Current time:0:00Total duration:7:30

# Graphing hyperbolas (old example)

## 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 google, 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.