Difficulty: Intermediate to Advanced | Prerequisites: Completing the square, distance formula in 3D, standard forms of conic sections
Big picture: Quadric surfaces are the 3D analogues of conic sections (ellipses, parabolas, hyperbolas). Being able to recognise and name them from their equations is a core skill in MATH 126. Many exam problems ask you to set up an equation from a geometric condition (often involving distances) and then identify the resulting surface. This connects to later topics like level surfaces, gradients, and surface integrals.
Quadric surfaces are second-degree equations in x, y, z. Common types include ellipsoids, hyperboloids (of one or two sheets), elliptic paraboloids, hyperbolic paraboloids, and cones. Exam problems often ask you to derive an equation from a distance condition and then identify which surface it describes by matching it to a standard form.
Quadric surface
Any surface in R³ described by a second-degree polynomial equation in x, y, z. The six standard types are: ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic paraboloid, hyperbolic paraboloid, and elliptic cone.
Think of it as: the 3D version of conic sections. Just as slicing a cone gives ellipses, parabolas, and hyperbolas in 2D, second-degree equations in 3D give these six surface types.
Hyperboloid of two sheets
A surface whose standard form is −x²/a² − y²/b² + z²/c² = 1 (or any permutation where exactly two variables have negative signs). It consists of two separate pieces that do not touch.
In simple terms, this means: picture two bowls facing away from each other with a gap between them. The surface exists only where the variable with the positive coefficient is large enough.
Locus
The set of all points satisfying a given geometric condition. For example, "all points equidistant from two given points" is a locus that turns out to be a plane.
Think of it as: a rule that picks out a specific shape. The shape is whatever the algebra tells you once you translate the rule into an equation.
Distance from a point to a plane
The perpendicular distance from point P₀ = (x₀, y₀, z₀) to the plane Ax + By + Cz = D is |Ax₀ + By₀ + Cz₀ − D| / √(A² + B² + C²).
In simple terms, this means: drop a perpendicular from the point to the plane and measure its length. The formula does this algebraically.
Ellipsoid: x²/a² + y²/b² + z²/c² = 1. All cross-sections are ellipses. Bounded surface.
Hyperboloid of one sheet: x²/a² + y²/b² − z²/c² = 1. Connected surface, shaped like a cooling tower. One variable has the negative sign.
Hyperboloid of two sheets: −x²/a² − y²/b² + z²/c² = 1. Two separate pieces. Two variables have negative signs.
Elliptic paraboloid: z = x²/a² + y²/b². Bowl-shaped, opens in the direction of the lone variable.
Hyperbolic paraboloid: z = x²/a² − y²/b². Saddle-shaped. One squared term positive, one negative.
Elliptic cone: x²/a² + y²/b² = z²/c². The limiting case between the two hyperboloids. The apex is a singular point.
The quick identification rule: count how many squared terms appear and check the signs. All positive with constant on the right = ellipsoid. Two positive, one negative, equals 1 = one sheet. One positive, two negative, equals 1 = two sheets. Mixed signs with no constant = cone. One variable linear (not squared) = paraboloid.
Find an equation for the surface consisting of all points P(x, y, z) for which the distance from P to the point (2, 0, 0) is √2 times the distance from P to the plane x = 1. Then identify the surface.
Distance from P to (2, 0, 0): √((x − 2)² + y² + z²)
Distance from P to the plane x = 1: |x − 1| (the plane is x = 1, so A = 1, B = 0, C = 0, D = 1)
Set up the equation: √((x − 2)² + y² + z²) = √2 · |x − 1|
Square both sides: (x − 2)² + y² + z² = 2(x − 1)²
Expand the left side: x² − 4x + 4 + y² + z²
Expand the right side: 2(x² − 2x + 1) = 2x² − 4x + 2
Set equal: x² − 4x + 4 + y² + z² = 2x² − 4x + 2
Simplify (the −4x terms cancel): x² + 4 + y² + z² = 2x² + 2
Rearrange: −x² + y² + z² = −2
Multiply by −1: x² − y² − z² = 2
This can be written as x²/2 − y²/2 − z²/2 = 1. Two variables (y and z) have negative signs, one variable (x) has a positive sign, and the right side is 1. This is a hyperboloid of two sheets, opening along the x-axis.
\text{Distance from } P(x_0,y_0,z_0) \text{ to point } Q(a,b,c): \quad \sqrt{(x_0-a)^2+(y_0-b)^2+(z_0-c)^2}\text{Distance from } P(x_0,y_0,z_0) \text{ to plane } Ax+By+Cz=D: \quad \frac{|Ax_0+By_0+Cz_0-D|}{\sqrt{A^2+B^2+C^2}}\text{Hyperboloid of two sheets (example): } \frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1Students often confuse hyperboloid of one sheet with hyperboloid of two sheets. The counting rule: for the form ±x²/a² ± y²/b² ± z²/c² = 1, count the negative signs. One negative sign = one sheet. Two negative signs = two sheets.
When squaring both sides of a distance equation, students sometimes forget to square the constant factor. If the condition is "distance to point = √2 times distance to plane," the √2 becomes 2 after squaring.
Students sometimes identify a cone as a hyperboloid. The difference: a cone has 0 on the right side of the standard form, while hyperboloids have 1.
When simplifying, students lose track of signs after expanding. Write each expansion on its own line and combine carefully.
⚠️ "Identify the surface" questions require you to name the surface, not just derive the equation. Partial credit is typically awarded for getting the equation right even if you misname the surface, but the naming is easy marks if you learn the sign-counting rule.
⚠️ Distance-based locus problems combine two skills: setting up distance expressions and simplifying quadratic equations. Practise both.
⚠️ This problem type often involves squaring both sides of an equation, which can introduce extraneous solutions. In this context, because distances are non-negative and you square a clean ratio, extraneous solutions do not arise, but keep the principle in mind.
True or false: x² + y² + z² = 4 is an ellipsoid. (True, it is a sphere, which is a special case of an ellipsoid with a = b = c.)
Fill in the blank: a hyperboloid of one sheet has ______ negative sign(s) in its standard form. (one)
True or false: a cone passes through the origin in its standard form. (True, the right side is 0.)
Fill in the blank: z = x² + y² is an ______ paraboloid. (elliptic)
True or false: z = x² − y² has a saddle shape. (True, it is a hyperbolic paraboloid.)
Q: Identify the surface x² + 4y² − z² = 4. Name the surface and describe its orientation.
A: Dividing by 4: x²/4 + y² − z²/4 = 1. One negative sign, constant = 1. This is a hyperboloid of one sheet, with the z-axis as the axis of symmetry.
Q: Find the equation of the set of all points equidistant from the point (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 opening upward.
Q: The surface −x² − y² + z² = 9 is what kind of surface?
A: Two negative signs, one positive, constant on the right. Dividing by 9: −x²/9 − y²/9 + z²/9 = 1. This is a hyperboloid of two sheets, opening along the z-axis.
Quadric surfaces return when you study level surfaces of functions of three variables (e.g. f(x, y, z) = c gives a quadric when f is a second-degree polynomial). They also appear in setting up limits of integration for triple integrals, where you need to visualise the boundary surface. The distance-to-plane formula used here reappears in computing the distance between parallel planes and between a point and a line.
quadric surfaces, ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic paraboloid, hyperbolic paraboloid, saddle surface, elliptic cone, distance formula 3D, locus, level surface, completing the square, conic sections in 3D, MATH 126, Calculus III