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### Course: Precalculus > Unit 5

Lesson 5: Foci of a hyperbola# Proof of the hyperbola foci formula

Sal proves why, for the general hyperbola equation x^2/a^2-y^2/b^2=1, the focal length f forms the equation f^2=a^2+b^2 with the parameters a and b. Created by Sal Khan.

## Want to join the conversation?

- At ~5:30when he squares the two radicals, why isn't it +/- the values beneath the radical?(5 votes)
- Because he took the distance, and distance isn't negative.(26 votes)

- At5:32Sal says
*"And then to square this we have to square the first term,*" Why do we multiply the two terms and then multiply by 2? It seems like we've doubled the operation, so I must be missing something. Can you explain with a less complex example? Thanks for any help.

which is 4a squared. Then we multiply the two terms and multiply that by 2, right?(5 votes)- This is required when you square a binomial. For example (a + b)^2 = a^2 + 2ab + b^2. He is referring to the middle term there when he says to multiply the two terms and then multiply that by 2. See https://www.khanacademy.org/math/algebra/polynomials/multiplying_polynomials/v/special-products-of-binomials(20 votes)

- 13:55Why is f^2-a^2=b^2?(8 votes)
- one way to think about it is: Both the equation of a hyperbola( the one with the b^2), and the equation that we have near the end of the proof equal one. We could make make a new equation with the equation we found on one side and the original(the b^2 one)on the other side. Then you could solve for b^2.(5 votes)

- Sal, Thanks for the proof. Just wanted to say that I enjoyed the ending: I'll leave you there, that was an exhausting problem. I have to go get a glass of water now. Thank you for the dry humor, haha. Thumbs up. :))))))))

I may use that at some point.....LOL(7 votes) - How can u apply the pythagorus theorum when they are not right angled triangles(2 votes)
- the distance formula makes a right triangle between two points. the difference between the x coordinates are one leg, the difference between the y coordinates are the other leg, the right angle is the angle between those legs, and the hypotenuse is the line between the original points(7 votes)

- I've proved the ellipse's foci by myself before this video, but the proofs were quite different. I definitely did more than just add the d's instead of subtracting them. I then attempted to prove it the way shown in the video, but the proof fell apart after9:07. When squaring both sides, I got (fx-a^2)^2 which is the same thing as (a^2-fx)^2, meaning an ellipse is a hyperbola?(5 votes)
- There is an error in your reasoning.

When you got (fx-a^2)^2, that is not the same as (a^2-fx)^2, because you did not follow PEMDAS (Parenthesis, Exponents, Multiplcation and Division, Addition and Subtraction).

You cannot say fx - a^2 is equal to a^2 - fx, because subtraction is not commutative, i.e.,

a - b does NOT equal b - a

hope this helps(2 votes)

- I understand how to "reverse engineer" the proof once you already know what you're looking for (that is, that the difference of the distances from the foci to a given point is 2a), but how do you figure that out to begin with? What is the process to find that this relationship is true when you don't know in the beginning what you are looking for?(2 votes)
- There is no one process for figuring these things out from scratch. It's a pretty massive creative effort to come up with new formulae that are likely to be true.

This is basically what mathematicians do all day: they look at some structure (like a hyperbola), notice a pattern, make a guess about some formula or fact that might hold true, then try to prove it rigorously.(6 votes)

- I'm having problems with the exercise right before this video, is there someone that can tell how to find the formula out of a graph? None of the videos helped me with that. Thanks(3 votes)
- What about the Practice exercises immediately preceding, on finding the foci from the equation: Did those make sense to you? You can use the same method here, in reverse. Did you do the earlier exercises for a circle, and then for a parabola? The method of solution is basically the same for all the conic sections. What method did you follow to find the equation of a circle?(4 votes)

- Can anyone explain5:55? Why did he multiply 2a by 2 when squaring the right side of the equation? In addition, at9:30how did he get -2a^2xf+a^4?(2 votes)
- Check out this video also in the algebra playlist: https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/multiplying-binomials/v/square-a-binomial

Recall the distributive property you may have learned, which states that n(a + b) = na + nb

Well then (a + b)(a + b) = a(a + b) + b(a + b) = a² + ab + ba = b²

and since ab = ba

(a + b)(a + b) really equals to a² + 2ab + b².

So for you first questions, to get to the middle tern, Sal multiplied the 2a and the square root together, then by another 2.

For the second question, since we are taking the square of xf - a², the middle term after expanded it out would be the**product**of xf and a² times 2. the a^4 is really the square of a².

