Introduction
A parabolic cylinder is a surface formed by extending a parabola in a straight direction. Its equation therefore combines two features: a parabolic relationship between two coordinates and an independent third coordinate.
Recognizing which coordinate is independent tells you the direction in which the surface extends. The squared and linear coordinates determine the shape of its parabolic cross-sections and the direction those cross-sections open.
Definition and Standard Form
Let \(u\) be the coordinate that is squared, let \(v\) be the coordinate that appears linearly, and let \(w\) be the coordinate that does not appear. A standard parabolic-cylinder equation is
Here \(a\), \(h\), and \(k\) are constants, and \(u,v,w\) are three different coordinates chosen from \(x,y,z\). Since \(w\) does not appear, it may take any real value.
For a fixed value of \(w\), the equation describes the same parabola in the \(uv\)-plane. For a fixed value of \(u\), the equation fixes \(v\), while \(w\) remains free; this gives a straight line parallel to the \(w\)-axis.
The point \((u,v)=(h,k)\) is the vertex of each parabolic cross-section. Because \(w\) is unrestricted, these vertices form a whole line: \(u=h,\ v=k\), with \(w\) arbitrary. This vertex line is parallel to the cylinder’s generating direction.
Axis Direction and Opening Direction
Two directions must be distinguished when describing the surface:
- Cylinder direction: the direction of the missing coordinate \(w\). The surface extends along straight lines parallel to that coordinate axis.
- Parabolic opening direction: the direction of the linear coordinate \(v\). The cross-sections open toward increasing \(v\) if \(a>0\), and toward decreasing \(v\) if \(a<0\).
The squared coordinate \(u\) runs across the width of each parabola. The vertex occurs at \(u=h\), where the squared term is zero. The coefficient \(a\) determines which way the parabola opens, since \((u-h)^2\) is nonnegative.
The equivalent form \((u-h)^2=4p(v-k)\), where \(p\ne 0\), gives the same information: the cross-sections open toward increasing \(v\) if \(p>0\), and decreasing \(v\) if \(p<0\). The coefficient form \(v=a(u-h)^2+k\) is often more convenient when reading an equation directly.
Why the Missing Coordinate Makes a Cylinder
The cylinder structure follows directly from the equation. If a point \((u,v,w)\) satisfies \(v=a(u-h)^2+k\), then for any real number \(t\), the point \((u,v,w+t)\) also satisfies it: neither \(u\) nor \(v\) has changed, and \(w\) does not occur in the equation. Thus every point on the surface belongs to a whole straight line parallel to the \(w\)-axis.
Conversely, fixing \(w=c\) leaves the equation \(v=a(u-h)^2+k\), which is a parabola in that coordinate plane. Every such cross-section has the same equation. Together, the repeated parabolas and the lines parallel to the missing coordinate produce the parabolic cylinder.
Set \(w=c\) for any constant \(c\). The remaining equation is a parabola in the other two coordinates.
Allow \(c\) to take all real values. The same parabola is repeated in parallel coordinate planes.
Points with the same \(u\) and \(v\) but different \(w\) form straight generators parallel to the missing coordinate axis.
Worked Examples
Worked Example: Reading a Cylinder That Extends Along the \(z\)-Axis
Analyze and sketch the surface
The variable \(z\) is absent, so the cylinder extends parallel to the \(z\)-axis. The squared coordinate is \(x\), and the linear coordinate is \(y\). In the form \(y=a(x-h)^2+k\), we have \(a=\frac14\), \(h=2\), and \(k=-1\).
Since \(a=\frac14>0\), each parabolic cross-section opens toward increasing \(y\). Its vertex is at \(x=2,\ y=-1\). The vertices form the line \(x=2,\ y=-1\), with \(z\) arbitrary.
To get points for a cross-section, take \(z=0\). At \(x=0\), \(y=\frac{(0-2)^2}{4}-1=\frac44-1=0\). At \(x=1\), \(y=\frac{(1-2)^2}{4}-1=\frac14-1=-\frac34\). At \(x=2\), \(y=-1\). At \(x=3\), \(y=\frac14-1=-\frac34\), and at \(x=4\), \(y=\frac44-1=0\). These points show a symmetric upward-opening parabola in the \(xy\)-plane.
To sketch the surface, draw that parabola and extend it in both directions parallel to the \(z\)-axis. As a check, the points \((0,0,0)\) and \((2,-1,5)\) satisfy the equation: for the first, the right side is \(\frac{(0-2)^2}{4}-1=0\); for the second, it is \(\frac{(2-2)^2}{4}-1=-1\). The different \(z\)-values do not affect either calculation.
Worked Example: A Cylinder Opening Toward Negative \(x\)
Identify the directions and key features of
where \(y\) does not appear. First isolate \(x\):
Now the form is \(x=a(z-h)^2+k\), with \(a=-\frac12\), \(h=-1\), and \(k=3\). The missing coordinate is \(y\), so the cylinder extends parallel to the \(y\)-axis. Since \(a<0\), the parabolic cross-sections open toward decreasing \(x\). Their vertices have \(z=-1,\ x=3\), giving the vertex line \(x=3,\ z=-1\), with \(y\) arbitrary.
