Parabola Focus and Directrix 1

Algebra

Quadratics

Graphing quadratics

Ex 3: Graphing a quadratic function
Example: Graphing a quadratic
Example: Roots and vertex of a parabola
Example: Parabola vertex and axis of symmetry
Graphs of Quadratic Functions
Examples: Graphing and interpreting quadratics
Applying Quadratic Functions 1
Applying Quadratic Functions 2
Applying Quadratic Functions 3
Graphing parabolas in standard form
Parabola Focus and Directrix 1
Focus and Directrix of a Parabola 2
Vertex of a parabola
Graphing parabolas in vertex form
Graphing parabolas in all forms
Parabola intuition 3

Parabola Focus and Directrix 1 Parabola as the locus of all points equidistant from a point and a line

Back

Parabola Focus and Directrix 1

⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.

Let's say I have a line, let me make it a straight line.
And the equation of that line, since it's running in the
horizontal direction is going to be y is equal
to some constant.
So let me write that.
So the equation of this line right here is y is equal to k.
And let's say I have some other point.
I'll call that-- well, we'll call that a focus because
it will be a focus.
And let's say that point-- let's put it right here-- is
it the coordinate, let's call it a comma b.
And let's think about the set of all points or the locus of
all points that are equidistant to this point right
here and this line.
Just so you get the terminology we'll call
this line a directrix.
Let me write that down.
Directrix, it's probably a word you've never heard before.
And frankly, I've never heard it outside of the context
of parabolas or conic sections in general.
So let's call that a directrix.
This is our focus.
And what we want to do is we want to find all of the points
in the xy-plane that are equidistant to this focus
and this directrix.
So there's one point that just by looking at it we could say
is going to be in our locus.
So it's going to be right there because clearly, this point is
equidistant; it's halfway between that point
and this line.
Now let's see what other points.
There will probably be this point right here because the
distance from there to there is this name as the distance
from if you just drop a line straight down.
Remember, we want to get the shortest distance between
this point and the line.
We could have said something like this distance, but this
distance wouldn't be the shortest distance between
this point and this line.
So you would go straight down to the line there.
Likewise, this point would be there.
So that distance is the same as that distance.
And you can already, I think, begin to imagine
the type of shape this is.
And if you look at the title of this video you can probably
guess where this is going.
That the shape is going to look something like this.
Or something pretty similar to what we know as a parabola.
And actually, it will be a parabola.
There's no mystery there.
But what we're doing this in this video is actually show
you that is a parabola.
Show it to you mathematically instead of this
little drawing here.
Where I show you that it kind of looks like it would be a
parabola because this distance right here looks about the same
as that distance right there.
I could keep going up and down the curve and keep doing that,
but that's not satisfactory.
Let's actually mathematically show that the locus of all
points that are equidistant to this point and this line-- this
focus and this directrix-- is in fact, a parabola.
So let's say I have some point that's in this locus and let me
call the point here-- let me do it in a different color because
it's the same color as my directrix-- x, y.
So I want to find all of the x, y's-- all of the points that
satisfy an equation where their distance to the focus is equal
to their distance to the directrix.
So what's the distance between x, y and the focus?
What's this distance right here?
d sub 1.
Well, we just used the distance formula.
It's x minus the x-coordinate of the focus.
So it's x minus a squared.
So the difference in the x's squared.
Plus the difference in the y's squared-- y minus b squared.
The square root of that.
So this is the distance to the focus.
So I'll call that d sub 1.
That's this distance right there.
And we want to find all of the x and y's where the distance to
the focus is equal to the distance to this
line right there.
So if we call this distance-- let's call this the d sub 2--
the distance to the directrix.
So the distance to the focus is going to be equal to d sub 2.
And d sub 2, what's this going to be equal to?
Well, it's just the difference in y because no matter where we
are we're just going to drop down straight to the directrix.
So the difference in y-- I could just say y minus k.
That would be the difference in y.
But in this case I did make it so that the coordinate up here
is higher than the k over here.
But what if we had a situation where this point was
below this line?
I mean, there's nothing here that I said that we can only
deal with points above this line.
