Source: Homework Set 2, §12.6
Tags: quadratic surfaces, ruled surfaces, hyperboloid, hyperbolic paraboloid, ellipsoid, paraboloid, conic sections, locus, equidistant
Difficulty: Intermediate to Challenging Prerequisites: Conic sections in 2D (circles, ellipses, parabolas, hyperbolas), completing the square, equations of planes (§12.5).
Quadratic surfaces are the 3D analogues of conic sections. Just as circles, ellipses, parabolas and hyperbolas are defined by second-degree equations in two variables, quadratic surfaces are defined by second-degree equations in three variables. Learning to recognise and classify these surfaces is essential for later work on surface integrals and flux. This section also introduces ruled surfaces, which have the remarkable property that a straight line passes through every point on the surface.
Quadratic surfaces are classified by looking at the signs and structure of the squared terms: ellipsoids have all positive squares, hyperboloids have mixed signs, paraboloids have one linear term. Ruled surfaces (like hyperboloids of one sheet and hyperbolic paraboloids) contain entire straight lines. Locus problems (equidistant from a point and a plane) produce these surfaces when you set up and simplify the distance equations.
Quadratic surface (quadric)
A surface in 3D defined by a second-degree polynomial equation in x, y, z. Think of it as the 3D version of a conic section.
Ellipsoid
x²/a² + y²/b² + z²/c² = 1. All cross-sections are ellipses. In simple terms, a stretched sphere.
Hyperboloid of one sheet
x²/a² + y²/b² – z²/c² = 1. One connected piece, shaped like a cooling tower. Contains straight lines on its surface.
Hyperboloid of two sheets
–x²/a² – y²/b² + z²/c² = 1. Two separate pieces (bowls facing away from each other).
Elliptic paraboloid
z = x²/a² + y²/b². A bowl that opens upward (or downward with a minus sign). Cross-sections parallel to the base are ellipses.
Hyperbolic paraboloid (saddle surface)
z = x²/a² – y²/b². A saddle shape, curving up in one direction and down in the other. This is a ruled surface.
Cone
z² = x²/a² + y²/b². The degenerate case between a hyperboloid of one sheet and two sheets.
Ruled surface
A surface through every point of which there passes at least one straight line that lies entirely on the surface. In simple terms, you could sweep a ruler to trace it out. Hyperboloids of one sheet and hyperbolic paraboloids are ruled surfaces.
Locus
The set of all points satisfying a given geometric condition (e.g. equidistant from a point and a plane).
The standard approach: put the equation in standard form (complete the square if needed), then compare with the known types.
Count the squared terms and check signs:
All three positive with = 1 on the right: ellipsoid.
Two positive, one negative, = 1: hyperboloid of one sheet.
One positive, two negative, = 1 (or equivalent): hyperboloid of two sheets.
Two squares equal to a linear term: elliptic paraboloid.
Two squares with opposite signs equal to a linear term: hyperbolic paraboloid.
Two squares equal to one square: cone.
Cross-sections (traces) help confirm: slice with planes parallel to coordinate planes and identify the resulting conic.
A ruled surface contains at least one straight line through every point.
For z = x² – y² (a hyperbolic paraboloid), factor: z = (x – y)(x + y). A line through (x₀, y₀, z₀) on the surface can be found by parameterising along a direction that keeps both factors linear.
For x² + y² – z² = c (a hyperboloid of one sheet), the line must satisfy the surface equation identically. Substitute r(t) = P + t v into the equation and require it to hold for all t, then solve for v.
Example (Q3a): on z = x² – y² through (2, 0, 4). Write the line as (2 + at, bt, 4 + ct). Substituting: 4 + ct = (2 + at)² – (bt)². Expanding: 4 + ct = 4 + 4at + a²t² – b²t². Matching coefficients: c = 4a and a² – b² = 0, so a = ±b. One solution: a = b = 1, c = 4, giving the line (2 + t, t, 4 + 4t).
"Find the surface of points equidistant from a point and a plane" is a classic definition of a paraboloid.
Set distance to point = distance to plane, square both sides, expand, and simplify.
Example (Q4a): equidistant from (0, 2, 0) and y = –3. Distance to point: √(x² + (y–2)² + z²). Distance to plane: |y + 3|. Setting equal and squaring: x² + (y–2)² + z² = (y+3)². Expanding: x² + y² – 4y + 4 + z² = y² + 6y + 9. Simplifying: x² + z² = 10y + 5, or y = (x² + z² – 5)/10. This is an elliptic paraboloid opening in the y-direction.
"Distance to A is always k more than distance to B" defines a hyperboloid (3D) or hyperbola (2D).
Example (Q4b): |r – (1,0,0)| = |r – (–1,0,0)| + 1. The constant difference in distances to two foci defines a hyperboloid of two sheets (with the foci at (±1, 0, 0) and the difference equal to 1).
