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.