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.
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- At ~5:30when he squares the two radicals, why isn't it +/- the values beneath the radical?(5 votes)
- At5:32Sal says "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?" 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.(4 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(19 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)
- 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)
- 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(5 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'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?(4 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'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)
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.