Transcription
In the last video, we showed that if you had a line, which we'll call a directrix-- we draw the directrix. That's our directrix, and it has the equation y is equal to k. And you have a point, that's our focus, and it's the coordinate a comma b. That the locus of all points in the xy plane that are equidistant to this focus and this directrix has a shape that looks something-- we know that this was a point-- that looks something like this.
And the actual equation, we actually just took an arbitrary xy in the coordinate plane, and we said that xy has to satisfy the condition that its distance to this focus is equal to-- that this distance right here is equal to this distance here. And we set that up using the distance formula to here, and then setting that equal to the distance to the line, to the directrix. And then we did a bunch of algebra and we got this equation right here. For all of the points in the xy plane that satisfy the conditions that its distance to the focus is equal to the distance to this line. And we hopefully satisfied ourselves that this is in fact a parabola. And although it looks a little bit hairier than most parabolas we see, I think if you look at it closely, you'll realize that it is indeed a parabola.
For example, the classic parabola is y is equal to x squared. How is this the same thing as this? Well, in this case, this coefficient out here is just equal to 1. So in this case, 1 over 2b minus k is equal to 1, right? Let me delete that little thing there, because it looks like a squared. All right, the coefficient in front of the x squared term is equal to a 1. What is a equal to? Well, this is the same thing-- actually, let me rewrite it that way. I could rewrite this equation right here as y minus 0 is equal to 1 times x minus 0 squared. And now hopefully you see that this has the same pattern as this, although the y's and the x's are just sitting on different sides of the equation. So here you see that the 1/2 over b minus k, that's the coefficient in front of the x minus a squared term. So that's where we got that that has to be equal to 1. a has to be equal to 0 in this situation. This curve right here is equal to x squared, y is equal to x squared, a is 0. And then b plus k over 2, this is going to be equal to that. And keep in mind, I actually swapped the left and the right-hand sides. But this is y minus something, this is y minus 0. So this something has to be equal to 0. So you get b plus k over 2 is equal to 0.
And now we should be able to use this information to figure out the actual coordinates of the focus and the directrix of what I would call the classic parabola, y is equal to x squared. So keep in mind what we did. In the last video, we said what is the equation of the line that is equidistant between this focus and this directrix? And we got this equation, which we said, hey, that's a parabola. Now we're going the other way around. We say we have a parabola, and we're saying that this parabola is the set of all points that's equidistant between some focus and some directrix. What is that focus and that directrix for this particular parabola? That's what we're trying to do. And the way we did that was, we pattern matched this formula to this one, to essentially set up these variables, which essentially define a focus and a directrix, right? b is the y-coordinate of the focus, k defines the horizontal line of the directrix. So if we can solve for b and k, we know what the focus and the directrix are. And for a, although a is pretty easy.
So let's do that. So if we know that 1 over 2b minus k is equal to 1-- let me do that over here. I'm going to do it in particular for this case right here, but you don't want to do that every time, so then we'll do a more general formula that we can just plug in next time you see a new parabola. So if we multiply both sides of this equation by 2b minus k, you get 1 is equal to 2 times b minus k. Actually, I should re-divide both sides by 2, so you get 1/2 is equal to b minus k. Fair enough. Let's see, this equation right here, b plus k over 2 is equal to 0. That means that b plus k must be equal to 0, right? This numerator has to be equal to 0. So let me write that down. b plus k is equal to 0. Two linear equations of two unknowns. Let's add them up. The k's cancel out. You get 2b is equal to 1/2. Divide both sides by 2, you get b is equal to 1/4. And now what's k? Well, we can just resubstitute back. Let's use this equation. We get 0 is equal to 1/4 plus k. You can figure that out by inspection, but you can subtract 1/4 from both sides. So this tells us that k is equal to minus 1/4.
