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In the last video, we showed that if you had a line, which

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we'll call a directrix-- we draw the directrix.

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That's our directrix and it has the equation y is equal to k.

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And you have a point, that's our focus, and it's the

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coordinate a comma b.

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That the locus of all points in the xy plane that are

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equidistant to this focus and this directrix has a shape that

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looks something-- we know that this was a point-- that

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looks something like this.

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And the actually equation, we actually just took an arbitrary

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xy in the coordinate plane, and we said that xy has to satisfy

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the condition that its distance to this focus is equal to--

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that this distance right here is equal to this distance here.

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And we set that up using the distance formula to here, and

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then setting that equal to the distance to the line,

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to the directrix.

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And then we did a bunch of algebra and we got this

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equation right here.

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For all of the points in the xy plane that satisfy the

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conditions that its distance to the focus is equal to the

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distance to this line.

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And we hopefully satisfied ourselves that this is

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in fact a parabola.

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And although it looks a little bit hairier than most parabolas

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we see, I think if you look at it closely, you'll realize

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that it is indeed a parabola.

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For example, the classic parabola is y is

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equal to x squared.

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How is this the same thing as this?

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Well in this case, this coefficient out here

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is just equal to 1.

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So in this case, 1 over 2 b minus k is equal to 1, right?

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Let me delete that little thing there, because it

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looks like a squared.

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All right, the coefficient in front of the x squared

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term is equal to a 1.

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What is a equal to?

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Well this is the same thing-- actually, let

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me rewrite it that way.

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I could rewrite this equation right here as y minus 0 is

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equal to 1 times x minus 0 squared.

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And now hopefully you see that this has the same pattern as

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this, although the y's and the x's are just sitting

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on different sides of the equation.

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So here you see that the 1/2 over b minus k, that's the

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coefficient in front of the x minus a squared term.

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So that's where we got that that has to be equal to 1.

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a has to be equal to 0 in this situation.

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This curve right here is equal to x squared, y is equal

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to x squared, a is 0.

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And then b plus k over 2, this is going to be equal to that.

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And keep in mind, I actually swapped the left and

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the right hand sides.

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But this is y minus something, this is y minus 0.

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So this something has to be equal to 0.

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So you get b plus k over 2 is equal to 0.

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And now we should be able to use this information to figure

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out the actual coordinates of the focus and the directrix of

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what I would call the classic parabola, y is equal

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to x squared.

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So keep in mind what we did.

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In the last video, we said what is the equation of the line

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that is equidistant between this focus and this directrix?

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And we got this equation, which we said, hey,

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that's a parabola.

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Now we're going the other way around.

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We say we have a parabola, and we're saying that this parabola

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is the set of all points that's equidistant between some

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focus and some directrix.

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What is that focus and that directrix for this

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particular parabola?

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That's what we're trying to do.

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And the way we did that was, we pattern matched this formula to

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this one, to essentially set up these variables, which

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essentially define a focus and a directrix, right?

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b is the y-coordinate of the focus, k defines the horizontal

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line of the directrix.

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So if we can solve for b and k, we know what the focus

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and the directrix are.

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And for a, although a is pretty easy.

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So let's do that.

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So if we know that 1 over 2 b minus k is equal to 1--

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let me do that over here.

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I'm going to do it in particular for this case right

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here, but you don't want to do that every time, so then we'll

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do a more general formula that we can just plug in next time

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you see a new parabola.

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So if we multiply both sides of this equation by 2 b minus k,

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you get 1 is equal to 2 times b minus k.

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Actually I should redivide both sides by 2, so you get 1/2

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is equal to b minus k.

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Fair enough.

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Let's see, this equation right here, b plus k

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over 2 is equal to 0.

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That means that b plus k must be equal to 0, right?

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This numerator has to be equal to 0.

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So let me write that down.

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b plus k is equal to 0.

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Two linear equations of to unknowns.

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Let's add them up.

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The k's cancel out.

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You get 2b is equal to 1/2.

