Welcome to a video on the hyperboloid of two sheets.
As we have seen from the previous videos, the degree and sign of the degree-two terms
as well as which terms are present, help us determine which of the six basic quadratic surfaces is given by an equation
To graph a quadric surface, it's often helpful to graph the XY, XZ, and YZ traces
and these are where the surface would intersect these three planes.
To determine the XY trace, we'll set Z to 0.
To determine the XZ trace, we'll set Y to 0.
To determine the YZ trace, we'll set X equal to 0.
Here's the form of an equation that represents a hyperboloid of two sheets.
Notice that all three degree-two terms are present.
However, when the equation is equal to positive 1, two of the terms will be negative and one will be positive.
Remember, when we had a hyperboloid of one sheet, only one of the terms was negative.
So if it fits this form where one is negative, it's a hyperboloid of one sheet.
If two of the terms are negative, it's a hyperboloid of two sheets.
Two of the traces will be hyperbolas, and one trace will be an ellipse
Then if it's centered at the origin, it will appear in a plane parallel to the XY plane.
And then lastly, the axis will be parallel to the positive-variable term.
Here the axis is parallel to the Z axis, and notice the Z term is positive.
Let's go ahead and take a look at our example.
We'll start by finding the XY trace, so we'll set Z equal to zero.
So we'll have X squared over sixteen
minus Y squared over nine
equals positive one.
This is a hyperbola, and because the X term is positive,
this hyperbola will open left and right along the X axis.
It is also centered at the origin,
so we know that A squared is equal to 16, so A is equal to 4.
So we're going to plot two points: one four units to the right, and one four units to the left.
Since B squared is equal to 9, B is equal to 3
We'll plot a point three units up, and three units down.
These four points allow us to construct a rectangle that will aid us in graphing the hyperbola.
Remember, the diagonals of the rectangle will be the asymptotes. So let's go ahead and sketch those.
And again, because we know this hyperbola opens left and right, it would look something like this.
Approaching the asymptotes from the vertices.
Now let's take a look at the XZ trace.
To find this trace, we'll set Y equal to zero.
So now we'll have X squared over sixteen
minus Z squared over four
equals one.
So we have another hyperbola, which is what we expect. Two hyperbolas and one ellipse.
Since the X term is positive, it will open left and right along the X axis again.
And since A squared is equal to sixteen, A is four.
So we'll plot a point four units to the right, four units to the left. Again, it's centered at the origin.
And C squared is equal to four, so C is two.
So we'll go up two units and make a point, and down two units and make a point.
This will give us our rectangle as before.
Then the diagonals.
and now we can sketch a nice hyperbola from the vertices approaching the asymptotes.
Here and here.
Now this last trace is going to be a little more interesting.
If we let X equal to zero and find the YZ trace, we're going to have negative Y squared over nine
minus Z squared over four
equals positive one.
There's no solution to this because both of these terms are going to be negative.
Therefore it will never be positive. So that tells us there is no YZ trace.
So now we're stuck with a little bit of a dilemma. We do want to have some form of a YZ trace.
We know there should be an elliptical trace, it just won't occur in the YZ plane.
But it will occur in a plane that's parallel to this plane.
What I mean by that is
If this was equal to negative one,
we would be okay because then we could multiply everything by negative one.
And then we would graph an ellipse as we normally would expect.
So going back up here where we let X equal zero, if we could find a value of X that would make this equal to positive two
then if we subtracted it on both sides, that would give us a negative one over here.
So if we want this to be a positive two, we could let X equal positive or negative square root thirty two.
Because if we square square root of thirty two, that's going to give us thirty two.
Thirty two divided by sixteen would give us two.
So that's what we're gonna do.
Let's go ahead and clear this information off.
And it's going to take a little bit more work so let's go ahead and start it over here.
We're going to let X equal the square root of thirty two.
And if we do that, it's going to give us two
minus Y squared over nine
minus Z squared over four equals positive one.
So now, if we were to move these terms to the right, they would be positive.
If we subtract one on both sides, we'd have a positive one on the left.
So what we're going to use for our trace is going to be
One equals Y squared over nine
plus Z squared over four
And, again, we subtracted the one to the other side. That's why this was positive one on the left.
So now this really isn't the YZ trace, it's the
X equals square root thirty two trace
Now the square root of thirty two, well, the square root of thirty six would be six
So this is somewhere between X equals five and X equals six.
But this will be parallel to the YZ plane, so let's go ahead and use this as our trace.
Again, we should now recognize this is an ellipse.
The major axis will be along the Y axis.
So we'll call this the Y axis, and this the Z axis.
So since B squared is equal to nine, B is three
So we'll make a point here, and a point here.
The endpoints of the major axis.
And since C squared is equal to four, C is two.
So we'll make a point up and down two units, and here is our elliptical trace.
Again, this is not the YZ trace, it's the X equals square root thirty two trace.
But it does tell us, that if we take a look at just the YZ plane, we should see an elliptical trace.
If we put all these traces together, we will produce a hyperboloid of two sheets.
And here it is. Let's go and take a look at this so we can rotate it.
If we take a look at just the XY plane
Looking down on it, it would look like this.
And you can see the hyperbolic trace here, and here.
If we now look at the XZ trace,
you can see the hyperbolic trace here, and here.
And then lastly, the one that we had some difficulty with. Looking at the YZ trace,
you can see the elliptical traces
from planes that are parallel to the YZ plane.
This is the hyperboloid of two sheets.
Ok that's going to do it for this video. I hope you found this helpful. Thank you for watching.