Hopefully that helps!(4 votes)

- At4:42, wouldn't it be most proper to write d1 - d2 = +-2a ? Or does this not matter? This difference may not be positive so the video seems wrong to me.(4 votes)
- I think the way u r writing is also correct. Sal mentions in the video that we should take the absolute value of d1 -d2 i.e. |d1 - d2| = 2a . So the video isn't wrong. He has clarified more about it in the previous video.(1 vote)

## Video transcript

In the last video, I told you
that if I had a hyperbola with the equation x squared over a
squared minus y squared over b squared is equal to 1, that the
focal distance for this hyperbola is just equal to the
square root of the sum of these two numbers. The square root of a
squared plus b squared. In this video I really just
want to show you that. And actually, just so you know,
this equation right here, this is a particular hyperbola that
opens to the left and the right. And that's because those
are the asymptote points. Those would be the axis. And that's because the
x term is positive. If the y term was positive and
the x term had a negative sign, then the hyperbola would open
upwards and downwards, like that. And the proof that I'm showing
you in this video, it's just a bunch of algebra, really, is
identical in the y case, you'd just switch around
x's and the y's. But I just wanted to make
sure you realize that. That I'm just doing a
particular case of hyperbola that opens the left
and the right. I could call it a horizontal
hyperbola, instead of a vertical one, but I wanted
to make it clear there is another type of hyperbola. But anyway, let's draw a
graphical representation of all this just to make sure
we understand, or we re-understand, or better
understand what the foci points are and where they sit on the
hyperbola So those are my axis. The asymptotes of this
hyperbola are the lines y is equal to plus
or minus b over a. Oh woops, not using
my line tool. So, that's one and that's
the other asymptote. then the hyperbola will
look something like this. It looks something like that. It's going to intersect at
a comma 0, right there. This is going to be a comma 0. And the intersect at
minus a comma 0. We saw this in the
previous video. It looks something like that. And then the focus
points are going to sit out here someplace. There and there. And the focal length this
a squared plus b square. The square root of a
squared plus b squared. That's just this
distance right here. That distance is
the focal length. So this is going to be the
point f,0 and this is going to be the point minus f,0. Now we learned in the last
video that one of the definitions of a hyperbola is
the locus of all points, or the set of all points, where if I
take the difference of the distances to the two foci, that
difference will be a constant number. So if this is the point x comma
y, and it could be any point that satisfies this equation,
it's any point on the byperbola, we know, or we are
told, that if we take this distance right here-- --let's
call that d1 --and subtract from that the distance to the
high other foci-- call that d2 --that that number is a
constant regardless of where we are on the hyperbola. In fact the locus of all points
are-- the hyperbole in fact is all of the points that
satisfy that condition. And we learned in last video
just by taking the difference of the distance, we picked this
point and we said, OK, what's that distance minus
that distance. And we figured out
that it's 2a. So d1 minus d2 is equal to--
I'm going off the video screen --d1 minus d2 is equal to 2a. So let's use this fact right
here that d1 minus d2 is equal to 2a, to try to prove this. Right there. So the first thing to do is
figure out what d1 and d2. Just using the
distance formula. So what's d1? d1 is the distance between this
point and this point minus f0. So what you do is we just
use the distance formula, which is really just
Pythagorean theorem. So it's the difference
of the x's. It's the x distance. So it's x minus f squared
plus the y distances. y minus 0, so that's
just y, squared. Take the square root of that. So that's d1, right there. d1. And we want to subtract
from that d2. the difference of the
distances, and in this case d1 is definitely bigger than d2. Or you could take the absolute
values if you didn't want to worry about that. And so here, we get the square
root of x minus f, x minus f squared, plus y squared. What does that equal to? Well, we said that equals
to 2a, that equals this distance right here. So that is equal to 2a. Now let's see if we can
simplify this at all. Well an interesting thing to do
my just to be the put this on the other side of the equation. And this just can get hairy,
so I really hope I don't make any careless mistakes. So this becomes-- and I might
write small to save space --this becomes x plus f, right,
minus minus, squared plus y squared is equal to 2a plus the
square root of x minus f squared plus y squared. Now, to get rid of these
radicals, let's square both sides of this equation. The left hand side, if you were
to square it just becomes x plus f squared plus y squared. And then to square this we have
to square the first term, which is 4a squared. Then we multiply the two terms
and multiply that by 2, right? We're just taking this whole
thing and squaring it, so that's-- and this is just a
review of kind of binomial algebra --so this is equal to
plus 2a times this times 2 is 4a times the square root of
x minus f squared plus y. y squared, I don't want to lose
that squared right there. And then we square this term. And this is just
multiplying a binomial. So that's equal to-- you just
get rid of the radical sign, and I'm just going to be
staying in that color for now --that's equal to x minus
f squared plus y squared. And already it looks like
there's some cancellation that we can do. We can cancel out-- there's a
y squared on both sides of this equation --so let's
just cancel those out. Subtract y squared from both
sides of the equation. And let's multiply
this term out. So this right here is x squared
plus 2xf plus f squared. And then that is equal to 4a
squared plus 4a times the square root of x minus f
