Difficulty: Intermediate | Prerequisites: Completing the square, conic sections in 2D, plane equations (Section 12.4)
Quadric surfaces are the 3D analogues of conic sections (ellipses, parabolas, hyperbolas). They show up whenever a second-degree equation in x, y, z describes a surface. Knowing how to classify them by inspection and sketch them from their equation is a core skill tested on exams. This section also introduces the idea of building a surface as an equidistant locus (the set of points equidistant from a point and a plane), which produces a paraboloid.
A quadric surface is any surface described by a degree-2 equation in three variables. The six standard types (ellipsoid, hyperboloid of one/two sheets, elliptic paraboloid, hyperbolic paraboloid, cone) are identified by the signs and powers in their equation. Sketch them by slicing with coordinate planes to find cross-sections.
Quadric surface
Any surface in 3D whose equation is a second-degree polynomial in x, y, z. Think of it as the 3D version of a conic section.
Ellipsoid
x^2/a^2 + y^2/b^2 + z^2/c^2 = 1. A closed, egg-shaped surface. All cross-sections are ellipses (or circles when two denominators match).
Hyperboloid of one sheet
x^2/a^2 + y^2/b^2 - z^2/c^2 = 1. A surface that looks like a cooling tower or an hourglass that never pinches shut. It is connected (one piece) and has hyperbolic cross-sections in some planes and elliptical ones in others.
Hyperboloid of two sheets
-x^2/a^2 - y^2/b^2 + z^2/c^2 = 1. Two separate bowl-shaped pieces, one above and one below. Think of it as the surface you get when the hourglass pinches past zero and disconnects.
Elliptic paraboloid
z = x^2/a^2 + y^2/b^2 (or similar, with one variable isolated linearly). A bowl shape opening along one axis. Cross-sections perpendicular to that axis are ellipses; cross-sections containing that axis are parabolas.
Hyperbolic paraboloid (saddle surface)
z = x^2/a^2 - y^2/b^2. A saddle or Pringle-chip shape. Cross-sections in one direction are upward-opening parabolas; in the other, downward-opening.
Cone (elliptic cone)
z^2/c^2 = x^2/a^2 + y^2/b^2. Two cones meeting at the origin, tip to tip. The cross-section at z = 0 is a single point.
Cylinder (in the quadric sense)
A surface generated by translating a 2D curve along an axis. x^2 + y^2 = 64 is a circular cylinder of radius 8. If one variable is missing from the equation, the surface extends infinitely along that variable's axis.
Paraboloid of revolution
An elliptic paraboloid where a = b, so every horizontal cross-section is a circle. x^2 + y^2 + 18z = 0 describes one opening downward.
Equidistant locus
The set of all points equidistant from a given point and a given plane (or line). When the reference objects are a point and a plane, the locus is a paraboloid. In simple terms, this is the 3D version of the focus-directrix definition of a parabola.
Put the equation in standard form: collect like terms, complete the square if needed, and divide so the right side is 0 or 1.
Count the signs of the squared terms on the side with the variables:
All positive, equals 1: ellipsoid.
Two positive, one negative, equals 1: hyperboloid of one sheet.
One positive, two negative, equals 1 (or equivalently two positive, one negative, equals -1): hyperboloid of two sheets.
Two squared terms equal a first-degree term (no constant on the right): paraboloid.
Two squared terms with opposite signs equal a first-degree term: hyperbolic paraboloid.
Two positive squared terms equal one negative squared term: cone.
A missing variable means the surface is a cylinder extending along that variable's axis.
The set of all points equidistant from the point (5, 0, 0) and the plane y = 2 is found by setting the distance to the point equal to the distance to the plane.
Distance to (5, 0, 0): sqrt((x-5)^2 + y^2 + z^2).
Distance to y = 2: |y - 2|.
Setting these equal and squaring: (x-5)^2 + y^2 + z^2 = (y-2)^2.
Expanding: x^2 - 10x + 25 + y^2 + z^2 = y^2 - 4y + 4.
Simplifying: x^2 - 10x + z^2 + 4y + 21 = 0, which rearranges to x^2 - 10x + z^2 = -4y - 21.
Completing the square in x: (x-5)^2 + z^2 = -4(y - 1) + 4, i.e. (x-5)^2 + z^2 = -4(y-2). Wait, let us redo carefully: (x-5)^2 + z^2 = -4y - 21 + 25 = -4y + 4 = -4(y - 1).
This is a circular paraboloid opening in the negative y-direction, with vertex at (5, 1, 0).
The standard form is x^2/a^2 + y^2/b^2 - z^2/c^2 = 1.
The minimum cross-section (the "waist") occurs where z gives the smallest radius. For the standard form with the z-axis vertical, the waist is at z = 0, where the cross-section is the ellipse x^2/a^2 + y^2/b^2 = 1.
If the base has diameter 80 m and the waist has diameter 40 m at height 30 m, place the origin at the waist: z = 0 at the waist, z = -30 at the ground.
At the waist (z = 0): the cross-section is a circle of radius 20, so a = b = 20.
At the ground (z = -30): the cross-section is a circle of radius 40, so 40^2/20^2 + 0 - (-30)^2/c^2 ... actually we use x^2/400 + y^2/400 - z^2/c^2 = 1, and at z = -30 the radius is 40, so 40^2/400 - 900/c^2 = 1, giving 4 - 900/c^2 = 1, so c^2 = 300, c = 10sqrt(3).
The equation is x^2/400 + y^2/400 - z^2/300 = 1.
A remarkable property: through every point on a hyperboloid of one sheet, there pass exactly two straight lines that lie entirely on the surface. These are called rulings.
