Parabolas as Parametrized Space Curves

From a Parabola in a Plane to a Space Curve

A parabola can be described by a quadratic equation in two coordinates. When it lies in a coordinate plane, its third coordinate is fixed at zero. A parametrization packages all three coordinates into a vector-valued function, even though the curve remains planar.

The useful choice is often immediate: take one of the coordinates in the plane to be the parameter, then express the other coordinate using the parabola’s equation. The fixed coordinate completes the vector.

Key idea: If a parabola lies in a coordinate plane, set its fixed coordinate to zero, let one coordinate that varies be \(t\), and use the equation of the parabola to write the remaining coordinate as a function of \(t\).

Throughout, vectors are written in the coordinate order \((x,y,z)\). A parametrization gives a point for each parameter value; its trace is the set of points obtained as the parameter ranges over its domain.

Vertical Parabolas in the Coordinate Planes

Start with a parabola whose dependent coordinate \(v\) is a quadratic function of the other coordinate \(u\):

v=a(u-h)^2+k,\qquad a\ne 0.

Here \((h,k)\) is the vertex in the \(uv\)-plane. The nonzero condition on \(a\) ensures that the equation is genuinely quadratic. For the full parabola, \(u\) can be any real number.

To parametrize it, set \(u=t\). The other coordinate is then \(v=a(t-h)^2+k\). If \(u\) and \(v\) are two coordinates from \(x,y,z\), put these expressions into the appropriate positions and put zero in the remaining position.

Coordinate plane	Equation	Parametrization, \(t\in\mathbb R\)
\(xy\)-plane	\(y=a(x-h)^2+k\)	\(\mathbf r(t)=(t,\ a(t-h)^2+k,\ 0)\)
\(xz\)-plane	\(z=a(x-h)^2+k\)	\(\mathbf r(t)=(t,\ 0,\ a(t-h)^2+k)\)
\(yz\)-plane	\(z=a(y-h)^2+k\)	\(\mathbf r(t)=(0,\ t,\ a(t-h)^2+k)\)

These formulas describe parabolas whose dependent coordinate is quadratic in the parameter coordinate. For example, in the \(yz\)-plane, \(x\) must stay zero, and \(y=t\) leaves the equation to determine \(z\).

Why the Parametrization Has the Correct Trace

The justification is direct but important. A parametrization must produce points on the stated parabola, and it must produce every point on that parabola.

Every generated point lies on the parabola.
For the \(yz\)-plane example, \(\mathbf r(t)=(0,t,a(t-h)^2+k)\). Its first coordinate is zero, and its last two coordinates satisfy \(z=a(y-h)^2+k\) because \(y=t\).

Every point on the parabola is generated.
If \((0,y,z)\) is any point satisfying \(z=a(y-h)^2+k\), choose \(t=y\). Then \(\mathbf r(t)=(0,y,a(y-h)^2+k)=(0,y,z)\).

Thus the parametrization produces exactly the full parabola when \(t\in\mathbb R\). This reasoning applies to the other coordinate-plane formulas as well: the parameter is the chosen free coordinate, and the equation supplies the other one.

Worked Example: A Shifted Parabola in the \(yz\)-Plane

Parametrize the full parabola \(z=2(y-1)^2-3\) in the \(yz\)-plane.

The \(yz\)-plane has \(x=0\). Choose \(y=t\), where \(t\in\mathbb R\). Substitute \(t\) for \(y\) in the equation to obtain \(z=2(t-1)^2-3\). Therefore, in \((x,y,z)\) order,

\mathbf r(t)=(0,\ t,\ 2(t-1)^2-3),\qquad t\in\mathbb R.

Check the vertex: at \(t=1\), \(y=1\) and \(z=2(1-1)^2-3=-3\), giving \((0,1,-3)\), as expected. At \(t=0\), \(z=2(0-1)^2-3=2-3=-1\), so the parametrization gives \((0,0,-1)\).

For a trace check, the vector has \(x=0\), \(y=t\), and \(z=2(t-1)^2-3\). Replacing \(t\) by \(y\) recovers \(z=2(y-1)^2-3\). Conversely, any point on that parabola is obtained by setting \(t\) equal to its \(y\)-coordinate.

Sideways Parabolas

A parabola can also open along the coordinate that appears as the dependent variable in the equation. In a plane with coordinates \(u,v\), a common sideways form is

u=a(v-k)^2+h,\qquad a\ne 0.

Here \(v\) is the coordinate to use as the parameter: set \(v=t\), then compute \(u=a(t-k)^2+h\). The vertex is \((h,k)\) in the \(uv\)-plane. If the plane is \(xy\), this becomes \(x=a(y-k)^2+h\), with parametrization \((a(t-k)^2+h,t,0)\). If the plane is \(xz\), it becomes \(x=a(z-k)^2+h\), with parametrization \((a(t-k)^2+h,0,t)\). In the \(yz\)-plane, the form \(y=a(z-k)^2+h\) gives \((0,a(t-k)^2+h,t)\).

The choice is not limited to the coordinate that appears first in a written equation. Choose the coordinate that makes the other coordinate easy to calculate. For a sideways form, using \(v=t\) avoids solving a quadratic equation for \(v\), which could otherwise require separate positive and negative branches.

Worked Example: A Sideways Parabola in the \(xz\)-Plane

Parametrize the full parabola \(x=\frac12(z-1)^2-2\) in the \(xz\)-plane.

Since the curve lies in the \(xz\)-plane, \(y=0\). Use \(z=t\), with \(t\in\mathbb R\). Then \(x=\frac12(t-1)^2-2\), so

\mathbf r(t)=\left(\frac12(t-1)^2-2,\ 0,\ t\right),\qquad t\in\mathbb R.

