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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 ~ when he squares the two radicals, why isn't it +/- the values beneath the radical?
• Because he took the distance, and distance isn't negative.
• At Sal 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.
• Why is f^2-a^2=b^2?
• 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.
• 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
• How can u apply the pythagorus theorum when they are not right angled triangles
• 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
• 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 after . 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?
• 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
• 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?
• 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.
• 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
• 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?
• Can anyone explain ? Why did he multiply 2a by 2 when squaring the right side of the equation? In addition, at how did he get -2a^2xf+a^4?
• 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!