1.3 Quadric Surfaces and Their Traces

Quadric surfaces are the graphs of second-degree equations in $x$ , $y$ , and $z$ — the three-dimensional relatives of ellipses, parabolas, and hyperbolas. There are only a handful of shapes, but memorizing pictures of them is fragile knowledge. The durable skill is slicing: intersect the surface with a plane on which one variable is constant and identify the resulting curve, called a trace. Fix $z = c$ and you get horizontal traces; fix $x$ or $y$ and you get vertical ones. The traces are ordinary conics, and the way the family changes as $c$ varies is the surface's fingerprint.

Our specimen is the hyperbolic paraboloid

$z = \frac{x^2}{4} - \frac{y^2}{9},$

the saddle. In the figure below it comes with a movable slicing plane: a slider sets the plane's height, the figure draws the cut the plane makes in the saddle, and the whole trace family plays like a film.

A translucent saddle-shaped surface, the hyperbolic paraboloid z equals x squared over 4 minus y squared over 9, cut by a translucent horizontal plane at height z equals c, where a slider named c starts at 0. The cut the plane makes on the saddle is drawn in amber: at height 0 it is a pair of straight lines crossing at the origin, and moving the slider up turns it into a hyperbola opening along the x-direction, moving it down into a hyperbola opening along the y-direction. Above the saddle floats a flat sky-blue contour map of 9 level curves: two families of hyperbolas separated by the same pair of crossing lines. A hidden wireframe hyperboloid of one sheet can be shown in the full editor as a second quadric. — The saddle $z = x^2/4 - y^2/9$ cut by a translucent horizontal plane $z = c$ , opening at $c = 0$ ; the slider $c$ moves the plane, and the cut it makes in the saddle is drawn in amber. A sky-blue contour map of the saddle floats flat above it. A second surface starts hidden — leave it hidden until the last exploration step.

Explore

Press the xy view button (or orbit) to look straight down the $z$ -axis, so the sky-blue level curves read as a flat map. You should see two families of curved lines and, separating them, something not curved at all. Which family occupies the east–west sectors, and which the north–south?
Predict the shape of the trace at $z = 1$ and at $z = -1$ : circle, ellipse, parabola, hyperbola, or lines? Which way does each open?
Drag the slider $c$ slowly from 0 up toward 1, then down through 0 to $-1$ , and watch the amber cut: it flips from one hyperbola family to the other. At exactly which height is the cut not a hyperbola at all?
Vertical traces next: predict what the planes $y = 0$ and $x = 0$ cut from the saddle, then use the xz and yz view buttons (or orbit) to look straight down the $y$ -axis and the $x$ -axis and read each answer off the surface's silhouette. Why does one parabola open up and the other down?
Press Open in XYZ 3D ↗ under the figure and, in the editor's object list, turn on the one surface that starts hidden: a wireframe. Its components involve $\cosh$ and $\sinh$ , and the identity $\cosh^2 u - \sinh^2 u = 1$ is the key. Predict its horizontal traces before orbiting, then check: at every height the cross-section is a circle, smallest at the waist. What is this surface called?

Reading the algebra of a slice

Setting $z = c$ in the saddle's equation gives $x^2/4 - y^2/9 = c$ . For $c > 0$ this is a hyperbola opening along the $x$ -direction; for $c < 0$ , dividing by $c$ flips the roles and the hyperbolas open along $y$ . The flip you drove through in the figure above happens at $c = 0$ , where the equation factors:

$\frac{x^2}{4} - \frac{y^2}{9} = 0 \quad\Longleftrightarrow\quad \left(\frac{x}{2} - \frac{y}{3}\right)\left(\frac{x}{2} + \frac{y}{3}\right) = 0,$

a degenerate pair of crossing lines $y = \pm\tfrac{3}{2}x$ . The vertical traces explain the name: $y = 0$ gives the upward parabola $z = x^2/4$ , and $x = 0$ gives the downward parabola $z = -y^2/9$ . Two parabolas of opposite temperament through one point — that is what a saddle is.

Example 1 Classifying a mystery quadric by traces

A surface satisfies $x^2 + y^2 - z^2 = 1$ . Describe its horizontal traces and its trace in the plane $x = 0$ , and name the surface.

Show solution

Fix $z = c$ : then $x^2 + y^2 = 1 + c^2$ , a circle of radius $\sqrt{1 + c^2}$ at every height — never empty, smallest at $c = 0$ , growing as $|c|$ increases. Fix $x = 0$ : then $y^2 - z^2 = 1$ , a hyperbola opening along the $y$ -direction. Circular horizontal traces with a waist plus hyperbolic vertical traces identify a hyperboloid of one sheet — the hidden wireframe in the figure above, which the parametrization $x = \cosh u \cos v$ , $y = \cosh u \sin v$ , $z = \sinh u$ builds directly from the identity $\cosh^2 u - \sinh^2 u = 1$ .

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.