Hyperbolas
Foci of a Hyperbola Introduction to the foci (focuses) of a hyperbola
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- In the last video, we learned that an ellipse can be defined
- as the locus of all points where the sum of the distances
- to two special points, called foci-- and let me draw this all
- out, so that's my x-axis-- the sum of the distance to these
- two special points, called focuses or foci, is a constant.
- So if this is my ellipse-- I'll draw it out, that looks about
- where I want it to be right around, that looks about
- right-- it's centered at the origin, doesn't have to be, but
- for our purposes, let's make it centered on the origin.
- If this is one focus point right here, and this is the
- other focus point, this ellipse could be defined as the set of
- all points, or the locus of all points, where if I take the
- distance of any one of these points that exist on the
- ellipse and take the distance to each of the locuses-- sorry,
- each of the focuses-- if I take that-- nope, I don't want to do
- that-- if I take that distance and add it to this distance, so
- let's call this d1-- nope, too thick-- let's call that d1,
- this is d2-- that that's going to be equal to a constant
- number along the whole ellipse.
- So if I take a random point along the ellipse, say I take
- this point right here, and if I were to sum this distance to
- that distance-- so let's call this d3, this is d4-- the sums
- of these distances to the focus along with this ellipse are
- going to be a constant.
- So in this case, d2 plus d1, this plus that, is going
- to be equal to d3 plus d4.
- And this would be true wherever you go along the whole ellipse,
- and we learned in the last video that this quantity is
- actually going to be equal to 2a, where a is the distance
- of the semi-major radius.
- If this is the formula for the ellipse, this is
- where the a comes from.
- x squared over a squared plus y squared over b
- squared is equal to 1.
- And we learned that the focus, the focal distance-- or the
- distance from the central of the ellipse, which is this
- distance right here-- that focal distance is just the
- square root of the difference of these two numbers.
- If this is the focal distance from here to here, it's just
- equal to the square root of-- if a is larger then it would be
- a squared minus b squared, which is the case
- in this ellipse.
- If we have a vertical ellipse, and I really didn't cover it in
- the last video, but let me just show you what it
- would look like.
- Let's say that the ellipse looks something like this.
- I'll use blue.
- Let's say the ellipse looks like that.
- In this case, our semi-major radius is now in the y
- direction, so this is b, this is a, and in this case, b is
- greater than a, because the ellipse is tall and skinny.
- In this case the, focuses are always going to lie
- along the major axis.
- In this case the major axis is the vertical axis, so the
- focuses are going to sit here and here, and in this case, the
- focal lengths are going to be vertical down from the origin
- and vertical up from the origin, and we get, instead of
- it being a squared minus b squared, now since b is larger
- than a, the focal length, which is this, is going to be equal
- to b squared minus a squared.
- Fair enough.
- Now I did all of that to kind of compare it to what we're
- going to cover in this video, which is the focus points or
- the foci of a hyperbola.
- And a hyperbola, it's very close to an ellipse, you could
- probably guess that, because if this is the equation of
- an ellipse, this is the equation of a hyperbola.
- x squared over a squared minus y squared over b
- squared is equal to 1.
- Or we could switch these around, where the minus
- is in front of the x instead of the y.
- And we could cover that in a second.
- But this hyperbola looks something like this.
- Let me see if I can draw it.
- If I draw the axes, and then I want to draw the asymptotes--
- you could prove it, you could look at the some of the
- previous videos, but the asymptotes for this hyperbola
- are going to be y is equal to plus or minus b over ax, so
- it's going to look-- I'll just draw them as kind of tilted
- lines, so it would look something like that,
- something like-- nope.
- I want to make it something like that.
- Those are the as-- but this one's centered at the origin,
- because it hasn't been shifted, and then that's those
- two lines right there.
- And then, this is what I call kind of a horizontal hyperbola.
- The way you can think about that is, well, if you solve for
- y, you'll see that you're always going to be a little
- bit lower than the asymptote.
- The other option is you say, OK, can x or y equal 0?
- Well if y is equal to 0, that puts us along the x-axis, and
- you get x squared over a squared is equal to 1, which
- means that x squared is equal to a squared, which means that
- x is equal to plus or minus a.
- So the points a,0, and the point minus a,0, are
- both on this hyperbola.
- And since they have to kind of be contained by these
- asymptotes, never go through it, you know that this is going
- to be a hyperbola that opens to the left and the right, so
- it'll look something like this.
- Let me use its color.
- So it'll look something-- this is the hard part-- it'll get
- closer and closer on that side, and then you kind of view this
- as one of the vertexes or vertices of the hyperbola,
- and it'll go like that.
- Where this distance-- and notice the similarity here with
- the ellipse-- this distance right here-- let me do it in a
- more vibrant color-- this distance right here, between
- these two I guess you could call them the elbows of the two
- ellipses-- that distance right there, that is
- a, and this is also a.
- So you have a 2a distance, which is very similar to this
- situation, where this distance is a and this distance is a.
- So your distance between the two left and right points in a
- horizontal ellipse is the same as the distance between the two
- left and right points on a hyperbola.
- It's just the hyperbola opens outward while the
- ellipse opens inward.
- Fair enough.
- But the whole point of this video is to discuss foci, and
- you might have guessed-- and I did touch on it in the last
- video-- that hyperbola also have foci, but they open.
- They're going to be to the right and the left of these two
- points, so this is a-- let me do them in a bright color,
- because I want you see them-- let's say that those are the
- two foci-- and a hyperbola, this is-- and notice
- the difference.
- An ellipse, one of the definitions of an ellipse, was
- the locus of all points or the set of all points where the
- distance from each of those points, the sum of the distance
- from each of those points to the two foci is a constant.
