Difficulty: Intermediate | Prerequisites: §12.5 (lines and planes), completing the square, conic sections from Calculus I/II
Quadric surfaces are the three-dimensional analogues of conic sections. Just as circles, ellipses, parabolas and hyperbolas are defined by second-degree equations in two variables, quadric surfaces are defined by second-degree equations in three variables. This section catalogues the standard forms and teaches you to identify a surface from its equation by examining traces (cross-sections). If you are comfortable with conic sections and the §12.5 material on planes, you are ready.
A quadric surface is any surface whose equation is a second-degree polynomial in x, y and z. There are six standard types. You identify them by looking at the signs and structure of the squared terms and by examining traces in coordinate planes. Many applied problems (cooling towers, satellite dishes, lenses) reduce to fitting a quadric surface to geometric constraints.
Quadric surface
Any surface in 3D whose equation is a polynomial of degree two in x, y and z. Think of it as the 3D version of a conic section.
Trace
The curve you get by intersecting a surface with a plane (usually a coordinate plane like z = k). In simple terms, it is a cross-sectional slice. Traces are the main tool for identifying and sketching quadric surfaces.
Ellipsoid
x²/a² + y²/b² + z²/c² = 1. A closed surface where every trace is an ellipse (or a circle, if two denominators match). Think of a squashed or stretched sphere.
Hyperboloid of one sheet
x²/a² + y²/b² - z²/c² = 1 (one negative term). A connected surface that looks like a cooling tower or a waist. Horizontal traces are ellipses; vertical traces are hyperbolas.
Hyperboloid of two sheets
-x²/a² - y²/b² + z²/c² = 1 (two negative terms). Two separate bowl-shaped pieces. Horizontal traces are ellipses, but only for |z| > c.
Elliptic paraboloid
z = x²/a² + y²/b². A bowl opening upward (or downward if the sign is flipped). Horizontal traces are ellipses; vertical traces are parabolas.
Hyperbolic paraboloid
z = x²/a² - y²/b². A saddle shape. Traces in one vertical direction are upward-opening parabolas; in the other, downward-opening parabolas. Also called a "saddle surface" or "Pringles shape."
Elliptic cone
x²/a² + y²/b² = z²/c². The boundary case between one-sheet and two-sheet hyperboloids. Horizontal traces are ellipses (except at z = 0, which is a point); vertical traces are pairs of lines through the origin.
Write the equation in standard form by completing the square if needed.
Count the number of squared terms with positive and negative signs.
All positive, equals a constant: ellipsoid.
Two positive, one negative, equals a positive constant: hyperboloid of one sheet.
One positive, two negative, equals a negative constant (equivalently two negative, one positive = 1): hyperboloid of two sheets.
Two positive squared terms equal to a first-degree term: elliptic paraboloid.
One positive and one negative squared term equal to a first-degree term: hyperbolic paraboloid.
Two positive squared terms equal to one positive squared term: cone.
Examine traces (set x = k, y = k, or z = k) to confirm and to sketch.
The set of all points equidistant from a point and a plane forms a paraboloid.
Method: let the point be F and the plane be L. Set distance(P, F) = distance(P, L), square both sides, expand, and simplify.
Example: equidistant from (5, 0, 0) and the plane x = -1. Setting √((x-5)² + y² + z²) = |x + 1|, squaring gives (x-5)² + y² + z² = (x+1)². Expanding: x² - 10x + 25 + y² + z² = x² + 2x + 1. Simplifying: y² + z² = 12x - 24, or y² + z² = 12(x - 2). This is a circular paraboloid opening in the positive x-direction, with vertex at (2, 0, 0).
A hyperboloid of one sheet x²/a² + y²/b² - z²/c² = 1 has circular cross-sections when a = b.
The minimum radius occurs at z = 0, where the radius equals a.
At height z = h, the radius is a√(1 + h²/c²).
To find parameters from physical constraints: use the known radii at known heights.
Example (cooling tower): circular base diameter 100 m, minimum diameter 60 m at height 120 m. Place the origin at the narrowest point, so a = b = 30 (minimum radius). At z = -120 (the base), the radius is 50. Then 50² = 30²(1 + 120²/c²), giving 2500/900 = 1 + 14400/c², so c² = 14400/(25/9 - 1) = 14400/(16/9) = 8100, hence c = 90.
A remarkable property: a hyperboloid of one sheet is a ruled surface, meaning straight lines lie entirely on it.
To find a line through (a, 0, 0) on x²/a² + y²/b² - z²/c² = 1: write (x, y, z) = (a, 0, 0) + t(d, e, f) and substitute into the hyperboloid equation. Since (a, 0, 0) satisfies the equation, the constant term is zero. Choose d, e, f so the linear and quadratic terms in t both vanish.
This yields a system from which you can find two families of lines (one for each sign choice).
\text{Ellipsoid: } \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\text{Hyperboloid of one sheet: } \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\text{Hyperboloid of two sheets: } -\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\text{Elliptic paraboloid: } \frac{z}{c} = \frac{x^2}{a^2} + \frac{y^2}{b^2}\text{Hyperbolic paraboloid: } \frac{z}{c} = \frac{x^2}{a^2} - \frac{y^2}{b^2}\text{Cone: } \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z^2}{c^2}\text{Radius at height } z \text{ on a hyperboloid (circular, } a = b\text{): } r(z) = a\sqrt{1 + z^2/c^2}Cooling towers at power plants are hyperboloids of one sheet: the shape is structurally strong and allows natural air circulation. Satellite dishes and car headlight reflectors are paraboloids, chosen because a parabolic cross-section focuses parallel rays to a single point. The saddle shape of a hyperbolic paraboloid appears in modern architecture (roofs of sports arenas) because it distributes loads efficiently.
