlight. So the back of the headlight on your car has this shape. Well, we've introduced the first of three conic sections. In the next episode, we'll look at ellipses. And in the final episode, or in the third episode of this series, we'll look at hyperbolas before we go on to other information in the next chapter. And I'll see you next time for episode 29. What's that?