In space, one equation in $x$, $y$, $z$ usually describes a surface. The second-degree ones are called quadric surfaces, the 3-D cousins of the conics you learned to identify from their equations. The trick for picturing them: slice. Setting one variable to a constant gives a curve in that plane, called a trace, and the traces are conics you already know.
Key idea
| Equation | Surface | Traces |
|---|---|---|
| $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ | ellipsoid | all ellipses |
| $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ | hyperboloid of one sheet | ellipses across the $z$-axis, hyperbolas along it |
| $-\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ | hyperboloid of two sheets | no trace for $|z| < c$; ellipses beyond |
| $\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z^2}{c^2}$ | elliptic cone | ellipses; lines through the vertex |
| $z = \frac{x^2}{a^2} + \frac{y^2}{b^2}$ | elliptic paraboloid | ellipses and parabolas |
| $z = \frac{y^2}{b^2} - \frac{x^2}{a^2}$ | hyperbolic paraboloid (saddle) | parabolas opening up and down; hyperbolas |
Why it works
Two more situations come up often. If one variable is missing, the surface is a cylinder: $x^2 + y^2 = 4$ has the same circle as its trace at every height, so it is a round pipe around the $z$-axis, and $z = y^2$ is a parabola swept along the $x$-axis, a trough. If the equation has first-degree terms in a squared variable, complete the square to find the center: $x^2 + y^2 + z^2 - 4y = 0$ becomes $x^2 + (y - 2)^2 + z^2 = 4$, a sphere of radius ${}2$ centered at $(0, 2, 0)$. A good routine: complete the squares, divide so the constant is ${}1$ (if there is one), count the squared terms and their signs, and check one or two traces to be sure.
Worked examples
Worked example: Divide to reach standard form
- Divide by 4: $x^2 + \frac{y^2}{4} - \frac{z^2}{4} = 1$.
- Two plus signs, one minus, right side $+1$: a hyperboloid of one sheet.
- The minus sign is on $z$, so its axis is the $z$-axis: traces $z = k$ are the ellipses $x^2 + \frac{y^2}{4} = 1 + \frac{k^2}{4}$.
Answer: hyperboloid of one sheet around the $z$-axis
Worked example: A saddle
- $z$ appears to the first power and the squares have opposite signs: a hyperbolic paraboloid.
- $x = k$: $z = y^2 - k^2$, parabolas opening up. $y = k$: $z = k^2 - x^2$, parabolas opening down.
- $z = k$: $y^2 - x^2 = k$, hyperbolas (two crossing lines when $k = 0$).
Answer: hyperbolic paraboloid
Worked example: A missing variable
- $y$ does not appear, so every trace $y = k$ is the same circle $x^2 + z^2 = 9$.
- Copying a circle along the $y$-axis gives a circular cylinder of radius 3.
Answer: a cylinder around the $y$-axis
Common mistake