squared plus y squared. And then multiply this out. Plus x squared minus
2xf plus f squared. And then let's see, what
can we cancel out. We have x squared on both
sides of this, we subtract x squared from both
sides of the equation. We have an f squared on both
sides of the equation so let's cancel that out. And let's see, what can
we do to simplify it. So we have a minus
2xf and a plus 2xf. Let's add 2xf to both sides
this equation or bring this term over here. So if you add 2xf to both sides
of this equation-- let's see, my phone is ringing, let me
just turn it off --if you add 2xf to both sides of this
equation, what do you get? You get 4xf-- remember I just
brought this term over this left hand side --is equal to
4a squared plus 4a times a square root of x minus f
squared plus y squared. It's easy to get lost
in the algebra. Remember all we're doing, just
to kind of remind you of what this whole point was, we're
just simplifying the difference of the distances between these
two points, and then see how it relates to the equation
of the hyperbola itself. The a's and the b's. Let's take this 4a put it on
this side, so you get 4xf minus 4a squared is equal to 4a times
the square root of-- well let's just multiply this out 'cause
we'll probably have to eventually --x squared minus
2xf plus f squared plus y squared. That's this just
multiplied out. That's the y squared
right there. We could divide both
sides of this by 4. All I'm trying to do is just
simplify this as much as possible, so then this becomes
xf minus a squared is equal to a times the square root
of this whole thing. x squared minus 2xf plus f
squared plus why squared. And now we could square
both sides of this equation right here. And then if you square both
sides, this side becomes x squared f squared minus 2a
squared xf plus a to the fourth. That's this side squared. And that's equal to, if you
square the right hand side, a squared times the square of
a square root is just that expression, x squared minus 2xf
plus f squared plus y squared. This really is quite hairy. And let's see what
we can do now. Let's divide both sides of this
equation by a squared, and then you get x squared-- I'm really
just trying to simplify this as much as possible --over a
squared minus-- so the a squareds cancel out --minus 2xf
plus a to the fourth divided a square well that's
just a squared. So a squared is equal to x
squared minus 2xf plus f squared plus y squared. Well good. There's something
to cancel out. There's a mine 2xf on both
sides of this equation so let's cancel that out. Simplify our situation
a little bit. And let's see, we have. so we could do is subtract
this x squared from this. So you get-- let me rewrite
it --so you get x squared f squared over a squared
minus x squared. And let's bring this y to this
side of the equation too. So minus y squared. That's all I did, I just
brought that to that side. And then let's bring-- and I'm
kind of skipping a couple of steps, but I don't want to take
too long --let's take this a and put it on that
side equation. So we took the x and the y,
we subtracted that from both sides of the equation. So they ended up on
the left hand side. And then if we subtract a
squared from both sides of this equation-- this is a fatiguing
problem --then you get f squared minus a squared. I think we're almost there. This can simplify to,
let's see we can factor out the x squared. This becomes f squared
over a squared minus 1 times x squared. I just factor out the x squared
there, minus y squared is equal to f squared, the focal length
squared, minus a squared. And let's see, let's divide
both sides of the equation by this expression right there,
and we get-- and this should start to look familiar --we get
f squared over a squared minus 1, x squared divided by f
squared minus a squared minus y squared over f squared mine
is a squared is equal 1. Right, I divided both sides
by this, so I just get a 1 on this right hand side. Let's see if I can
simplify this. If I multiply the numerator
and the denominator by a squared, right? As long as I multiply the
numerator and the denominator by the same number I'm just
multiplying my 1, so I'm not changing anything. So if I do that, the numerator
becomes f-- if I multiply it, it becomes f squared
minus a squared. I'm just multiplying
that times a squared. And the denominator
becomes a squared times f squared minus a squared. And all that times x squared. Minus y squared minus
a squared mine is a squared is equal to 1. This cancels with this. And we get something to
starting to look like the equation of a hyperbola. My energy is coming back! It seems like I see the light
at the end of the tunnel. We get x squared over a squared
minus y squared over f minus a squared is equal to 1. Now this looks a lot like our
original equation of the hyperbola, which was x squared
over a squared minus y squared over b squared is equal to 1. In fact this is equation of the
hyperbola but instead set of writing b squared, since we
wrote it, we essentially said, what is the locus of all points
where the difference of the distances to those two
foci is equal to 2a? And we just played with
the algebra for while. It was pretty tiring, and I'm
impressed if you've gotten this far into the video, and we got
this equation, which should be the equation of the hyperbola,
and it is the equation of the hyperbole. It is this equation. So this is the same
thing is that. So f squared minus a square. Or the focal length squared
minus a squared is equal to b squared. You add a squared to both
sides, and you get f squared is equal to b squared plus
a squared or a squared plus b squared. Which tells us that the focal
length is equal to the square root of this. Of a squared plus b squared. And that's what we set out to
figure out in the beginning. So hopefully you're now
satisfied that the focal length of a hyperbola is the sum of
these two denominators. And it's also truth if it is an
upward or vertical hyperbola. And if we're dealing with an
ellipse, it's the difference of these two-- the square root
of the difference of these two numbers. Anyway, I'll leave you there. That was an exhausting problem. I have to go get a
glass of water now.