Hyperbolas
Conic Sections: Hyperbolas 2 Continuation of the intro to hyperbolas
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- 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.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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