To find one through (0, b, 0) = (0, 20, 0), parametrise as (x, y, z) = (0, 20, 0) + t(l, m, n) and substitute into the surface equation, then choose (l, m, n) so the equation holds for all t.
Set each variable to a constant in turn to find cross-sections (traces).
z = k gives the horizontal cross-section at height k. x = k and y = k give vertical cross-sections.
Identify the shape of each trace (circle, ellipse, parabola, hyperbola, or lines).
Note how the size of the trace changes as k changes: this tells you whether the surface opens outward, closes in, or stays constant.
Mark key points: vertices, intercepts, the waist of a hyperboloid, the vertex of a paraboloid.
Connect the traces into a 3D sketch.
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \quad \text{(ellipsoid)}\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \quad \text{(hyperboloid of one sheet)}-\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \quad \text{(hyperboloid of two sheets)}z = \frac{x^2}{a^2} + \frac{y^2}{b^2} \quad \text{(elliptic paraboloid)}z = \frac{x^2}{a^2} - \frac{y^2}{b^2} \quad \text{(hyperbolic paraboloid)}\frac{z^2}{c^2} = \frac{x^2}{a^2} + \frac{y^2}{b^2} \quad \text{(cone)}Cooling towers at power plants are built as hyperboloids of one sheet because the shape is structurally strong (doubly ruled, so it can be built from straight steel beams) and aerodynamically efficient for drawing air upward. Satellite dishes and car headlight reflectors are paraboloids of revolution: they focus incoming parallel signals (or outgoing light) to a single point at the focus.
Students frequently confuse the hyperboloid of one sheet with the hyperboloid of two sheets. One sheet has the form "two positive, one negative = 1" and is connected. Two sheets has "one positive, two negative = 1" (or equivalently two positive, one negative = -1) and is disconnected.
Students sometimes forget that a "cylinder" in this context does not have to be circular. x^2 + z^2 = 9 is a circular cylinder, but y^2 - z^2 = 1 is a hyperbolic cylinder. The key feature is that one variable is missing from the equation.
When sketching, students often skip the step of finding cross-sections and try to draw from memory. The homework explicitly warns that sketching from equations is an exam skill. Take the cross-sections.
The equidistant-locus derivation requires squaring both sides. Students sometimes forget to expand correctly or lose signs. Write out every step.
⚠️ Sketching a surface from its equation is explicitly flagged in this homework ("hint hint") as an exam-relevant skill. Practise identifying the surface type, finding traces, and drawing.
⚠️ The equidistant-locus problem (point vs. plane) is a classic derivation question. Know the setup: set sqrt(...) = |...|, square, simplify.
⚠️ Hyperboloid of one sheet problems (cooling tower) combine equation-finding, sketching, and the ruling-line property. All three sub-parts are fair game.
⚠️ Stewart Section 12.6 exercises 23 to 30 are assigned. These are identification and matching problems for quadric surfaces.
True or false: x^2/4 + y^2/9 - z^2/16 = 1 is a hyperboloid of two sheets. (False: it is a hyperboloid of one sheet, since two terms are positive and one is negative.)
What type of surface is z = x^2 + y^2? (Elliptic paraboloid, specifically a paraboloid of revolution.)
If the equation contains x and z but not y, the surface is a ______ extending along the ______ axis. (Cylinder; y-axis.)
True or false: through every point on a hyperboloid of one sheet, exactly one straight line on the surface passes. (False: exactly two.)
The set of points equidistant from a point and a plane is a ______. (Paraboloid.)
Q: Find the equation of the surface consisting of all points equidistant from (5, 0, 0) and the plane y = 2. Identify the surface type.
A: Set sqrt((x-5)^2 + y^2 + z^2) = |y - 2|. Squaring: (x-5)^2 + y^2 + z^2 = (y-2)^2. Expand and simplify: (x-5)^2 + z^2 = -4y + 4, i.e. (x-5)^2 + z^2 = -4(y - 1). This is a circular paraboloid opening in the -y direction with vertex at (5, 1, 0).
Q: A cooling tower has equation x^2/a^2 + y^2/b^2 - z^2/c^2 = 1 with circular cross-sections. The base diameter is 80 m, the waist diameter is 40 m, and the waist is 30 m above the ground. Find a, b, and c (placing the origin at the waist).
A: Circular cross-sections mean a = b. At z = 0 (waist), radius = 20, so a = b = 20. At z = -30 (ground), radius = 40: 1600/400 - 900/c^2 = 1, giving 3 = 900/c^2, so c^2 = 300 and c = 10sqrt(3). The equation is x^2/400 + y^2/400 - z^2/300 = 1.
Q: Identify the surface: 4x^2 - y^2 + 4z^2 = -4.
A: Divide by -4: -x^2 + y^2/4 - z^2 = 1. One positive term and two negative terms equal 1. This is a hyperboloid of two sheets, opening along the y-axis.
Q: What are the traces of the surface z = x^2/9 + y^2/4 in the planes z = 1 and y = 0?
A: z = 1: x^2/9 + y^2/4 = 1, an ellipse. y = 0: z = x^2/9, an upward-opening parabola in the xz-plane.
Quadric surfaces extend the conic sections you studied in Calculus I/II. The traces (cross-sections) of quadric surfaces are themselves conics, so the classification skills transfer directly. In Chapter 14, you will find tangent planes to these surfaces using partial derivatives. The equidistant-locus idea connects to the focus-directrix definition of conics and reappears in optics and physics.
Tags: quadric surface, ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic paraboloid, hyperbolic paraboloid, saddle surface, cone, cylinder, paraboloid of revolution, traces, cross-sections, equidistant locus, focus-directrix, cooling tower, rulings, doubly ruled surface, MATH 215, Calculus III, Stewart Section 12.6