At \(t=1\), \(x=\frac12(1-1)^2-2=-2\); the vertex is \((-2,0,1)\). At \(t=3\), \(x=\frac12(3-1)^2-2=\frac12(4)-2=0\), giving \((0,0,3)\).

To check the trace, the components give \(z=t\) and \(x=\frac12(t-1)^2-2\). Substituting \(t=z\) recovers \(x=\frac12(z-1)^2-2\), while the middle component is always zero. Every point satisfying the stated equation in the \(xz\)-plane is reached by choosing \(t=z\).

Shifts and Reflections

The constants in a quadratic component control where the vertex is and which way the parabola opens. In \(z=a(y-h)^2+k\), changing \(h\) moves the vertex in the \(y\)-direction, while changing \(k\) moves it in the \(z\)-direction. The coefficient \(a\) determines the direction of opening relative to the vertex: positive \(a\) opens toward increasing \(z\), and negative \(a\) opens toward decreasing \(z\).

More precisely, changing \(a\) to \(-a\) while keeping \(h\) and \(k\) fixed reflects the graph across the horizontal line \(z=k\), which passes through its vertex. It does not, in general, reflect the graph across \(z=0\). Reflection across \(z=0\) changes each \(z\)-value to its negative, so the reflected equation is \(z=-a(y-h)^2-k\).

Worked Example: Reflecting a Shifted Parabola Across \(z=0\)

Start with \(z=2(y-3)^2+5\) in the \(yz\)-plane and reflect it across \(z=0\). Then write a parametrization of the reflected parabola.

Reflection across \(z=0\) replaces every value of \(z\) by \(-z\). Equivalently, if the original value is \(2(y-3)^2+5\), the reflected value is its negative:

z=-\bigl(2(y-3)^2+5\bigr)=-2(y-3)^2-5.

The reflected vertex is \((y,z)=(3,-5)\). Since the parabola lies in the \(yz\)-plane, \(x=0\). Set \(y=t\), so the parametrization is

\mathbf r(t)=(0,\ t,\ -2(t-3)^2-5),\qquad t\in\mathbb R.

Check the vertex by setting \(t=3\): \(z=-2(3-3)^2-5=-5\), giving \((0,3,-5)\). At \(t=4\), \(z=-2(4-3)^2-5=-2-5=-7\). The component equation satisfies \(z=-2(y-3)^2-5\) after substituting \(y=t\), so the parametrization has the required trace.

A Common Reflection Mistake

A tempting but incorrect shortcut is to say that reflecting \(z=a(y-h)^2+k\) across \(z=0\) only changes \(a\) to \(-a\). That operation gives \(z=-a(y-h)^2+k\), which reflects across \(z=k\), not across \(z=0\). The constant term must also change sign for reflection across \(z=0\).

For instance, starting with \(z=2(y-3)^2+5\), the incorrect calculation produces \(z=-2(y-3)^2+5\). Its vertex is still \((3,5)\), so it cannot be the reflection across \(z=0\), which must send the vertex \((3,5)\) to \((3,-5)\). The corrected equation is \(z=-2(y-3)^2-5\), whose vertex has the required reflected coordinate.

Check before changing a sign: Identify the line or axis of reflection and test what happens to the vertex. A reflection across \(z=0\) negates the whole \(z\)-value; changing only the quadratic coefficient reflects across the horizontal line through the vertex.

A Practical Construction Method

For a full parabola in a coordinate plane, the following sequence keeps the coordinate placement and algebra organized.

Identify the plane.
Set its fixed coordinate to zero: \(z=0\) in the \(xy\)-plane, \(y=0\) in the \(xz\)-plane, or \(x=0\) in the \(yz\)-plane.

Choose a parameter coordinate.
For \(v=a(u-h)^2+k\), choose \(u=t\). For \(u=a(v-k)^2+h\), choose \(v=t\).

Calculate the remaining coordinate.
Substitute \(t\) into the given equation, keeping all shifts and signs inside the expression.

Write the vector in \(x,y,z\) order.
Place the fixed zero and the two varying components in their correct coordinate positions.

Verify the trace.
Substitute the components into the original equation and confirm that choosing the parameter as the specified coordinate reaches every point.

For the full parabola, use \(t\in\mathbb R\), since the coordinate chosen as the parameter can take every real value. Restricting \(t\) to an interval would describe only the portion of the parabola corresponding to that interval. The vector components still need to satisfy the original equation; a parameter interval does not change the equation of the curve.

Why These Parametrizations Are Useful

A vector-valued parametrization makes the geometry of a planar curve available in the three-coordinate language used for space curves. It records which coordinate stays fixed, identifies the vertex through a particular parameter value, and tells how the points are traced as \(t\) varies. It also gives a direct way to check whether a proposed vector has the intended curve as its trace.

The main care required is consistency. Keep the coordinate order \((x,y,z)\), put zero in the correct position, and substitute the parameter into the equation without dropping a shift or a constant. When a reflection is involved, transform the coordinate values according to the stated reflecting line or plane rather than changing a coefficient by memory.

Check Your Understanding

For each question, identify the fixed coordinate and the coordinate chosen as the parameter before writing or checking a vector.

Write a parametrization of \(y=3(x+1)^2-2\) in the \(xy\)-plane, using \(x=t\).
For \(x=-2(z-4)^2+1\) in the \(xz\)-plane, which coordinate is fixed, and what parametrization results from choosing \(z=t\)?
What is the vertex of \(z=-(y-2)^2+6\), and in which direction does the parabola open?
Reflect \(z=3(y-1)^2+4\) across \(z=0\). Write the reflected equation and explain why changing only the coefficient of the square would be wrong.
How can you check that a proposed parametrization produces every point of a full parabola, rather than just points lying on it?