For a cross-section with \(y=0\), choose \(z=-3,-1,1\). At \(z=-3\), \(x=3-\frac12(-2)^2=1\). At \(z=-1\), \(x=3-\frac12(0)^2=3\). At \(z=1\), \(x=3-\frac12(2)^2=1\). The vertex is the rightmost point in the \(x\)-direction, and the two other points confirm the opening toward negative \(x\).
Check the original equation at \((1,0,-3)\): \(2(1)+(-3+1)^2=2+4=6\). At the vertex point \((3,0,-1)\), \(2(3)+(-1+1)^2=6+0=6\). Thus both points lie on the surface.
Worked Example: Completing the Square to Reveal the Cylinder
Classify and sketch
The variable \(z\) is absent, but the \(x\)-terms are not yet written as a squared expression around the vertex. Complete the square:
Divide both sides by \(2\):
This is a parabolic cylinder extending parallel to the \(z\)-axis. Since the coefficient of the square is \(\frac12>0\), its cross-sections open toward increasing \(y\). The vertex line is \(x=3,\ y=-2\), with \(z\) arbitrary.
For points in the \(xy\)-plane, set \(z=0\). When \(x=1\), \(y=\frac12(1-3)^2-2=2-2=0\). When \(x=3\), \(y=\frac12(0)^2-2=-2\). When \(x=5\), \(y=\frac12(2)^2-2=0\). The points are symmetric about \(x=3\).
Check the original equation at \((1,0,0)\): its left side is \(2(0)=0\), and its right side is \(1^2-6(1)+5=1-6+5=0\). At \((3,-2,4)\), the left side is \(2(-2)=-4\), and the right side is \(3^2-6(3)+5=9-18+5=-4\). Both points satisfy the original equation, including a point with a different \(z\)-coordinate.
A Reliable Sketching Procedure
When an equation has one squared coordinate, one linear coordinate, and one missing coordinate, use the following sequence. If necessary, first isolate the linear coordinate and complete the square in the squared coordinate.
- Find the missing coordinate. This identifies the direction of the cylinder’s generators.
- Write the parabolic relation clearly. Aim for \(v=a(u-h)^2+k\), with the squared coordinate \(u\) and linear coordinate \(v\).
- Read the vertex line. Set \(u=h\) and \(v=k\); allow the missing coordinate to vary freely.
- Determine the opening direction. A positive \(a\) means increasing \(v\); a negative \(a\) means decreasing \(v\).
- Draw a representative cross-section. Choose a convenient fixed value for the missing coordinate, plot the vertex and symmetric points, then extend the curve along the missing-coordinate direction.
A genuine parabolic-cylinder equation in this form requires \(a\ne0\). If \(a=0\), the equation reduces to \(v=k\), which describes a plane rather than a parabolic cylinder. Also, if the equation has not been rearranged correctly, its appearance alone can hide which coordinate plays each role; completing the square can expose the vertex and sign.
Common Mistake: Confusing the Two Directions
A frequent error is to say that the cylinder opens in the direction of its missing coordinate. These are different geometric features. In \(y=\frac14(x-2)^2-1\), the missing coordinate is \(z\), so the surface extends parallel to the \(z\)-axis. But the positive coefficient means that its parabolic cross-sections open toward increasing \(y\), not toward \(z\).
A second error is to read the sign before isolating the linear coordinate. For example, \(2x+(z+1)^2=6\) has a positive squared term on the left, but isolating \(x\) gives \(x=3-\frac12(z+1)^2\). The relevant coefficient is \(-\frac12\), so the cross-sections open toward negative \(x\). Always put the equation in the form of the linear coordinate equal to a coefficient times a square, plus a constant, before deciding the opening direction.
Why These Features Matter
The equation encodes the three-dimensional shape without requiring a separate formula for every point. The missing coordinate identifies the straight direction in which the surface continues; the squared relation determines the parabolic cross-section; and the sign of its coefficient determines the opening. Once these features are identified, a sketch can show the surface’s essential geometry rather than a single isolated curve.
For a quick final check, verify that your sketch has all three properties: each fixed value of the missing coordinate gives the same parabola, each fixed value of the squared coordinate gives a line parallel to the missing-coordinate axis, and the parabolic cross-sections open in the direction indicated by the coefficient.
Check Your Understanding
For each question, identify the relevant coordinates and explain how they determine the surface.
- For \(z=2(x+1)^2-3\), which coordinate is missing, what is the cylinder direction, and which way do the cross-sections open?
- For \(y=-\frac14(z-2)^2+5\), state the vertex line and the opening direction.
- Rewrite \(x=y^2-4y+1\) in vertex form. Identify its cylinder direction and opening direction.
- Why does fixing the missing coordinate produce a parabola, while fixing the squared coordinate produces a straight line?
- What geometric surface results from \(v=0(u-h)^2+k\), and why is it not a parabolic cylinder?