 Previously in this lecture series, we've learned about quadratic functions, and as part of that, we've learned about graphs of quadratic functions, which we often call parabolas. Now, long before the invention of algebra, mathematicians have been studying parabolas, so how do they study it if they didn't have a formula like y equals x squared or something more complicated than that? Well, to the ancient geometers, a parabola was considered a collection of all points in the plane that are the same distance from some fixed point f, which we'll call the focus of the parabola, and some fixed line d, which we call the directorics of the parabola. So as an example, you see a parabola illustrated here on the screen, where the focus would be a point like this right here, and then the directorics will be a line as illustrated here in blue. And hence, the parabola is the locus of all points, which is equidistant from the point to the focus and from the point to the directorics. Now, just so we're clear, the distance from a point to a line is actually the shortest path from that point to the line, which is necessarily going to be the right angled path, like so. And so then the parabola is just a collection of all points, which are equidistant from this line to the focus. And you see several examples of that right here. Now, let me discuss a little bit of the anatomy of a parabola. We've mentioned already the focus and the directorics. If we take the line that passes through the focus and is perpendicular to the directorics, there's a unique such line, and this line is called the axis of symmetry of the parabola. It gets its name because the parabola will always be reflective. It'll be symmetric across the axis of symmetry, and it can always pass us to the focus, which will sit inside of the parabola over here. This, if we take the midpoint between the midpoint between the focus and the directorics, which will live on the axis of symmetry here, because it's the midpoint, the distance between the focus at that point and the distance from that point to the directorics will be the same. And therefore, this is a point on the parabola, which we call the vertex of the parabola. And this does coincide with the previous notion of vertex of parabola, as we mentioned before. The next thing I want to mention is that if we take the line that's parallel to the directorics that goes through the focus. And we look at the points on the parabola associated to this line. We form a chord of the parabola, which is referred to as the focal diameter of the parabola. It kind of measures how big the parabola is going to be. Now, so say that we say that the distance between the vertex and the focus is some distance p p for parabola, I guess, and because of the equidistance condition the distance from the vertex the directorics would likewise be p. So if we look at the endpoints of the focal diameter, then the distance between those endpoints to the directorics will be the same as the distance from the focus of the directorics, which would be 2p. And therefore, the distance from the focus to the endpoints of the focal diameter would likewise be 2p. And because of symmetry, we see that on both sides. And therefore, when you put the focal diameter together, this tells us that the diameter has a has a girth of 4p, which that'll come up again in just a second. So what I want to do next is actually connect this geometric notion of equidistance between a point and a line to the usual algebraic formula of parabolas that we've seen previously. So we say that a parabola is horizontal if its axis of symmetry is a horizontal line, like the one we see illustrated below. Similarly, we'll say a parabola is vertical if its axis of symmetry is a vertical line. And so for simplicity, suppose that a parabola is horizontal and it passes its center, its vertex of the parabola is the origin. Again, like the picture you see right here. Now, if the origin is the vertex of the parabola, that actually says that the axis of symmetry will be the x axis itself, again, illustrated in the diagram below. In such a situation, the equation for this parabola will actually be given as y squared equals 4px, where x and y have the usual meaning and 4p, like we saw before, was the focal diameter of the parabola. Now in this situation, the focus would be located at the point p comma 0, the directorics would be located at the line x equals negative p, and then we're going to derive an equation using this relationship right here. So by the equidistance of the parabola, we see that the distance from any point on the parabola from the point to the focus is the same as the distance from the directorics to that point. Now let's say that point has the coordinates x comma y, just kind of a generic point there. Then by the distance formula, the distance from the focus to the point would be x minus p squared plus y minus 0 squared. That's the distance from the focus to the point. This will be the same thing as the distance from the point to the directorics, for which, just to be clear, we're looking at this point right here, which is going to be negative p comma y. So we get x minus a negative p quantity squared plus y minus y squared. Squaring both sides and simplifying this thing, we're going to end up with x minus p squared plus y squared is equal to x plus p squared. Of course, you can y minus y, which is 0, so that disappears. If we foil out the x minus p squared and the x plus p squared, we end up with x squared plus, excuse me, minus 2px plus p squared plus y squared, and this is equal to x squared plus 2px plus p squared. Combining like terms, we're going to see that the x squared actually cancel out and the p squared is likewise cancel out. And then if we add 2px to both sides, we skip the formula we had above, we see that y squared is going to equal 4px. That's giving us the formula we had above. That gives us a formula for the horizontal parabola centered at the origin. By similar argument, if you have a vertical parabola, that is, it centers the origin and its focus is at the point 0 comma p, this will actually recapture the usual notion of a parabola that we're used to, x squared equals 4py. Now to see that, we would solve for y and we get 1 over 4px squared. This is, this would be a usual function, right? And so recognizing here that p just equal, like, if you take the standard parabola, I should say, y equals x squared, this is just the situation where you have 1 over 4 times 1 fourth x squared. So the standard parabola that we're used to is just the parabola who's the distance between the vertex and the focus is 1 fourth. Now, as