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### Course: Algebra (all content)>Unit 17

Lesson 8: Introduction to hyperbolas

# Graphing hyperbolas (old example)

Given the hyperbola equation y^2/4-x^2/9=1, Sal determines the direction to which it opens and its vertices in order to draw its graph. Created by Sal Khan.

## Want to join the conversation?

• This may be a silly question but I am just beginning to understand this....do the asymptotes always pass through the origin? (like at )
• Just like a circle, a hyperbola can be shifted.

A "normal" or "unshifted" hyperbola:
``x^2/a^2 - y^2/b^2 = 1``

A "shifted" hyperbola:
``(x-h)^2/a^2 - (y-k)^2/b^2 = 1``

where h and k specify the amount of horizontal and vertical shift respectively. In Sal's examples so far, h and k have effectively been zero, so the asymptotes have gone through the origin.
• I was taught that the equation of a hyperbola is (ax+b)/(cx+d) in its most developed form, or just simply 1/x, what's the difference?
• y=1/x is a rectangular hyperbola, where the asymtotes are the x and y axies, this a conic hyperbola, efectivly a 45 degree rotation of a rectangular hyperbola
• What are rectangular hyperbolas? and why are they called so??
• Great question! Rectangular hyperbolas are hyperbolas that have perpendicular asymptotes.
So a rectangular hyperbola's asymptotes intersect each other at a ninety degree angle.
• When the y^2/4 is multiplied out by 4 I don't understand how the 4/9x^2 part of 4/9x^2+4 is made. Can anyone point me to any materials show: where moving the denominator to the other side results in it knocking the top part of the fraction off & then be multiplied by that top part.
• Probably a good idea to review fraction notation.
(1/9)*4 = 4/9 right?
so when Sal writes (4/9)x^2,
(4/9)*x^2 = (4x^2)/9

You can always "knock off" a number in the numerator. It doesn't affect the value. It's useful for grouping terms, and can make the equation easier to see. In this example,
(some number) times (x^2)
... is a useful idea.
• At he says that it gets really close but never touches, and I thought of Gabriel's Horn. It gets really close to 0 but never touches there and goes for infinity. How does it have an infinite surface area, but a finite volume?
• Just to clarify: Gabriel's Horn is the 3D shape that forms the shape of y = 1/x rotated about the x axis from x = 1 to infinity. http://en.wikipedia.org/wiki/Gabriel's_Horn

The answer has to do with integral calculus. Simply put, the surface area at any point is proportional to 1/x, which, when summed from 1 to infinity, is infinite. On the other hand, the volume at any point on the x axis is proportional to 1/x^2 (because there's an extra dimension), which does have a finite upper limit when summed from 1 to infinity. More detail would require integrals themselves. Sal probably mentions this in his definite integral videos, as it is a notable paradox of integral calculus... https://www.khanacademy.org/math/calculus/integral-calculus
• The +4 disappears just because it "becomes irrelevant" compared to the other term?
• The +4 is reverent to the equation, Its what makes the asymptotes that are the basis of the hyperbolas shape but when finding the asymptotes we want to find the line it get increasingly closer to. Since the +4 is what separates the function from the asymptotes if you remove the +4 you get the asymptotes
• What happens when b is 1? Then wouldn't the b^2 under the square root be gone, therefore the hyperbola would touch the asymptotes?
• If 𝑏 were 1, you are correct that you could write the equation without it. You could write 𝑦²/1 or simply 𝑦² because dividing something by 1 does not change it. However, it would not touch the asymptotes, because 1² is 1, not 0.

If a 𝑏 of 1 is under the positive term, then it indicates the vertices are 1 unit from the center (where the asymptotes meet) whether you write the 1 in the equation or not.

(1 vote)
• Can't we solve +/- sqrrt ( (4/9)x^2 + 4 )?
Shouldn't it go to either (2/3)x + 2 or -(2/3)x - 2?
(1 vote)
• No... we can't do that. You are trying to take and apply the square root individually to each term. It doesn't work. Consider this scenario.
1) Calculate: sqrt( 9 + 16) = sqrt(25) = 5
2) Calculate: sqrt(9) + sqrt(16) = 3 + 4 = 7.
Notice: 5 does NOT = 7. I basically applied the same techniques as you did, but with a numeric example. We can clearly see that the process does not work.

Hope this helps.
• What is the intuition behind having a negative sign in hyperbola equations X^2/a^2 - Y^2/b^2 = 1. This is the only difference between an ellipse and a hyperbola equation. Can anyone please explain me the reason of the sign intuition changing the shape of a ellipse curve to a hyperbola ?
(1 vote)
• If we insist that the difference of two number is 1, those two numbers can be as large as we like. If one number is 1000000, we can set the other as 999999. So in the graph of a hyperbola, we get point with large x- and y-values and so the branches of the graph go off to infinity.

If we insist that the sum of two nonnegative numbers is 1 (nonnegative because both terms are squares), that means that neither term can exceed 1. If one term is too large, the other term cannot bring the sum back down to 1. So the set of solutions is bounded, and the ellipse is a bounded, finite curve.
• I know hyperbolas can be shifted (up, right, left, down), but can it be rotated?
• Hello Mr. Beast,

Absolutely. All of these conic section can be rotated to any position you desire. Its an interesting application of trig. For some worked out problems see this link. This man has amazing penmanship!