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# Foci of an ellipse from equation

Sal explains how the radii and the foci of an ellipse relate to each other, and how we can use this relationship in order to find the foci from the equation of an ellipse. Created by Sal Khan.

Video transcript

Let's say we have an ellipse
formula, x squared over a squared plus y squared over
b squared is equal to 1. And for the sake of our
discussion, we'll assume that a is greater than b. And all that does for us is, it
lets us so this is going to be kind of a short
and fat ellipse. Or that the semi-major axis,
or, the major axis, is going to be along the horizontal. And the minor axis is
along the vertical. And let's draw that. Draw this ellipse. I want to draw a
thicker ellipse. Let's say, that's my ellipse,
and then let me draw my axes. OK, this is the
horizontal right there. And there we have the vertical. And we've studied an ellipse
in pretty good detail so far. We know how to figure out
semi-minor radius, which in this case we know is b. That's the same b right there. And that's only the
semi-minor radius. Because b is smaller than a. If b was greater, it would
be the major radius. And then, of course,
the major radius is a. And that distance is
this right here. Now, another super-interesting,
and perhaps the most interesting property of an
ellipse, is that if you take any point on the an ellipse,
and measure the distance from that point to two special
points which we, for the sake of this discussion, and not
just for the sake of this discussion, for pretty much
forever, we will call the focuses, or the foci,
of this ellipse. And these two points,
they always sit along the major axis. So, in this case, it's
the horizontal axis. And they're symmetric around
the center of the ellipse. So let's just call
these points, let me call this one f1. And this is f2. And it's for focus. Focuses. f2. So the super-interesting,
fascinating property of an ellipse. And it's often used as the
definition of an ellipse is, if you take any point on this
ellipse, and measure its distance to each of
these two points. So, let's say that I have
this distance right here. Let's call this distance d1. And then I have this distance
over here, so I'm taking any point on that ellipse, or this
particular point, and I'm measuring the distance to
each of these two foci. And this is d2. We'll do it in a
different color. So this is d2. This whole line right here. That's d2. So when you find these two
distances, you sum of them up. So this d2 plus d1, this
is going to be a constant that it actually turns
out is equal to 2a. But it turns out that it's true
anywhere you go on the ellipse. Let me make that point clear. And I'm actually going to prove
to you that this constant distance is actually 2a, where
this a is the same is that a right there. So, just to make sure you
understand what I'm saying. So let me take another
arbitrary point on this ellipse. Let's take it right there. And if I were to measure the
distance from this point to this focus, let's call that
point d3, and then measure the distance from this point to
that focus -- let's call that point d4. d4. If I were to sum up these two
points, it's still going to be equal to 2a. Let me write that down. d3 plus d4 is still going
to be equal to 2a. That's just neat. And, actually, this is often
used as the definition for an ellipse, where they say that
the ellipse is the set of all points, or sometimes they'll
use the word locus, which is kind of the graphical
representation of the set of all points, that where the sum
of the distances to each of these focuses is
equal to a constant. And we'll play with that a
little bit, and we'll figure out, how do you figure out
the focuses of an ellipse. But the first thing to do is
just to feel satisfied that the distance, if this is true,
that it is equal to 2a. And the easiest way to figure
that out is to pick these, I guess you could call them, the
extreme points along the x-axis here and here. We're already making the claim
that the distance from here to here, let me draw that
in another color. That this distance plus this
distance over here, is going to be equal to some
constant number. And using this extreme point,
I'm going to show you that that constant number is equal to 2a,
So let's figure out how to do that. So one thing to realize is that
these two focus points are symmetric around the origin. So, whatever distance this is,
right here, it's going to be the same as this distance. Right there. Because these two points are
symmetric around the origin. So this, right here, is the
same distance as that. And, of course, we have -- what
we want to do is figure out the sum of this distance and this
longer distance right there. Well, what's the sum of this
plus this green distance? Well, this right here
is the same as that. So this plus the green --
let me write that down. So let me write down these, let
me call this distance g, just to say, let's call that g,
and let's call this h. Now, if this is g and this is
h, we also know that this is g because everything's symmetric. So what's g plus h? Well, that's the same
thing as g plus h. Which is the entire major
diameter of this ellipse. Which is what? Well, we know the minor
radius is a, so this length right here is also a. So the distance, or the sum of
the distance from this point on the ellipse to this focus, plus
this point on the ellipse to that focus, is equal to g plus
h, or this big green part, which is the same thing as the
major diameter of this ellipse, which is the same thing as 2a. Fair enough. Hopefully that that is
good enough for you. Now, the next thing, now that
we've realized that, is how do we figure out where
these foci stand. Or, if we have this equation,
how can we figure out what these two points are? Let's figure that out. So, the first thing we realize,
all of a sudden is that no matter where we go, it was easy
to do it with these points. But even if we take this point
