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Let's say I have a line, let me make it a straight line.

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And the equation of that line, since it's running in the

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horizontal direction is going to be y is equal

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to some constant.

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So let me write that.

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So the equation of this line right here is y is equal to k.

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And let's say I have some other point.

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I'll call that-- well, we'll call that a focus because

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it will be a focus.

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And let's say that point-- let's put it right here-- is

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it the coordinate, let's call it a comma b.

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And let's think about the set of all points or the locus of

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all points that are equidistant to this point right

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here and this line.

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Just so you get the terminology we'll call

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this line a directrix.

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Let me write that down.

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Directrix, it's probably a word you've never heard before.

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And frankly, I've never heard it outside of the context

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of parabolas or conic sections in general.

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So let's call that a directrix.

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This is our focus.

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And what we want to do is we want to find all of the points

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in the xy-plane that are equidistant to this focus

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and this directrix.

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So there's one point that just by looking at it we could say

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is going to be in our locus.

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So it's going to be right there because clearly, this point is

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equidistant; it's halfway between that point

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and this line.

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Now let's see what other points.

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There will probably be this point right here because the

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distance from there to there is this name as the distance

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from if you just drop a line straight down.

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Remember, we want to get the shortest distance between

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this point and the line.

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We could have said something like this distance, but this

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distance wouldn't be the shortest distance between

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this point and this line.

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So you would go straight down to the line there.

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Likewise, this point would be there.

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So that distance is the same as that distance.

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And you can already, I think, begin to imagine

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the type of shape this is.

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And if you look at the title of this video you can probably

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guess where this is going.

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That the shape is going to look something like this.

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Or something pretty similar to what we know as a parabola.

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And actually, it will be a parabola.

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There's no mystery there.

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But what we're doing this in this video is actually show

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you that is a parabola.

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Show it to you mathematically instead of this

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little drawing here.

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Where I show you that it kind of looks like it would be a

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parabola because this distance right here looks about the same

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as that distance right there.

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I could keep going up and down the curve and keep doing that,

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but that's not satisfactory.

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Let's actually mathematically show that the locus of all

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points that are equidistant to this point and this line-- this

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focus and this directrix-- is in fact, a parabola.

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So let's say I have some point that's in this locus and let me

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call the point here-- let me do it in a different color because

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it's the same color as my directrix-- x, y.

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So I want to find all of the x, y's-- all of the points that

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satisfy an equation where their distance to the focus is equal

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to their distance to the directrix.

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So what's the distance between x, y and the focus?

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What's this distance right here?

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d sub 1.

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Well, we just used the distance formula.

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It's x minus the x-coordinate of the focus.

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So it's x minus a squared.

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So the difference in the x's squared.

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Plus the difference in the y's squared-- y minus b squared.

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The square root of that.

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So this is the distance to the focus.

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So I'll call that d sub 1.

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That's this distance right there.

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And we want to find all of the x and y's where the distance to

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the focus is equal to the distance to this

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line right there.

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So if we call this distance-- let's call this the d sub 2--

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the distance to the directrix.

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So the distance to the focus is going to be equal to d sub 2.

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And d sub 2, what's this going to be equal to?

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Well, it's just the difference in y because no matter where we

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are we're just going to drop down straight to the directrix.

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So the difference in y-- I could just say y minus k.

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That would be the difference in y.

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But in this case I did make it so that the coordinate up here

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is higher than the k over here.

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But what if we had a situation where this point was

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below this line?

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I mean, there's nothing here that I said that we can only

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deal with points above this line.

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We haven't proven that to ourselves yet.

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So to make sure that this distance is positive we could

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take the absolute value or we could just square it and

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then take the square root.

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And then that ensures that we're not dealing with

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negative distances.

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So here we set up the equation for all of the x's and y's

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where the distance to the focus is equal to the distance

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to the directrix.

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And let's see if this actually turns out to be a parabola.

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So the first thing we can do is we can square both sides and

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get rid of the radicals.

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So we get x minus a squared.

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Radicals aren't good.

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Well, I don't want to make any social commentary.

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So x minus a squared plus y minus b squared is equal to

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the square root of the other side-- y minus k squared.

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And right now, at least, it looks like there's going to be

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some y squared terms and I mean there's definitely

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an x squared term.

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That doesn't seem like it's a parabola right now.

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But let's keep going forward.

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This is x minus a squared.

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And then let me square these-- actually, expand

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these two binomials.

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So plus y squared minus 2yb plus b squared is equal to y

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squared minus 2yk plus k squared.

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Now what can we do to simplify this?

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Well, we have a y squared here and we have a y squared

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on the right-hand side.

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So we can subtract y squared from both

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sides of the equation.

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They cancel out, and then we got rid of the y squared so all

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of a sudden this is starting to look a lot more like a parabola

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because this is now an equation that relates y's

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to an x squared.

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So let's clean this up and actually make sure that we can

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get it in a form that we normally associate

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with a parabola.

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So we now have-- I'll just rewrite it-- x minus a squared

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minus 2yb plus b squared is equal to minus 2yk

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plus k squared.

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Now what can we do?

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Let's take all the y's and the constants on one

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side of the equation.

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So if we move both of these terms onto the right-hand

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side, so we want to add 2yb to both sides.

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So we get x minus a squared is equal to-- I'm moving both of

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these terms to the right-hand side.

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So if we add 2yb to both sides then we'll have a positive

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2yb on the right hand side.

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So 2yb, and then we'll have that minus 2yk right there.

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And then we're subtracting b squared from both sides.

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So it's minus b squared plus k squared.