We haven't proven that to ourselves yet.
So to make sure that this distance is positive we could
take the absolute value or we could just square it and
then take the square root.
And then that ensures that we're not dealing with
negative distances.
So here we set up the equation for all of the x's and y's
where the distance to the focus is equal to the distance
to the directrix.
And let's see if this actually turns out to be a parabola.
So the first thing we can do is we can square both sides and
get rid of the radicals.
So we get x minus a squared.
Radicals aren't good.
Well, I don't want to make any social commentary.
So x minus a squared plus y minus b squared is equal to
the square root of the other side-- y minus k squared.
And right now, at least, it looks like there's going to be
some y squared terms and I mean there's definitely
an x squared term.
That doesn't seem like it's a parabola right now.
But let's keep going forward.
This is x minus a squared.
And then let me square these-- actually, expand
these two binomials.
So plus y squared minus 2yb plus b squared is equal to y
squared minus 2yk plus k squared.
Now what can we do to simplify this?
Well, we have a y squared here and we have a y squared
on the right-hand side.
So we can subtract y squared from both
sides of the equation.
They cancel out, and then we got rid of the y squared so all
of a sudden this is starting to look a lot more like a parabola
because this is now an equation that relates y's
to an x squared.
So let's clean this up and actually make sure that we can
get it in a form that we normally associate
with a parabola.
So we now have-- I'll just rewrite it-- x minus a squared
minus 2yb plus b squared is equal to minus 2yk
plus k squared.
Now what can we do?
Let's take all the y's and the constants on one
side of the equation.
So if we move both of these terms onto the right-hand
side, so we want to add 2yb to both sides.
So we get x minus a squared is equal to-- I'm moving both of
these terms to the right-hand side.
So if we add 2yb to both sides then we'll have a positive
2yb on the right hand side.
So 2yb, and then we'll have that minus 2yk right there.
And then we're subtracting b squared from both sides.
So it's minus b squared plus k squared.
I just arbitrarily decided to write whatever I transferred
from the left-hand side to the right-hand side in magenta
just so maybe it makes it a little bit clearer.
And let's see if we can simplify this further.
And remember, just so you know what I'm doing this whole time.
In the back of my mind I'm trying to get it into the
format that I normally associate with a parabola.
And I'll do it in the corner here, just so you know
where the algebra's going.
I'm trying to get into a format x minus a squared
is equal to y minus b.
Because if we have it in this form, or actually ax.
Let me write that because I went too far to the right.
A times x minus-- well, actually I already used
the A so let me write x minus-- I don't know.
I'll make up x minus-- I'll call that v squared.
I'm picking maybe v for x value of the vertex.
I'm making up variables on the fly.
A times x minus v squared is equal to y minus b.
And you know, the actual letters I chose don't matter.
But this is a general form of a parabola.
And this is where I'm trying to go.
If I can get this thing that we're working on to this
form then we know we're dealing with a parabola.
And then, we can actually relate what we normally
associate with a parabola to these terms up here.
And obviously, these are different letters
than these letters.
And then we can try to figure out-- if we have a parabola,
how do we figure out its focus and is directrix.
But anyway, that was a bit of an aside just so you
know where we're going.
So let's try to simplify this a little bit more.
So we get--
[PHONE RINGING]
--my phone is ringing.
Let me silent it.
So we get x minus a squared is equal to-- let's see.
If we factor out a 2 and a y we get-- let me
factor out a y first.
We get 2b minus 2k times y.
And then I want to make it minus some constant.
So this is a constant right here.
This is one constant squared plus another constant.
So I could make this equal to minus-- let
me think about this.
This would be minus b squared minus k squared.
If I expand this minus out you would get minus b
squared plus k squared.
So this is the same thing right there.
And let's see.
Can I factor out-- yeah, I can factor out a 2.
The whole thing becomes x minus a squared is equal to 2 times b
minus k times y minus-- and now this is the difference of--
yeah, this is b plus k times b minus k.
You might want to review factoring polynomials if
that doesn't look familiar.
But this is b plus k times b minus k.
And I want to do that because I saw this b minus k out there so
that looked interesting to get it on both terms right there.
Now let's see.
If we divide everything times-- let's divide everything by 1
over 2 b minus k because I just want a y here, so I have some
coefficient in front of the y.
So let's divide everything by 1 over 2 b minus k.