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Divide both sides by 2, you get b is equal to 1/4.

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And now what's k?

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Well, we can just resubstitute back.

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Let's use this equation.

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We get 0 is equal to 1/4 plus k.

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You can figure that out by inspection, but you can

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subtract 1/4 from both sides.

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So this tells us that k is equal to minus 1/4.

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So let's draw this.

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So if this is my coordinate axis-- I'll do it in

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a different color.

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That's the x-axis, that's the y-axis.

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We're dealing with y is equal to x squared, which you should

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hopefully be reasonably familiar with at this point.

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It's kind of the classic U-shaped parabola that

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looks just like that.

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Let me write that down.

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This is y is equal to x squared, which we could have

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rewritten as y minus 0 is equal to 1 times x minus 0 squared.

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Same thing, that's this line.

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It's equidistant from a focus and a directrix, where

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this is the focus.

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And how do I know that the focus is right there, instead

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of right here or right there?

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Well one, you could kind of think about it.

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But the other thing to realize is, what is the vertex

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of this parabola?

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And I think we've gone over that already.

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But the vertex are the coordinates given by this

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x value and this y value.

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Or what x values make this whole term 0 and make

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this whole term 0?

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And that's essentially 0,0.

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So this is the vertex of this parabola.

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This is the vertex, 0,0 is the vertex.

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I'll say v for vertex.

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In an upward opening parabola, you can view that as

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the minimum point.

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If it was a downward opening, it would be the topmost

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point, or the maximum point.

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That's the vertex.

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And we saw from when we solved this, when we just did our

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pattern matching, that the x-coordinate of our focus is

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equivalent to, if we just pattern match right here, x

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minus a squared, here you have x minus 0 squared.

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This is always going to be the same as this.

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So the x-coordinate of our focus is always going to

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be the same thing as the x-coordinate of our vertex.

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So we figured out that a was equal to 0.

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So the coordinate here is 0, and then what's b?

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This was the coordinate 0,b.

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We figured out that b is equal to 1/4, 0, 1/4.

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And then the directrix is the line y is equal to k.

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In this case, we figure out k is equal to minus 1/4.

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So our directrix is going to be right down here.

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Right down here, like that.

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And it's going to be line y is equal to minus 1/4.

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And that first of all looks about right.

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And it also gels with what we know.

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Because this point right here, our vertex, is equidistant,

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it's 1/4 below our focus and it's 1/4 above our directrix.

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So at least that little reality check holds.

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But it's was a particular circumstance, and you don't

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want to have to go through this whole situation every time you

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have to figure out the focus or the directrix of a parabola.

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So let's see if we can come up with a more general solution.

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And actually what I'm going to do is, I'm going

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to erase all of this.

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Just because I want to reuse that, and I haven't committed

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that formula to heart yet.

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Let me erase all of this, and erase all the work

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we did down here.

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So let's do a general form.

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And actually, I've abused this so much, let me rewrite it.

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So I'll put the y's on the left hand side, instead

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of keep telling you that I've switched them.

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So this is the same thing as-- I'll do it in-- the color

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choosing is the hard part.

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I'll do it in this light, off-white color.

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So you get y minus b plus k over 2 is equal to 1 over 2

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times b minus k, times x minus a squared.

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This is the locus of all points that are between the directrix,

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y is equal to k-- well, they're equidistant to the directrix,

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y is equal to k, and the focus a comma b.

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So let's say that I have a parabola.

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And I'm going to try to use different letters, so

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we don't get confused.

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I give you the parabola y minus y1.

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And let's say this parabola has a vertex at x1 comma y1.

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So this parabola I'm drawing has a vertex right there.

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So its formula will be y minus y1 is equal to some constant,

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let's make that a capital A.

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This capital A is different than that lowercase a.

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This is what I was going to embark on in the previous

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video, and then I realized that I was trying to load that video

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up too much and probably confusing you.

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So some constant factor, some type of scaling factor, times

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x minus the x value of our vertex.

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So x1, all of that, squared.