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I just arbitrarily decided to write whatever I transferred

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from the left-hand side to the right-hand side in magenta

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just so maybe it makes it a little bit clearer.

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And let's see if we can simplify this further.

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And remember, just so you know what I'm doing this whole time.

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In the back of my mind I'm trying to get it into the

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format that I normally associate with a parabola.

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And I'll do it in the corner here, just so you know

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where the algebra's going.

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I'm trying to get into a format x minus a squared

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is equal to y minus b.

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Because if we have it in this form, or actually ax.

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Let me write that because I went too far to the right.

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A times x minus-- well, actually I already used

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the A so let me write x minus-- I don't know.

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I'll make up x minus-- I'll call that v squared.

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I'm picking maybe v for x value of the vertex.

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I'm making up variables on the fly.

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A times x minus v squared is equal to y minus b.

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And you know, the actual letters I chose don't matter.

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But this is a general form of a parabola.

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And this is where I'm trying to go.

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If I can get this thing that we're working on to this

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form then we know we're dealing with a parabola.

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And then, we can actually relate what we normally

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associate with a parabola to these terms up here.

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And obviously, these are different letters

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than these letters.

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And then we can try to figure out-- if we have a parabola,

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how do we figure out its focus and is directrix.

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But anyway, that was a bit of an aside just so you

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know where we're going.

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So let's try to simplify this a little bit more.

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So we get--

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[PHONE RINGING]

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--my phone is ringing.

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Let me silent it.

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So we get x minus a squared is equal to-- let's see.

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If we factor out a 2 and a y we get-- let me

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factor out a y first.

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We get 2b minus 2k times y.

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And then I want to make it minus some constant.

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So this is a constant right here.

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This is one constant squared plus another constant.

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So I could make this equal to minus-- let

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me think about this.

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This would be minus b squared minus k squared.

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If I expand this minus out you would get minus b

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squared plus k squared.

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So this is the same thing right there.

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And let's see.

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Can I factor out-- yeah, I can factor out a 2.

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The whole thing becomes x minus a squared is equal to 2 times b

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minus k times y minus-- and now this is the difference of--

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yeah, this is b plus k times b minus k.

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You might want to review factoring polynomials if

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that doesn't look familiar.

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But this is b plus k times b minus k.

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And I want to do that because I saw this b minus k out there so

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that looked interesting to get it on both terms right there.

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Now let's see.

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If we divide everything times-- let's divide everything by 1

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over 2 b minus k because I just want a y here, so I have some

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coefficient in front of the y.

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So let's divide everything by 1 over 2 b minus k.

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Let me do it down here so I have some space.

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1 over 2 b minus k times x minus a squared is equal to--

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I'm dividing this by 2 b minus k, so that goes away.

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So it's y minus-- so if I divide this by 2 b minus k the

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b minus k's cancel out and I'm just left with b plus k over 2.

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And I think I have gotten it in the form that I wanted.

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In this case, the a that I wrote up here is now

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1 over 2 b minus k.

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This v is this a that I did here.

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The y is the y and then my constant term is out here.

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So I have hopefully shown to you that the locus of all

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points that are equidistant to this point-- which is really

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just an arbitrary point.

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I said it's the point a, b-- and some line, is

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in fact, a parabola.

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And that's one of the ways that people define a parabola.

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And we'll actually see soon that when you actually change

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this relation-- in this case, we found that all of the points

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that are equidistant to the focus and the directrix.

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If we changed that ratio, if we find all of the points that are

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1/2 as far away from the directrix as the focus, we

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start getting another conic section.

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If we said twice as far we would get another

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conic section.

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So this is a very interesting way of relating-- sometimes the

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conic sections in the first video I showed how they're all

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related to cutting the three-dimensional cone.

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But they're also related in this way and eventually I want

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to relate this to the three-dimensional cone so you

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can see how it all fits together.

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Which is always the fun thing about mathematics.

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So this so far, all we've shown to you is if we do take the

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locus of all points that are equidistant to a point and a

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directrix that it is a parabola.

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Now, the interesting thing to do is let's

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say I have a parabola.

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Let me give a general form.

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And I'll try to use different letters than I used up here.

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So let's say I have a parabola y minus--

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let's call it y sub 1.

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Where it's just a particular point on the parabola.

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Y minus y sub 1 or you could call it some constant-- is

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equal to-- let's call it, I want to use a capital A just

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because that tends to be-- I'll use a capital A and just

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know that it's different than this lowercase A.

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Some capital A times x minus x sub 1.

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squared.

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Whenever you see a y sub 1 or an x sub 1, that means you're

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actually dealing with a constant term.

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If you see just an x or y that generally means that you're

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dealing with a variable term.

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So these are constants.

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I could have put letters there.

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I could have put a and b there, but then that would've

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gotten confused with that.

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But let's see if we can relate these letters because this is

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something that we normally see.

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And then, we want to find the focus or the directrix

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of a parabola like this.

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How can we relate these to these values right here?

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How does k and a and b relate to these right there?

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So let's see if we can set that up.

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So if we just pattern match that is A.

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If we say x sub 1 is lowercase a, actually that's a straight

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up pattern match, so we can already write that on the side.

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x sub 1 is equal to a.

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So whatever x value I have here, this is also the x

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value of the focus point.

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Which makes sense because in this case, we already learned

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this in previous video.

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But we learned that the vertex of this parabola is the

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point x sub 1 comma y sub 1.

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Anyway, I don't want to confuse you.

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Actually, you know what I'm going to do?

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I don't want to go too far off into this topic, and I think

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this video has already gotten long enough.

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I will continue this in the next video.

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See you soon.

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