Given a parametric curve r(t), to find a quadratic surface it lies on, eliminate t from the components.
Use trigonometric identities (cos²t + sin²t = 1, double-angle formulas) to combine the components into a surface equation.
Example (Q5a): r₁(t) = ⟨cos t, cos(2t)sin t, sin(2t)sin t⟩. Note x = cos t, and y² + z² = sin²t(cos²(2t) + sin²(2t)) = sin²t. So x² + y² + z² = cos²t + sin²t = 1. The curve lies on the unit sphere.
\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: } z = \frac{x^2}{a^2} + \frac{y^2}{b^2}\text{Hyperbolic paraboloid: } z = \frac{x^2}{a^2} - \frac{y^2}{b^2}\text{Cone: } \frac{z^2}{c^2} = \frac{x^2}{a^2} + \frac{y^2}{b^2}\text{Equidistant locus: } \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2} = |ax+by+cz-d|/\sqrt{a^2+b^2+c^2}Cooling towers at power stations are hyperboloids of one sheet, chosen because the ruled-surface property means they can be built from straight steel beams arranged at angles. Satellite dishes and car headlamp reflectors are paraboloids, exploiting the focus property to concentrate or direct signals. Saddle surfaces (hyperbolic paraboloids) appear in modern architecture, notably in roof designs like the Pringle-shaped canopies over stadiums.
Students confuse hyperboloids of one sheet and two sheets. One sheet has the form “two positive, one negative = 1” and is connected. Two sheets has “one positive, two negative = 1” (or equivalent) and has two separate pieces.
Assuming that a cone is always z² = x² + y². The cone can open along any axis. The key feature is that the right-hand side is zero (not 1), so it passes through the origin.
Forgetting to complete the square before classifying. An equation like x² + y² + z² – 4y = 0 looks unfamiliar until you rewrite it as x² + (y–2)² + z² = 4, which is a sphere.
When finding lines on a ruled surface, students sometimes try to find all lines at once. You only need one line through the given point.
⚠️ Surface classification is guaranteed exam material. Practise until you can identify the type from an equation in under 30 seconds.
⚠️ The equidistant-locus technique (set distance to point = distance to plane, square both sides) appears in multiple problem sets. The algebra is mechanical but easy to mess up; be careful expanding squared binomials.
⚠️ Ruled-surface problems test whether you understand that a line through a point on the surface must satisfy the surface equation for all parameter values, not just at the point.
True or False: x² + y² + z² = 9 is an ellipsoid. (True, specifically a sphere.)
Fill in the blank: A hyperboloid of one sheet has ______ connected piece(s). (one)
True or False: A hyperbolic paraboloid is a ruled surface. (True)
Fill in the blank: The set of points equidistant from a point and a plane is a(n) ______. (paraboloid)
True or False: Every cone passes through the origin. (True, in standard position.)
Q: Classify the surface x² + 4y² – z² = 4.
A: Divide by 4: x²/4 + y² – z²/4 = 1. Two positive squared terms, one negative, equals 1. This is a hyperboloid of one sheet.
Q: Find a line on z = x² – y² through the origin.
A: At the origin, z = 0. Try the line (t, t, 0). Check: z = 0 and x² – y² = t² – t² = 0. It works for all t.
Q: Find the surface of points equidistant from (0, 0, 1) and the plane z = –1.
A: √(x² + y² + (z–1)²) = |z + 1|. Squaring: x² + y² + z² – 2z + 1 = z² + 2z + 1. Simplifying: x² + y² = 4z. This is a circular paraboloid (a special elliptic paraboloid with a = b).
Q: The parametric curve r(t) = ⟨2 + 8t², 2t, 2t⟩ lies on what quadratic surface?
A: Let x = 2 + 8t², y = 2t, z = 2t. Note y = z, and t = y/2. Then x = 2 + 8(y/2)² = 2 + 2y². So x – 2 = 2y², and since y = z, this gives x = 2 + 2y², or also x = 2 + 2z². The curve lies on both parabolic cylinders, and also on the plane y = z.
Quadratic surfaces reappear throughout the rest of the course. Tangent planes to surfaces (§14.4) generalise the tangent-line idea to these shapes. Surface area and surface integrals (§16.6–16.7) require you to parameterise patches of these surfaces. The classification skill also helps in optimisation (§14.7–14.8), where the nature of a critical point depends on whether the local surface looks like a bowl (paraboloid) or a saddle.
Related Terms / Search Tags
quadratic surface, quadric surface, ellipsoid, hyperboloid one sheet, hyperboloid two sheets, elliptic paraboloid, hyperbolic paraboloid, saddle surface, cone, ruled surface, lines on surface, locus, equidistant point plane, completing the square 3D, surface classification, MATH 215 surfaces, Calc III quadric