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So if you're given this parabola, or if you can get a

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parabola to this form, how do you figure out the a's, the

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b's, and the k's so you know the coordinates of the

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directrix and the focus?

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So the easy thing is to figure out a, because you just

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do a pattern match.

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This is going to be equal to that.

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So you know that a is equal to x1.

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So the x-coordinate of our focus is the same as the

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x-coordinate of our vertex.

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You can pattern match that this scaling factor right here is

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going to be the same thing as this 1 over 2 b minus k.

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So let's write that down.

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A is equal to 1 over 2 b minus k.

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And then we have, finally, this.

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Let me do another color.

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I'll do a magenta.

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The b plus k over 2, we have y minus that thing.

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We have y minus y1.

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So that's going to be equal to y1.

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So you have y1 is equal to b plus k over 2.

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And now we have two equations with two unknowns.

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Remember, this is going to be a given.

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The A is going to be given, and the y1 is going to be given.

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So now we can solve for b and k in terms of the

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numbers that we have.

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So let's see if we can do that.

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So if we multiply both sides of this equation times b minus k,

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you get b minus k times A is equal to 1/2.

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Divide both sides by A, you get b minus k is

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equal to 1 over 2A.

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And on the right hand side, let's use this other

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formula right here.

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If we multiply both sides by 2-- I'll use a different

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color-- you get 2 times y1 is equal to b plus k.

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Two equations with two unknowns.

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Let's take this and bring it down here, so you get b plus

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k is equal to 2 times y1.

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And so you add them, and you get 2b is equal to-- the k's

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cancel out; I'm just adding these two equations--

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1 over 2A plus 2 y1.

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Or b-- which is, remember, what was b?

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That was the y-coordinate of our focus-- is equal to, divide

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everything by 2, is equal to 1 over 4A plus y1.

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Which is interesting.

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It tells us that we take the y-coordinate of our vertex

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and we add 1 over 4A to it.

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Which is exactly what happened last time, right?

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I actually erased what I did last time.

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But last time, when we had y equal x squared,

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this value was 0.

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The scaling factor was 1.

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So we saw that the y-coordinate of our focus was 1/4.

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So this is gelling with what we know.

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So that's our b.

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And let's see if we can solve back for k.

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So we know that-- let's do it up here.

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We know that 2 y1 is equal to b, which is 1 over

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4A plus y1 plus k.

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We can subtract y1 from both sides, so this goes away and

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we just have a y1 there.

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Subtract 1 over 4A from both sides and you get y1 minus

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1 over 4A is equal to k.

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So then we are done.

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So this is interesting.

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Right now these might look a little, these little

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hairy formulas.

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But if we actually graph it, I think they'll become a

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little bit more intuitive.

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So once again, we had the parabola y minus y1 is equal to

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A times x minus x1 squared.

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And so the graph of that parabola will look

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something like this.

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I don't know where it is relative to the x- and y-axis.

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It will look something like this.

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It'll be this U shape.

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It could actually be downward pointing, but I'll just assume

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an upward pointing parabola for now.

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That's this formula.

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Its vertex, right there, its vertex is the

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point x1 comma y1.

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Right?

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What y value makes this expression equal 0?

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It's y1.

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What x value makes that equal 0?

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It's x1.

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Now notice, what is this directrix?

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It's the line ok, y equal to k, but it's 1/4 less than

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this value right here.

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So you literally just go down 1/4-- not just 1/4, 1/4

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times this scaling factor.

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So it's 1/4 A.

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This distance is 1/4 A.

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And then you get your directrix, right?

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It's y1 minus 1/4 A is equal to k. y1 minus

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00:15:37,899 --> 00:15:40,549
1/4 A is equal to k.

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Or 1 over 4A.

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So your directrix is going to be right there.

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Let me write this.

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00:15:49,600 --> 00:15:55,070
It's y is equal to y1, which is just this

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00:15:55,070 --> 00:15:58,290
level, minus 1 over 4A.

