# Quadric Surface: The Hyperbolic Paraboloid

Welcome to a blog on graphing the hyperbolic paraboloid. As mentioned in the previous artcile, to graph a quadric surface it’s often helpful to look at the XY, XZ, and YZ traces. Which is where the surface intersects each of these three planes. And that’s what we’ll do in this post. We’re going to determine the XY trace by setting Z equal to zero. Then we’ll find the XZ trace by setting Y equal to zero. And then, lastly, we’ll find the YZ trace by setting X equal to zero. Remember, when we have a quadric surface, the form of the equation is what tells us what type of a quadric surface we have. Notice for a hyperbolic paraboloid, we have two terms of degree-two where one of them is positive and one of them is negative. And we also have one degree-one term.

When we take a look at the traces, one trace will be a hyperbola, and then we’ll have two traces that are parabolas. And the axis of the surface will be parallel to the degree-one variable. Let’s go and take a look at an example. Here we want to graph Z equals X squared divided by four, minus Y squared divided by two. This equation fits the pattern of a hyperbolic paraboloid, so we know what the general shape will be. Let’s go ahead and determine the three traces. So to determine the XY trace, we’ll set Z equal to zero.

So we’ll have zero equals X squared over four, minus Y squared over two. I’m going to go ahead and move the Y squared term to the left side of the equation so we’ll have Y squared over two equals equals X squared over four Now we’re going to go ahead and solve this equation for Y so we’ll multiply both sides by two That’ll give us Y squared equals If we multiply the right side by two, we’re going to have two in the numerator and four in the denominator That’ll give us one-half X squared Now we’ll take the square root of both sides, and we have Y equals plus or minus one over the square root of two X.

So here, we have two lines One line that has a positive slope of one over square root two The other line has a slope of negative one over square root two. Both pass through the origin. So, here’s the origin Square root two is approximately one point four So for the positive slope, we’ll go up one and then right one point four Notice this is scaled by two’s. So the first line will go up one and over maybe about here. And there’s the line with a positive slope. And then for the second line, we’ll go down one and right one point four. And here’s the second line. Now even though this graph is linear, this trace does fit the form of a hyperbola And we’ll take a look at another trace that’s parallel to this axis after we take a look at these other two traces. To determine the XZ trace, we’ll set Y equal to zero. So we’ll have Z, X squared divided by four This will be a parabola. And again we’ll call this the X axis and now this will be the Z axis. So we know it’s a parabola where the vertex is at the origin. We also know this opens up because the leading coefficient is positive. And then lastly, when X is equal to two we’ll have two squared which is four divided by four. Z will be one.

So, again, this is scaled by two’s so we’ll go right two and up one for that point. And then X is negative two, we’ll have the same Z coordinate. So here’s the XZ trace. Now to determine the YZ trace, we’re going to set X equal to zero. That’ll give us Z equals negative Y squared over two. Again this will be another parabola. But now it’ll open down. We’ll call this the Y axis, this the Z axis. Passes through the origin. So now when Y is equal to two, we’ll have two squared divided by two, that’s gonna give us two. And then it will be negative. And then when Y is negative two, we’ll have the same value for Z So we have another parabola which opens down And it’s also narrower. So, putting these three traces together, we should be able to sketch a three-dimensional surface. But before we do that, let’s go ahead and look at one more XY trace. Because as we said before, a hyperbolic paraboloid should have two traces that are parabolas and one that’s a hyperbola. So for this next trace, let’s go ahead and set Z equal to one. This plane would be one unit above the XY plane. So if we let Z equal one We would have one equals X squared over four. minus Y squared over two And this fits the form of a hyperbola that we’re used to. Here we have A squared is equal to four and B squared equals two. So this hyperbola has a major axis that’s horizontal It’s centered at the origin, and since A squared equals four that means A would be two. So from the origin, we’ll go right two units and left two units. And we’re going to make this rectangle so we can sketch the asymptotes from the diagonals.

Next, B squared is equal to two, so B is equal to the square root of two. So from the origin, we’ll go up one point four units, or approximately. And then down one point four units. We can use these four points to sketch the rectangle. It aids us in constructing the hyperbola. Next, the diagonals will be our asymptotes. So the sketch of this trace, which is a hyperbola. Notice the horizontal axis is the major axis, so it’ll look like this. Now, instead of trying to make a surface by hand, we’ll go ahead and use Maple to get a better sketch. Here’s a still of the surface. But let’s go ahead and take a look at a dynamic graph of this. Now if we take a look at just the three planes, we can see the three traces. If we look down on the XYplane you can see the hyperbolic trace. Now if we take a look at the XZ trace in this direction here. You can see the parabolic trace, or the parabola. And if we take a look at the YZ trace you can see the parabolic trace, that opens down. So putting those three traces together, we could try to sketch this by hand but you get a much better feel of the surface looking at it with graphing software. Okay, that’s going to do it for this article. I hope you found this example helpful, thank you for watching.

#### Professor Cram

Professor Lawrence Cram is a Visiting Fellow at the Australian National University working in the Department of Applied Mathematics. His interests include astronomy, mathematics, engineering, computing, and physics. Due to his extensive expertise, professor Cram has worked as a Professor of Physics at the University of Sydney and as the Deputy Vice-Chancellor at ANU during 2004-2012. In 2013, he retired as a Master, University House and Graduate House. In January 2014, he was appointed as an acting Deputy Vice-Chancellor at Charles Darwin University. Professor Cram is also a Fellow at the Royal Astronomical Society and the Australian Institute of Physics.
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