Students often confuse the hyperboloid of one sheet with the hyperboloid of two sheets. The one-sheet version has one negative sign and is a single connected surface. The two-sheet version has two negative signs and consists of two separate pieces.
A common mistake is forgetting to complete the square before trying to identify the surface type. The equation x² + y² + z² - 4x + 6z = 0 is a sphere, not an ellipsoid with a stray linear term.
Students sometimes assume a cone has an opening that curves. It does not. The sides of a cone are straight lines, which is why it is called a ruled surface.
The hyperbolic paraboloid is neither a hyperboloid nor a paraboloid in the usual sense. It is its own distinct type. Do not try to force it into one of the other categories.
⚠️ Identifying a quadric surface from a given equation is a near-certain exam question. Practise until you can classify by inspection.
⚠️ Completing the square to put an equation in standard form is tested often. A surface that looks unfamiliar is usually just a shifted or rotated standard form.
⚠️ Trace-sketching (setting one variable to a constant and identifying the resulting conic) is the expected method for showing your work when asked to "describe and sketch."
⚠️ Applied problems (cooling towers, reflectors, lenses) test whether you can set up a quadric equation from physical dimensions. Know how to place the origin strategically.
True or false: the equation x² + 4y² + 9z² = 36 represents an ellipsoid.
Fill in the blank: horizontal traces of a hyperboloid of one sheet are ______.
True or false: the equation z = x² - y² represents an elliptic paraboloid.
Fill in the blank: the minimum radius of the hyperboloid x²/a² + y²/a² - z²/c² = 1 occurs at z = ______.
True or false: a cone is a ruled surface (straight lines lie on it).
Answers: 1. True (divide through by 36 to get standard form). 2. Ellipses. 3. False (it is a hyperbolic paraboloid). 4. z = 0. 5. True.
Q: Find an equation for the surface consisting of all points equidistant from (5, 0, 0) and the plane x = -1. Identify and sketch the surface.
A: Set the distance from P = (x, y, z) to (5, 0, 0) equal to the distance from P to x = -1. Squaring: (x-5)² + y² + z² = (x+1)². Expand both sides and simplify to get y² + z² = 12(x - 2). This is a circular paraboloid with vertex at (2, 0, 0), opening in the positive x-direction.
Q: Describe and sketch the part of the first octant where x + 2y + 3z ≤ 6.
A: This is a tetrahedron (triangular pyramid) with vertices at the origin and the intercepts (6, 0, 0), (0, 3, 0) and (0, 0, 2). The boundary is the plane x + 2y + 3z = 6 and the three coordinate planes. Every point inside satisfies all four inequalities.
Q: A cooling tower has the shape of a hyperboloid of one sheet. The base diameter is 100 m, and the minimum diameter (120 m above the base) is 60 m. Place the origin at the narrowest point and find a, b, c.
A: With the origin at the waist, a = b = 30 (half the minimum diameter). The base is at z = -120, where the radius is 50. Substituting: 50²/30² = 1 + 120²/c², so 25/9 - 1 = 14400/c², giving c² = 14400/(16/9) = 8100 and c = 90. The equation is x²/900 + y²/900 - z²/8100 = 1.
Q: Find a line that lies entirely on the hyperboloid x²/a² + y²/b² - z²/c² = 1, passing through (a, 0, 0).
A: Write (x, y, z) = (a, 0, 0) + t(d, e, f). Substitute into the hyperboloid equation. Since (a, 0, 0) satisfies it, the constant term equals 1. Collect terms in t and t². For the equation to hold for all t, both the coefficient of t and the coefficient of t² must vanish. From the t-coefficient: 2d/a = 0 gives d = 0. From the t²-coefficient: e²/b² - f²/c² = 0, so e/b = ±f/c. Choosing e = b and f = c (or f = -c) gives two lines: (a, bt, ct) and (a, bt, -ct).
Q: Identify the surface 4x² - y² + 4z² = 4.
A: Divide by 4: x² - y²/4 + z² = 1, which is x²/1 + z²/1 - y²/4 = 1. Two positive squared terms, one negative, equals 1. This is a hyperboloid of one sheet with axis along the y-direction.
This connects to §12.5 (lines and planes) because planes are the simplest quadric surfaces (degree-one equations), and several problems here build on plane equations. Quadric surfaces reappear in §15.7 and §15.8 when you set up triple integrals in cylindrical and spherical coordinates, since the integration boundaries are often quadric surfaces. The ruled-surface property of hyperboloids connects to §13.1 (vector-valued functions), where you parametrise curves and lines on surfaces.
quadric surfaces, ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic paraboloid, hyperbolic paraboloid, saddle surface, elliptic cone, traces, cross-sections, completing the square, ruled surface, cooling tower, paraboloid, standard form, Math 215, Calculus III, Stewart §12.6, multivariable calculus, conic sections in 3D