Let me do it down here so I have some space.
1 over 2 b minus k times x minus a squared is equal to--
I'm dividing this by 2 b minus k, so that goes away.
So it's y minus-- so if I divide this by 2 b minus k the
b minus k's cancel out and I'm just left with b plus k over 2.
And I think I have gotten it in the form that I wanted.
In this case, the a that I wrote up here is now
1 over 2 b minus k.
This v is this a that I did here.
The y is the y and then my constant term is out here.
So I have hopefully shown to you that the locus of all
points that are equidistant to this point-- which is really
just an arbitrary point.
I said it's the point a, b-- and some line, is
in fact, a parabola.
And that's one of the ways that people define a parabola.
And we'll actually see soon that when you actually change
this relation-- in this case, we found that all of the points
that are equidistant to the focus and the directrix.
If we changed that ratio, if we find all of the points that are
1/2 as far away from the directrix as the focus, we
start getting another conic section.
If we said twice as far we would get another
conic section.
So this is a very interesting way of relating-- sometimes the
conic sections in the first video I showed how they're all
related to cutting the three-dimensional cone.
But they're also related in this way and eventually I want
to relate this to the three-dimensional cone so you
can see how it all fits together.
Which is always the fun thing about mathematics.
So this so far, all we've shown to you is if we do take the
locus of all points that are equidistant to a point and a
directrix that it is a parabola.
Now, the interesting thing to do is let's
say I have a parabola.
Let me give a general form.
And I'll try to use different letters than I used up here.
So let's say I have a parabola y minus--
let's call it y sub 1.
Where it's just a particular point on the parabola.
Y minus y sub 1 or you could call it some constant-- is
equal to-- let's call it, I want to use a capital A just
because that tends to be-- I'll use a capital A and just
know that it's different than this lowercase A.
Some capital A times x minus x sub 1.
squared.
Whenever you see a y sub 1 or an x sub 1, that means you're
actually dealing with a constant term.
If you see just an x or y that generally means that you're
dealing with a variable term.
So these are constants.
I could have put letters there.
I could have put a and b there, but then that would've
gotten confused with that.
But let's see if we can relate these letters because this is
something that we normally see.
And then, we want to find the focus or the directrix
of a parabola like this.
How can we relate these to these values right here?
How does k and a and b relate to these right there?
So let's see if we can set that up.
So if we just pattern match that is A.
If we say x sub 1 is lowercase a, actually that's a straight
up pattern match, so we can already write that on the side.
x sub 1 is equal to a.
So whatever x value I have here, this is also the x
value of the focus point.
Which makes sense because in this case, we already learned
this in previous video.
But we learned that the vertex of this parabola is the
point x sub 1 comma y sub 1.
Anyway, I don't want to confuse you.
Actually, you know what I'm going to do?
I don't want to go too far off into this topic, and I think
this video has already gotten long enough.
I will continue this in the next video.
See you soon.

Questions
Tips & Feedback

Be specific, and indicate a time in the video:

At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

Have something that's not a question about this content?

Post a tip or feedback
General discussion about the site
Report a technical problem with the site
Request a video or feature

This discussion area is not meant for answering homework questions.

Formatting tips

Cancel or

( total)

Share a tip

When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...

Suggest a fix

At 2:33, Sal says "double bonds" but should say "single bonds."

Have something that's not a tip or feedback about this content?

Ask a question
General discussion about the site
Report a technical problem with the site
Request a video or feature

This discussion area is not meant for answering homework questions.

Formatting tips

Cancel or

Discuss the site

For general discussions about Khan Academy, visit our Reddit discussion page.

Flag inappropriate posts

Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.

abuse

disrespectful or offensive
an advertisement

not helpful

low quality
not about the video topic
soliciting votes or seeking badges
a homework question
a duplicate answer
repeatedly making the same post

wrong category

a tip or feedback in Questions
a question in Tips & Feedback
an answer that should be its own question

about the site

a question about Khan Academy
(Visit our FAQ)
a post about badges
a technical problem with the site
(Report a problem)
a request for videos or features