these are graphs of equations in the plane, we can shift and transform these graphs. It's vertical stretching and reflecting will be represented in this parameter p for the focus. But in terms of shifting it, right, if we have x squared is equal to 4py or 4py, we could shift it so we could move it horizontally by replacing x with x minus h squared, and we could replace y with y minus k. And this would shift the graph so that its vertex becomes h comma k. This is, we wanted to shift a horizontal parabola, we wanted to shift a vertical parabola, the same basic idea. I'm sorry, this one right here is the vertical parabola. This one now is the horizontal parabola. Take y squared equals 4p, 4px, you could shift it. So you get y minus k squared is equal to 4p times x minus h squared. Then the vertex of this parabola would be h comma k. And if you know the p-value, which you can factor out the 4 here, you can then locate the focus, the directorics, and the focal diameter from this information. So this right here gives us the general formula of a parabola, whether it's horizontal or vertical. So let's take a look at an example of such a thing. Consider the parabola x squared plus 4x minus 4y equals 0. I can see for a fact that because we have an x squared, this means that the axis of symmetry is going to have to be vertical. We're going to have a vertical parabola because you only have the one, the only the one square, which in this case is going to be x squared. Now to determine the vertex and the focal constant p, we're going to have to complete the square here. So to complete the square, we're going to separate the x's from the y's, and I'm actually just going to move y to this side so we get 4y right here. Then we have to add the guest of honor here. So take half of the 4, we see 2, we then square that, we then add a 4 in there. Make sure you do so to both sides. The left hand side then factors as x plus 2 squared. And then the right hand side would also factor as 4 times y plus 1. So this tells us that the vertex is going to be negative 2 comma negative 1. You do have to take the opposite of the sign when you consider the shifting right here. So the vertex is going to be negative 2 and negative 1. The focal constant will just be a 1. Notice when we look at here, this is 4 times p, so p would have to be equal 1. And so because our parabola is vertical, we can get started on graphing this thing. We're going to get our focus right here, which is going to be negative 2 and negative 1. Since the parameter p is 1, we would go 1 above the focus to 1 above the vertex to find the focus. If you had a negative p, you would actually go below, but the directorics will be below and the focus will be above like so. And then we can find the focal diameter here by taking the distance of 2 to the left and right. And so let's actually consider doing that from scratch without seeing the graph here. That is without seeing the graph already made. If I was to graph this thing, I'd probably put like my x-axis right there, my y-axis right here, something like that. Let's then graph the focal, let's drop the vertex and the focus. The vertex we saw was at negative 2, negative 1. So that would be like a point right here. That's the focus, excuse me, that's the vertex. The focus would then be 1 point above it. So if we list this here, you get negative 2, negative 1. And then the focus would be at negative 2 and 0. We want to graph the directorics, which would then be below. It would be at the point y equals negative 2. And so then from there, we want to draw the points on the focal diameter. So we're going to find points, which we're going to go 2 to the right, which gives us the origin. And then 2 to the left, which would give us the point negative 4, 0. And then using these three points, the vertex and the focal diameter, you probably can get a pretty good picture. Ooh, that was pretty bad. Let me try that again. You get something like the following. The best drawing picture ever know, but that gives us the parabola in consideration here. Now, why might one desire this representation of the parabola as opposed to our previous exposure with y equals ax squared plus bx plus c? Well, it turns out that the parabola has a powerful and useful reflective property. Suppose that a mirror has a parabolic shape. If light is placed at the focus of the parabola, like you can see right here. Then all rays emanating from the light will reflect off of the mirror and lines parallel to the axis of symmetry. So if you have like a light source at the focus as as photons leave the focus, eventually they'll hit the reflective mirror. And then they'll reflect at an angle, which is parallel to the axis of symmetry. And this doesn't matter which line, which path you take, all of them will reflect outward in the same direction. This essentially magnifies the luminescence of the light source here. Conversely, if you have, for example, a ray of light or sound or whatever that's entering a parabola, once it hits the parabolic surface, it'll reflect and thus come to the focus. And this is actually why the focus gets its name is that as all lines enter the parabola, they'll reflect and meet at the focus. And so because of this reflective property, one can utilize this reflection of a parabola for many applications in engineering. For example, we could build headlights in our car, which you'll notice that your headlights in your car, you have a light source with some of the lights coming out, right? But some of the light is reflecting inward where it hits a reflective parabolic mirror and therefore it reflects outward, which this intensifies the light. One can construct telescopes using this reflective property. Satellite dish also work in this basic idea. If you have a direct TV satellite on your house or something and you probably have a satellite that looks just like this, where the dish is a parabola and then the transmitter slash receiver is located at the focus of that parabola. So that as particles, you know, information comes from outer space where the satellites are located, they'll bounce off of your dish and then reflect to this receiver where the signal then becomes strong enough that you can watch your favorite TV shows. And so this gives you some idea why we care about the focus of a parabola. And it comes down from this very simple observation here that y, I should say y squared is equal to 4px or if you prefer x squared equals 4p, 4py. However you want to orient this as a vertical horizontal, but focusing on this focal distance is important because of the engineering applications we can construct from parabolas.