- Now the definition of a hyperbola, one of the
- definitions of a hyperbola, can be the locus of all points
- where you take the difference-- not the sum, you take the
- difference of the distances between the two foci, so if--
- let me write that down.
- So this is d1, and this is d2, so you have a situation-- and
- we could take the absolute value of the difference,
- because they might be, they might at some points d1
- will be longer than d2 if you're on this curve.
- If you're on this curve, d1 will be shorter than d2-- So d1
- minus d2, the absolute value is going to be equal to constant.
- In the ellipse situation, d1 plus d2 was a constant.
- So they're very closely related.
- In the ellipse, you're taking the sum of the distances
- to the focus points and saying it's a constant.
- In a hyperbola, you're taking the difference of the distances
- to the focus points and saying that's a constant.
- So this number right here is going to be the exact same
- thing as if I took a point right here-- and I'm picking
- these points arbitrarily, as long as they're on the
- hyperbola, and if I called these two points d3 and d4, the
- difference between d1 and d2 is the same thing as the
- difference, between d3 minus d4.
- This is going to be a constant the entire way
- around the ellipse.
- And so the next question is, what is this constant
- going to be equal to?
- And this is where it's useful to find a point where you can
- kind of get the intuition, and we did it with the ellipses,
- where we said if we take these points, we used logic in the
- last video to say oh, the sum of the distance between this
- and this, that sum, is going to be equal to, we saw, is going
- to be equal to 2a, or the distance of the
- semi-major axis.
- Because this distance was the same as this distance, so this
- plus this is the same thing as this plus that, which is 2a.
- So the entire time, the contents, the sum of the
- distances to the two foci was equal to 2a.
- Now in the hyperbola, what is the difference of the
- distances to the two foci?
- So let's take this point right here on the hyperbola, and
- we're saying so what is-- let me do a good color-- what is
- the magenta distance-- that's the distance to that foci--
- minus this-- let me pick another color-- minus this
- light blue distance?
- This magenta distance minus this light blue distance.
- So we can make a very similar argument that we made in
- the ellipse situation.
- This light blue distance is the same, the distance from this
- vertex or from this leftmost point of this rightward opening
- hyperbola to this focus is the same as this distance.
- Because a hyperbola is symmetric around the origin or
- the focal length is the same on either side of the center of
- the hyperbola depending on how you may view it, but I think
- that's not too much of a stretch of a statement for
- you to for you to accept.
- So if this distance is the same as this distance, then the
- magenta distance minus this blue distance is going to be
- equal to this green distance.
- And this green distance is what?
- That's 2a.
- We saw that at the beginning of this video.
- So this, once again, is also equal to 2a.
- Anyway, I'll leave you there right now.
- Actually, let's actually just do one problem, just because
- I like to make one concrete.
- Because I told you at the beginning that if you wanted to
- find the-- so if you have an ellipse-- so if you have-- this
- is an ellipse, x squared over a squared plus y squared over b
- squared is equal to 1, we learned that the-- that's over
- b squared-- this is an ellipse.
- We learned that the focal length is equal to the
- square root of a squared minus b squared.
- Now for a hyperbola, you kind of see that there's a very
- close relation between the ellipse and the hyperbola,
- but it is kind of a fun thing to ponder about.
- And a hyperbola's equation looks like this.
- x squared over a squared minus y squared over b squared, or it
- could be y squared over b squared minus x squared over
- a square is equal to 1.
- It turns out, and I'll prove this to you in the next video,
- it's a little bit of a hairy math problem, that the focal
- length of a hyperbola is equal to the square root of the sum
- of these two numbers, is equal to the sum of a squared
- plus b squared.
- So if I were to give you-- so notice the difference.
- It's just a difference in sign.
- You're taking the difference of those two denominators, and now
- you're taking the sum of the two denominators.
- So if I were to give you the following hyperbola.
- x squared over 9 plus y squared over 16 is equal to 1.
- Well, the first thing you do is-- though I've never-- well,
- we could just figure out the focal length just by
- plugging into the formula.
- The focal length is equal to the square root of a
- squared plus b squared.
- This is squared, right? a is three.
- b is 4.
- So 9 plus 16 is 25, which is equal to 5.
- And so if we were to graph this-- that's my y-axis, that's
- my x-axis-- and the focal length is the distance to, in
- this case, to the left and the right of the origin.
- If it was kind of an up and down opening hyperbola, it
- would be above and below the origin, so this is a-- oh
- sorry, this should be a plus.
- We're doing with a hyperbola, that should be a minus.
- Don't want to confuse you.
- What I had written before, with a plus, that would
- have been an ellipse.
- A minus is the hyperbola.
- So the two asymptotes-- this is centered at the origin, it
- hasn't been shifted-- are going to be 16 over 9, so it's going
- to be fairly steep asymptotes, it's going to look something
- like that and that, those are the two asymptotes--
- The two vertex points are at 2 times a. a is three, right?
- a squared is equal to 9, b squared is equal to 16, so this
- is the center, so the two vertex points, this is 3 minus
- 3, and then the focal points are going to be at, from the
- center, 5 to the right, so that's going to be right here,
- so that's 5,0, and minus 5,0.
- This is minus 3, and this is 3, so if we were to graph it, it
- would look something like this.
- There you go.
- And if you were to take an arbitrary point on that
- hyperbola and take this distance and subtract that from
- that distance, that will be a constant number that would be
- exactly equal to 2a or exactly equal to 6, in this
- particular example.
- Anyway.
- Hopefully I didn't confuse you too much with that little sign
- error near the end of the video, but in the next video,
- I'll prove to you this formula, which is a little bit of hairy
- algebra but it's fun to do regardless.
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