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## Precalculus

# 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 at3:23)(45 votes)
- 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.(71 votes)

- 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?(12 votes)
- 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(18 votes)

- What are rectangular hyperbolas? and why are they called so??(6 votes)
- Great question! Rectangular hyperbolas are hyperbolas that have perpendicular asymptotes.

So a rectangular hyperbola's asymptotes intersect each other at a ninety degree angle.(10 votes)

- 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.(4 votes)
- 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.(6 votes)

- At5:35he 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?(4 votes)
- 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(4 votes)

- 1:55The +4 disappears just because it "becomes irrelevant" compared to the other term?(3 votes)
- 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(2 votes)

- 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?(4 votes)
- 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.

Have I answered your question, or did I misunderstand you?(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.

When simplifying radicals, we can split/apply the radical to individual factors. We can't split/apply the radical to individual terms inside the radical.

Hope this helps.(5 votes)

- I know hyperbolas can be shifted (up, right, left, down), but can it be rotated?(2 votes)
- 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!

https://www.youtube.com/watch?v=hFtNJQIi--k

Regards,

APD(2 votes)

- I get it the easy way, but then they find a even harder way to teach it. is it me?(2 votes)

## Video transcript

In the last hyperbola video
I didn't get a chance to do some concrete examples. So I'll do that right now. So, let's say I had the
hyperbola y squared over 4 minus x squared over,
I don't know, let me think of a good number. Let's say, x squared
over 9 is equal to 1. So the first thing to figure
out about this hyperbola is, what are its asymptotes? And, once again, I always
forget the formulas. And I just try to solve for y
and see what happens when x approaches positive or
negative infinity. So if you solve for y,
you can add x squared over 9 to both sides. And you get y squared
over 4 is equal to x squared over 9 plus 1. Now, I can multiply
4 times both sides. And you get y squared is
equal to 4 over 9 times x squared plus 4. I distribute the 4, take
the positive and negative square root both sides. y is equal to the plus or
minus square root of 4 over 9x squared plus 4. And you can't really
simplify this anymore. But we can think about, what
does this approach as x approaches positive or
negative infinity. So, as x approaches plus
or minus infinity, what does this roughly equal? What does this approximate? What does the graph
get a lot closer to? Well, then, y is approximately
equal to just the square root of this term. Because this becomes super huge
and relative to this term, this starts to matter less
and less and less. And that's why we get closer
and closer to the asymptotes. Because when this number
is, like, a trillion, or a googol, then this number
is almost insignificant. You take the square root,
you're pretty much taking the square root of this, and
you'll just be a little bit above the graph. Because you have this
extra plus-4 there. So as you approach positive
or negative infinity, this equation is approximately equal
to the plus or minus square root of 4 over 9x squared. And so, that is -- so y would
be approximately equal to the plus or minus. We can take the
square root of this. Plus or minus the square
root of 4/9 is 2/3, right? Square root of 4 over
square root of 9, times x. So, these are the asymptotes. There's two lines here. There's y is equal to 2/3 x. And then there's y is
equal to minus 2/3 x. So let's draw those two lines. Let me draw my axes. Let's make that my y axis. Make that the x axis. Let me switch some colors, just
to make things interesting. So let me draw the first one. See, y is equal to 2/3 x. So, you rise 2 for
every 3 that you run. So let me draw that. So if this is 1, 2, 3, 1, 2. So that would be a
point on the line. Let me draw the line now. Actually go to the origin. No, that's not it. Let me draw it like this. This way I can make sure
it goes to the origin. This is going to go
through like that. Then I can go from here. And then go like that. So that's one asymptote. And the other asymptote is
y is going to be equal to minus 2/3 x, right? Because plus or minus 2/3 x. So, minus 2/3 x, you go down 2
for every 3 that you go out. So that point will show
up, see if I do, 1, 2. So you go down 2 for
every 3 that you go out. So it'll go 3. So if I draw that asymptote,
it'll look something like there. Go out there. And then go from here. Go out there. And we've drawn our asymptotes. Now the question is, is it
going to open up to the left or the right, or up and down? There's two ways we
can think about it. And I'll do it the way that
might be more intuitive for you is, can x -- what happens
when x is equal to 0? Well, when x is equal to
0, when x is equal to 0, this disappears. And we're just left with,
I'll do it here, y squared over 4 is equal to 1. Or, y squared is equal to 4. Or, y is equal to
plus or minus 2. So, we know that the point
0, the points, 0 plus or minus 2, is on this graph. So x can be equal to 0
so 0 plus or minus 2. So 0 plus 2 is this
point right here. And 0 minus 2 is this
point right there. So that, by itself, actually,
is enough of a clue to know that it opens down here. And up here. Because it will never,
a hyperbola will never cross the asymptotes. It's not like it can go out
here and across this asymptote. So? We already know that the graph
of this parabola -- and you can try other points, if you
want, just to verify. It's going to look
something like this. It's going to go and then
-- nope, I want to make it so it never touches. It's going to get really
close, but no, I touched it. It's going to get really
close but never touch. And then on this side
it's going to get really close, but never touch. And I don't want to touch it. And then on the top side it's
going to do the same thing, it's going to get really close,
and as you approach infinity it's never going to touch it. And as you get reall close,
it'll get infiniitely close but never touch it. So that's what this parabola
-- this hyperbola -- is going to look like. And I did it by just trying to
see if x could be equal to 0. And I encourage you to try
what happens when y equals 0. And you'll get no solution. And that makes sense because
this hyperbola never crosses y equals 0, right? It never crosses the x axis. And this should also be
intuitive, because if we saw here when we did ths
approximation, as x approaches positive or negative infinity,
we saw that we always did have this plus 4 sitting here. We said, oh, well, as x gets
super large or super negative, this starts to matter
less and less. But we will always be slightly
larger than this number. Especially in the positive
quadrant, right? We're always going to be --
so the positive quadrant is always going to be slightly
larger than the asymptote. And even when we take the
positive square root, I guess, is the best way to say it. When we take the positive
square root, we'll always be larger than either
of the asymptotes. And, likewise, when you take
the negative square root, you're always going to be
a little bit smaller than either of the asymptotes. Because this number is
going to be a little bit bigger than this number. Then we take the negative
square root, you're going to be a little bit smaller,
and that's why we're a little bit below. I don't know which one's
more intutive for you. Maybe just the -- trying it
when x is equal to 0 and when y equals 0, and see what points
you get and say, oh, then I'm in kind of a vertical hyperbola
as opposed to a horizontal one. So let's see if I have time
for -- I'll leave that video right there. And then I'll do another
video where I actually shift the hyperbola. And shifting it is actually
no different than shifting an ellipse a circle. You just have, you know, y
minus something squared, and x plus something, or x minus
something squared and that just tells you where you
shift the origin. This hyperbola, of course, is
just centered at the origin. Anyway, see you in
the next video.