So let's draw this. So if this is my coordinate axis-- I'll do it in a different color. That's the x-axis, that's the y-axis. We're dealing with y is equal to x squared, which you should hopefully be reasonably familiar with at this point. It's kind of the classic U-shaped parabola that looks just like that. Let me write that down. This is y is equal to x squared, which we could have rewritten as y minus 0 is equal to 1 times x minus 0 squared. Same thing, that's this line. It's equidistant from a focus and a directrix, where this is the focus. And how do I know that the focus is right there, instead of right here or right there? Well, one, you could kind of think about it. But the other thing to realize is, what is the vertex of this parabola? And I think we've gone over that already. But the vertex are the coordinates given by this x value and this y value. Or what x values make this whole term 0 and make this whole term 0? And that's essentially 0,0. So this is the vertex of this parabola. This is the vertex, 0,0 is the vertex. I'll say v for vertex. In an upward opening parabola, you can view that as the minimum point. If it was a downward opening, it would be the topmost point, or the maximum point. That's the vertex. And we saw from when we solved this, when we just did our pattern matching, that the x-coordinate of our focus is equivalent to, if we just pattern match right here, x minus a squared, here you have x minus 0 squared. This is always going to be the same as this. So the x-coordinate of our focus is always going to be the same thing as the x-coordinate of our vertex. So we figured out that a was equal to 0. So the coordinate here is 0, and then what's b? This was the coordinate 0,b. We figured out that b is equal to 1/4, 0, 1/4. And then the directrix is the line y is equal to k. In this case, we figure out k is equal to minus 1/4. So our directrix is going to be right down here. Right down here, like that. And it's going to be line y is equal to minus 1/4. And that first of all looks about right. And it also gels with what we know. Because this point right here, our vertex, is equidistant, it's 1/4 below our focus and it's 1/4 above our directrix. So at least that little reality check holds.
But it's was a particular circumstance, and you don't want to have to go through this whole situation every time you have to figure out the focus or the directrix of a parabola. So let's see if we can come up with a more general solution. And actually what I'm going to do is, I'm going to erase all of this. Just because I want to reuse that, and I haven't committed that formula to heart yet. Let me erase all of this, and erase all the work we did down here.
So let's do a general form. And actually, I've abused this so much, let me rewrite it. So I'll put the y's on the left-hand side, instead of keep telling you that I've switched them. So this is the same thing as-- I'll do it in-- the color choosing is the hard part. I'll do it in this light, off-white color. So you get y minus b plus k over 2 is equal to 1 over 2 times b minus k, times x minus a squared. This is the locus of all points that are between the directrix, y is equal to k-- well, they're equidistant to the directrix, y is equal to k, and the focus a comma b. So let's say that I have a parabola. And I'm going to try to use different letters, so we don't get confused. I give you the parabola y minus y1. And let's say this parabola has a vertex at x1 comma y1. So this parabola I'm drawing has a vertex right there. So its formula will be y minus y1 is equal to some constant, let's make that a capital A. This capital A is different than that lowercase a. This is what I was going to embark on in the previous video, and then I realized that I was trying to load that video up too much and probably confusing you. So some constant factor, some type of scaling factor, times x minus the x value of our vertex. So x1, all of that, squared. So if you're given this parabola, or if you can get a parabola to this form, how do you figure out the a's, the b's, and the k's so you know the coordinates of the directrix and the focus? So the easy thing is to figure out a, because you just do a pattern match. This is going to be equal to that. So you know that a is equal to x1. So the x-coordinate of our focus is the same as the x-coordinate of our vertex. You can pattern match that this scaling factor right here is going to be the same thing as this 1 over 2b minus k. So let's write that down. A is equal to 1 over 2b minus k. And then we have, finally, this. Let me do another color. I'll do a magenta. The b plus k over 2, we have y minus that thing. We have y minus y1. So that's going to be equal to y1. So you have y1 is equal to b plus k over 2. And now we have two equations with two unknowns. Remember, this is going to be a given. The A is going to be given, and the y1 is going to be given. So now we can solve for b and k in terms of the numbers that we have.
