Multivariable Calculus, Interactive EditionXYZ Homework Edition

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1.3 Quadric Surfaces and Their Traces

Quadric surfaces are the graphs of second-degree equations in xx, yy, and zz — 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=cz = c and you get horizontal traces; fix xx or yy and you get vertical ones. The traces are ordinary conics, and the way the family changes as cc varies is the surface's fingerprint.

Our specimen is the hyperbolic paraboloid

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

the saddle. In the figure below it comes with a movable slicing plane: a sheet of level curves whose height you can drag, so the whole trace family plays like a film.

A translucent saddle-shaped surface (hyperbolic paraboloid z = x squared over 4 minus y squared over 9) with a horizontal plane of 15 dark level curves passing through it at z = 0: two families of hyperbolas separated by a pair of straight lines crossing at the origin. A hidden wireframe hyperboloid of one sheet waits to be revealed as a second quadric.Explore in 3D (opens in a new tab)
The saddle z=x2/4y2/9z = x^2/4 - y^2/9 with a draggable plane of level curves, opening at height z=0z = 0. A second surface hides in the object list — leave it hidden until the last exploration step.

Explore

  1. Orbit to look straight down the zz-axis so the 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?
  2. Predict the shape of the trace at z=1z = 1 and at z=1z = -1: circle, ellipse, parabola, hyperbola, or lines? Which way does each open?
  3. Select the slicing plane and drag its height slowly from 0 up toward 1, then down through 0 to 1-1. Watch the cut flip from one hyperbola family to the other. At exactly which height is the cut not a hyperbola at all?
  4. Vertical traces next: predict what the planes y=0y = 0 and x=0x = 0 cut from the saddle, then orbit to look straight down the yy-axis and the xx-axis and read each answer off the surface's silhouette. Why does one parabola open up and the other down?
  5. Reveal the hidden wireframe surface. Its components involve cosh\cosh and sinh\sinh, and the identity cosh2usinh2u=1\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=cz = c in the saddle's equation gives x2/4y2/9=cx^2/4 - y^2/9 = c. For c>0c > 0 this is a hyperbola opening along the xx-direction; for c<0c < 0, dividing by cc flips the roles and the hyperbolas open along yy. The flip you drove through in the figure above happens at c=0c = 0, where the equation factors:

x 2 4 y 2 9 = 0 ( x 2 y 3 ) ( x 2 + y 3 ) = 0 , \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=±32xy = \pm\tfrac{3}{2}x. The vertical traces explain the name: y=0y = 0 gives the upward parabola z=x2/4z = x^2/4, and x=0x = 0 gives the downward parabola z=y2/9z = -y^2/9. Two parabolas of opposite temperament through one point — that is what a saddle is.

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.

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