Elliptic Paraboloids

Introduction

An elliptic paraboloid is a bowl-shaped surface in three-dimensional space. Its horizontal cross-sections are ellipses, while suitable vertical cross-sections are parabolas. These two kinds of traces make the surface straightforward to recognize and sketch once its vertex and opening direction are identified.

The word “elliptic” describes the shape of the horizontal sections; it does not require every section to be a non-circular ellipse. If the two squared terms have equal coefficients, those sections are circles, and the surface is a circular paraboloid, a special case of an elliptic paraboloid.

Standard Form, Vertex, and Opening

A common standard form, with the axis of the surface parallel to the \(z\)-axis, is

The constants \(h\), \(\ell\), and \(k\) locate the vertex. The positive squared terms mean that \(z\) increases as the point moves away from \((h,\ell)\), so the surface opens in the positive \(z\)-direction.

In the first equation, setting \(x=h\) and \(y=\ell\) makes both squared terms zero, giving the point \((h,\ell,k)\). All other points on the surface have \(z>k\), so this point is the lowest point of the surface. In the second equation, all other points have \(z<k\), making the vertex its highest point.

The same pattern applies when the axis is parallel to another coordinate axis: the variable alone on one side is the axis variable, and the other two variables appear as positive or negative squared terms. For example, \(x=h+\frac{(y-\ell)^2}{a^2}+\frac{(z-m)^2}{b^2}\) opens in the positive \(x\)-direction and has vertex \((h,\ell,m)\).

Why the Horizontal Traces Are Ellipses

For a surface whose axis is parallel to the \(z\)-axis, a horizontal plane has equation \(z=c\), where \(c\) is constant. Substituting \(z=c\) into the standard equation that opens upward gives

If \(c>k\), divide both sides by \(c-k\). The resulting equation is an ellipse centered at \((h,\ell)\) in the plane \(z=c\). If \(c=k\), both squared terms must be zero, so the trace is just the vertex. If \(c<k\), the equation would set a nonnegative sum equal to a negative number, which is impossible, so the plane does not meet the surface.

This classification follows directly from the equation, not from a visual guess. For a surface that opens downward, the same reasoning gives ellipses when \(c<k\), a single vertex when \(c=k\), and no intersection when \(c>k\).

Why Vertical Traces Are Parabolas

A vertical plane is parallel to the \(z\)-axis. For the upward-opening standard form, the coordinate planes \(x=h\) and \(y=\ell\) are vertical planes through the axis. Setting \(x=h\), for example, gives

This is a parabola in the plane \(x=h\), with vertex \((h,\ell,k)\). Setting \(y=\ell\) instead gives a parabola in the plane \(y=\ell\). More generally, fixing \(y\) at any constant produces a vertical-plane trace that is also a parabola, though its vertex may sit above the surface’s vertex.

The distinction between horizontal and vertical planes matters. A plane \(z=c\) is horizontal; it is not a vertical plane. The following short derivation shows how these sections let us reconstruct the surface’s shape.

Together, the traces explain the bowl shape: moving away from the vertex in either horizontal direction raises \(z\) for an upward-opening surface, and each higher horizontal plane cuts a larger ellipse.

Worked Examples

A Common Trace Mistake

A frequent error is to call the plane \(z=0\) a vertical plane. It is horizontal: it is parallel to the \(xy\)-plane. For example, on \(z=\frac{x^2}{4}+\frac{y^2}{9}\), setting \(z=0\) gives \(\frac{x^2}{4}+\frac{y^2}{9}=0\). Since both terms are nonnegative, the only solution is \(x=y=0\). This is the single-point trace at the vertex in a horizontal plane, not a vertical parabolic trace.

To obtain a vertical trace, fix \(x\) or \(y\), not \(z\). In the same example, the vertical plane \(y=1\) gives

This is a parabola in the plane \(y=1\). Its vertex is \((0,1,\frac{1}{9})\), which is above the surface’s vertex \((0,0,0)\). Checking \(x=0\) gives \(z=\frac{1}{9}\), and substitution into the original equation gives \(\frac{0^2}{4}+\frac{1^2}{9}=\frac{1}{9}\). Thus the trace and the surface equation agree.