right here and we say, OK, what's this distance, and then
sum it to that distance, that should also be equal to 2a. And we could use that
information to actually figure out where the foci lie. So, let's say I have --
let me draw another one. So that's my ellipse. And then we want
to draw the axes. For clarity. Let me write down
the equation again. Just so we don't lose it.
x squared over a squared plus y squared over b
squared is equal to 1. Let's take this
point right here. These extreme points are always
useful when you're trying to prove something. Or they can be, I don't
want to say always. Now, we said that we have these
two foci that are symmetric around the center
of the ellipse. This is f1, this is f2. And we've already said that an
ellipse is the locus of all points, or the set of all
points, that if you take each of these points' distance from
each of the focuses, and add them up, you get a
constant number. And we've figured out that
that constant number is 2a. So we've figured out that if
you take this distance right here and add it to this
distance right here, it'll be equal to 2a. So we could say that if we
call this d, d1, this is d2. We know that d1 plus
d2 is equal to 2a. And an interesting thing
here is that this is all symmetric, right? This length is going to be the
same, d1 is is going to be the same, as d2, because everything
we're doing is symmetric. These two focal lengths
are symmetric. This distance is the
same distance as this distance right there. So, d1 and d2 have
to be the same. There's no way that you
could -- this is the exact center point the ellipse. The ellipse is symmetric
around the y-axis. So if d1 is equal to d2, and
that equals 2a, then we know that this has to be equal to a. And this has to be equal to a. I think we're making progress. And the other thing to think
about, and we already did that in the previous drawing of
the ellipse is, what is this distance? This distance is the
semi-minor radius. Which we already learned is b. And this of course is the
focal length that we're trying to figure out. This should already pop into
your brain as a Pythagorean theorem problem. So we have the focal length. And we could do it on this
triangle or this triangle. I'll do it on this
right one here. This focal length is f. Let's call that f. f squared plus b squared is
going to be equal to the hypotenuse squared, which
in this case is d2 or a. Which is equal to a squared. And now we have a nice
equation in terms of b and a. We know what b and a are,
from the equation we were given for this ellipse. So let's solve for
the focal length. The focal length, f squared,
is equal to a squared minus b squared. So, f, the focal length, is
going to be equal to the square root of a squared
minus d squared. Pretty neat and clean, and
a pretty intuitive way to think about something. So you just literally take
the difference of these two numbers, whichever is larger,
or whichever is smaller you subtract from the other one. You take the square root, and
that's the focal distance. Now, let's see if we can use
that to apply it to some some real problems where they might
ask you, hey, find the focal length. Or find the coordinates
of the focuses. So let's add the equation x
minus 1 squared over 9 plus y plus 2 squared over
4 is equal to 1. So let's just graph
this first of all. This could be interesting. So I'll draw the axes. That's the x-axis. This is the y-axis. And we immediately see,
what's the center of this? The center is going to be at
the point 1, negative 2. And if that's confusing, you
might want to review some of the previous videos. Center's at 1, x is equal to 1. y is equal to minus 2. That's the center. And then, the major axis is the
x-axis, because this is larger. And so, b squared is -- or
a squared, is equal to 9. And the semi-minor radius
is going to be equal to 3. So, if you go 1, 2, 3. Go there. Than you have 1, 2, 3. No. 1, 2, 3. 1, 2, 3. I think this -- let's see. 1, 2, 3. You go there, roughly. And then in the y direction,
the semi-minor radius is going to be 2, right? The square root of that. So b is equal to 2. So you go up 2, then
you go down 2. And this ellipse is going
to look something like -- pick a good color. It's going to look
something like this. And what we want to do is,
we want to find out the coordinates of the
focal points. So, the focal points
are going to sit along the semi-major axis. And we need to figure out
these focal distances. And then we can essentially
just add and subtract them from the center. And then we'll have
the coordinates. What we just showed you, or
hopefully I showed you, that the the focal length or this
distance, f, the focal length is just equal to the square
root of the difference between these two numbers, right? It's just the square
root of 9 minus 4. So the focal length is equal
to the square root of 5. So, if this point right here is
the point, and we already showed that, this is the point
-- the center of the ellipse is the point 1, minus 2. The coordinate of this focus
right there is going to be 1 plus the square root
of 5, minus 2. And the coordinate of this
focus right there is going to be 1 minus the square
root of 5, minus 2. And all I did is, I took the
focal length and I subtracted -- since we're along the major
axes, or the x axis, I just add and subtract this from the x
coordinate to get these two coordinates right there. So, anyway, this is the really
neat thing about conic sections, is they have these
interesting properties in relation to these foci or in
relation to these focus points. And in future videos I'll show
you the foci of a hyperbola or the the foci of a -- well, it
only has one focus of a parabola. But this is really starting
to get into what makes conic sections neat. Everything we've done up to
this point has been much more about the mechanics of graphing
and plotting and figuring out the centers of conic sections. But now we're getting into
a little bit of the the mathematical interesting
parts of conic sections. Anyway. See you in the next video.