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But this is the intuitive part.

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Just remember, you're going 1 over 4A below the vertex.

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And then the focal point, the x value of the focal point is

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going to be x1, the same as the x value of the vertex.

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And then the y value is just b, and we just go

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1 over 4A above it.

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Which makes sense because we need to be equidistant from

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the directrix and the focus.

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00:16:22,059 --> 00:16:26,069
So this is y1 plus 1 over 4A.

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00:16:26,070 --> 00:16:30,620
But the intuitive part is, we just took 1 over 4A above it.

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00:16:30,620 --> 00:16:33,179
So in general, if I gave you-- so just think

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00:16:33,179 --> 00:16:33,929
about this a second.

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00:16:33,929 --> 00:16:37,289
If I gave you an equation, an arbitrary equation now.

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00:16:37,289 --> 00:16:49,099
Say, y minus 1 is equal to 2 times x minus 3 squared, we can

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00:16:49,100 --> 00:16:54,700
graph this and draw the focus fairly-- and actually

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00:16:54,700 --> 00:16:57,580
it's foci, if there was more than one focus.

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00:16:57,580 --> 00:17:00,540
Let's see, that's the x-axis, that's the y-axis.

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Where's its vertex?

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The vertex is at the point 3 comma 1.

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00:17:05,200 --> 00:17:09,110
So we go 1, 2, 3 comma 1.

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00:17:09,109 --> 00:17:10,689
So that's its vertex.

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00:17:10,690 --> 00:17:12,320
It's an upward opening parabola, because this

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00:17:12,319 --> 00:17:13,789
is a positive number.

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00:17:13,789 --> 00:17:15,059
Actually, we don't even have to know that.

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00:17:15,059 --> 00:17:17,210
If you just draw the focus and the directrix, you might be

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00:17:17,210 --> 00:17:18,970
able to figure that out.

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00:17:18,970 --> 00:17:20,900
But where's the focus point?

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00:17:20,900 --> 00:17:25,960
The focal point is going to be 1 over 4A above this point.

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00:17:25,960 --> 00:17:30,740
So 1 over 4 times 2 is equal to 1/8.

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00:17:30,740 --> 00:17:32,339
So the focal point is going to be really close.

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00:17:32,339 --> 00:17:35,339
It's going to be right up here.

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00:17:35,339 --> 00:17:38,209
It's going to be 1/8 above our vertex.

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00:17:38,210 --> 00:17:40,970
And then our directrix is going to be 1/8 below it.

340
00:17:40,970 --> 00:17:42,130
So it's going to be right there.

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00:17:42,130 --> 00:17:44,000
The directrix is going to look like this.

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00:17:44,000 --> 00:17:46,579


343
00:17:46,579 --> 00:17:47,809
The directrix is going to be there, the

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00:17:47,809 --> 00:17:49,159
focal point is there.

345
00:17:49,160 --> 00:17:53,279
And then the graph of this parabola is going to look

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00:17:53,279 --> 00:17:56,089
something like this.

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00:17:56,089 --> 00:17:59,049
So anyway, hopefully I didn't confuse you too much.

348
00:17:59,049 --> 00:18:01,579
And the whole point of doing all of this is just to realize

349
00:18:01,579 --> 00:18:04,599
that it's pretty easy to find the focus and the directrix of

350
00:18:04,599 --> 00:18:07,359
a parabola if you have its equation in this form.

351
00:18:07,359 --> 00:18:11,599
You just take whatever is multiplying times the x minus

352
00:18:11,599 --> 00:18:13,569
whatever squared term.

353
00:18:13,569 --> 00:18:18,269
And you essentially divide 1 by 4 times this value.

354
00:18:18,269 --> 00:18:21,355
And then you know the distance from the vertex to the focus,

355
00:18:21,355 --> 00:18:24,210
and you know the distance from the vertex to the directrix,

356
00:18:24,210 --> 00:18:25,529
and you're all done.

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00:18:25,529 --> 00:18:27,450
See you in the next video.

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