So let's see if we can do that. So if we multiply both sides of this equation times b minus k, you get b minus k times A is equal to 1/2. Divide both sides by A, you get b minus k is equal to 1 over 2A. And on the right-hand side, let's use this other formula right here. If we multiply both sides by 2-- I'll use a different color-- you get 2 times y1 is equal to b plus k. Two equations with two unknowns. Let's take this and bring it down here, so you get b plus k is equal to 2 times y1. And so you add them, and you get 2b is equal to-- the k's cancel out; I'm just adding these two equations-- 1 over 2A plus 2y1. Or b-- which is, remember, what was b? That was the y-coordinate of our focus-- is equal to, divide everything by 2, is equal to 1 over 4A plus y1. Which is interesting. It tells us that we take the y-coordinate of our vertex and we add 1 over 4A to it. Which is exactly what happened last time, right? I actually erased what I did last time. But last time, when we had y equal x squared, this value was 0. The scaling factor was 1. So we saw that the y-coordinate of our focus was 1/4. So this is gelling with what we know. So that's our b. And let's see if we can solve back for k. So we know that-- let's do it up here. We know that 2y1 is equal to b, which is 1 over 4A plus y1 plus k. We can subtract y1 from both sides, so this goes away and we just have a y1 there. Subtract 1 over 4A from both sides and you get y1 minus 1 over 4A is equal to k. So then we are done. So this is interesting. Right now these might look a little, these little hairy formulas. But if we actually graph it, I think they'll become a little bit more intuitive.
So once again, we had the parabola y minus y1 is equal to A times x minus x1 squared. And so the graph of that parabola will look something like this. I don't know where it is relative to the x- and y-axis. It will look something like this. It'll be this U shape. It could actually be downward pointing, but I'll just assume an upward pointing parabola for now. That's this formula. Its vertex, right there, its vertex is the point x1 comma y1. Right? What y value makes this expression equal 0? It's y1. What x value makes that equal 0? It's x1. Now notice, what is this directrix? It's the line ok, y equal to k, but it's 1/4 less than this value right here. So you literally just go down 1/4-- not just 1/4, 1/4 times this scaling factor. So it's 1/4A. This distance is 1/4A. And then you get your directrix, right? It's y1 minus 1/4A is equal to k. y1 minus 1/4A is equal to k. Or 1 over 4A. So your directrix is going to be right there. Let me write this. It's y is equal to y1, which is just this level, minus 1 over 4A. But this is the intuitive part. Just remember, you're going 1 over 4A below the vertex. And then the focal point, the x value of the focal point is going to be x1, the same as the x value of the vertex. And then the y value is just b, and we just go 1 over 4A above it. Which makes sense because we need to be equidistant from the directrix and the focus. So this is y1 plus 1 over 4A. But the intuitive part is, we just took 1 over 4A above it.
So in general, if I gave you-- so just think about this a second. If I gave you an equation, an arbitrary equation now. Say, y minus 1 is equal to 2 times x minus 3 squared, we can graph this and draw the focus fairly-- and actually it's foci, if there was more than one focus. Let's see, that's the x-axis, that's the y-axis. Where's its vertex? The vertex is at the point 3 comma 1. So we go 1, 2, 3 comma 1. So that's its vertex. It's an upward opening parabola, because this is a positive number. Actually, we don't even have to know that. If you just draw the focus and the directrix, you might be able to figure that out. But where's the focus point? The focal point is going to be 1 over 4A above this point. So 1 over 4 times 2 is equal to 1/8. So the focal point is going to be really close. It's going to be right up here. It's going to be 1/8 above our vertex. And then our directrix is going to be 1/8 below it. So it's going to be right there. The directrix is going to look like this. The directrix is going to be there, the focal point is there. And then the graph of this parabola is going to look something like this. So anyway, hopefully I didn't confuse you too much. And the whole point of doing all of this is just to realize that it's pretty easy to find the focus and the directrix of a parabola if you have its equation in this form. You just take whatever is multiplying times the x minus whatever squared term. And you essentially divide 1 by 4 times this value. And then you know the distance from the vertex to the focus, and you know the distance from the vertex to the directrix, and you're all done. See you in the next video.