How to Change a Parabola to Canonical Form

A parabola may be written in coordinates that are rotated or shifted relative to its usual position. In those coordinates, its equation can contain an $xy$ term and linear terms, so it will not look like the familiar $U^2=4pV$. You can put it into canonical form by rotating the axes to remove the cross term, completing the square, and translating the origin to the vertex.

The method below applies to a general second-degree equation in two variables. It also explains an important caveat: the usual discriminant test identifies a parabolic-type quadratic part, but does not by itself prove that the whole equation describes a nondegenerate parabola.

General equation and parabolic condition

Write the conic as

$Ax^2+Bxy+Cy^2+Dx+Ey+F=0.$

For this convention, the quadratic part is parabolic when

$B^2-4AC=0.$

This condition means the quadratic part has one zero direction. It is necessary for a nondegenerate parabola, but the equation could instead describe parallel lines, a repeated line, or have no real points. To get an ordinary parabola, the linear part must vary along the zero direction of the quadratic part.

Procedure for reaching canonical form

1. Rotate to remove $xy$

If $B\ne 0$, the squared terms are generally not aligned with the coordinate axes. Rotate the axes so that the quadratic part has no cross term. For example, use

$x=u\cos\theta-v\sin\theta,\qquad y=u\sin\theta+v\cos\theta.$

With this convention, choose $\theta$ so that $\tan(2\theta)=B/(A-C)$ when that ratio is defined. In special cases, it is simpler to identify the directions directly. After rotation, the equation has the form

$\lambda u^2+\alpha u+\beta v+\gamma=0,$

where $\lambda\ne 0$ and there is no $v^2$ term. If the original equation already has no cross term, this rotation may not be needed.

2. Complete the square

Group the terms in $u$ and complete the square:

$\lambda\left(u+\frac{\alpha}{2\lambda}\right)^2+\beta v+\gamma-\frac{\alpha^2}{4\lambda}=0.$

The square is now centered at the vertex’s coordinate in the $u$ direction.

3. Translate the origin

Define

$U=u+\frac{\alpha}{2\lambda},\qquad V=v+\frac{\gamma-\alpha^2/(4\lambda)}{\beta},$

provided $\beta\ne 0$. The equation becomes

$U^2=-\frac{\beta}{\lambda}V=4pV,$

where $p=-\beta/(4\lambda)$. This is the canonical form of a parabola whose axis is the $V$-axis. If the squared coordinate is $v$ instead, the corresponding form is $V^2=4pU$.

The sign of $p$ gives the direction of opening: in $U^2=4pV$, the parabola opens toward increasing $V$ if $p>0$ and decreasing $V$ if $p<0$.

Worked example

Put

$(x+y)^2-2x+2y=0$

into canonical form. The quadratic terms are $x^2+2xy+y^2$, so the cross term means a rotation is needed. Choose rotated coordinates

$u=\frac{x+y}{\sqrt{2}},\qquad v=\frac{x-y}{\sqrt{2}}.$

Then $x+y=\sqrt{2}u$ and $x-y=\sqrt{2}v$. Rewrite the original equation using these expressions:

(x+y)^2-2x+2y&=0,\\ 2u^2-2(x-y)&=0,\\ 2u^2-2\sqrt{2}v&=0,\\ u^2&=\sqrt{2}v. \end{aligned}$$ This is already in canonical form. Comparing it with $U^2=4pV$ gives $U=u$, $V=v$, and $4p=\sqrt{2}$, so $p=\sqrt{2}/4$. The vertex is at $(u,v)=(0,0)$, and the axis is the $v$-axis in the rotated coordinates. ## What if the equation is degenerate? In the rotated equation $\lambda u^2+\alpha u+\beta v+\gamma=0$, a proper parabola requires $\lambda\ne0$ and $\beta\ne0$. If $\beta=0$, completing the square leaves an equation involving only $u^2$ and a constant. Depending on that constant, the result may be two parallel lines, a repeated line, or no real points—not a parabola opening along $v$. For instance, $u^2-1=0$ is the pair of parallel lines $u=1$ and $u=-1$. Its quadratic part has the same parabolic-type discriminant, but the full equation is not a parabola. ## Common mistakes - **Stopping at $B^2-4AC=0$.** This tests the quadratic part; check the linear terms too to rule out degeneracy. - **Trying to complete the square before removing a cross term.** With an $xy$ term, rotation is usually the clean first step. - **Forgetting to translate after completing the square.** Completing the square centers the squared coordinate, but a constant may remain in the other coordinate. - **Mixing canonical-form conventions.** Some books write $U^2=4pV$; others use a parameter such as $a$ instead of $p$. Compare coefficients to identify the parameter rather than assuming the symbols mean the same thing.