Rotation of axes

Every conic so far had its axes lined up with the ${x}$- and ${y}$-axes. A tilted conic has an equation with an ${xy}$ term, like ${xy=2}$. This page shows how to turn the coordinate axes so that the conic is straight again, and how to tell which conic an equation describes without any turning at all.

The general equation

Every conic has an equation of the form

$\begin{align*}&Ax^2+Bxy+Cy^2\\&\quad+Dx+Ey+F=0\end{align*}$

If ${B=0}$, the conic's axes are parallel to the coordinate axes, and we can complete the square as in the earlier lessons. If ${B\ne 0}$, the conic is tilted.

Rotating the axes

Draw new axes, ${X}$ and ${Y}$, by turning the ${x}$- and ${y}$-axes counterclockwise through an angle ${\varphi}$ (phi).

The ${XY}$-axes are the ${xy}$-axes turned by the angle ${\varphi}$. A point ${P}$ has coordinates ${(x,y)}$ and also ${(X,Y)}$.

Suppose ${P}$ is at distance ${r}$ from the origin, at angle ${\alpha}$ from the ${X}$-axis. Then ${X=r\cos\alpha}$ and ${Y=r\sin\alpha}$. From the ${x}$-axis, its angle is ${\alpha+\varphi}$, so ${x=r\cos(\alpha+\varphi)}$ and ${y=r\sin(\alpha+\varphi)}$. Expanding with the sum formulas gives the rotation formulas:

${x=X\cos\varphi-Y\sin\varphi}$

${y=X\sin\varphi+Y\cos\varphi}$

To remove the ${xy}$ term, choose the angle ${\varphi}$ (between ${0^\circ}$ and ${90^\circ}$) with

${\cot 2\varphi=\dfrac{A-C}{B}}$

In particular, if ${A=C}$, then ${\cot 2\varphi=0}$, so ${2\varphi=90^\circ}$ and ${\varphi=45^\circ}$. With ${\varphi=45^\circ}$, the rotation formulas become

${x=\dfrac{X-Y}{\sqrt{2}}}$${y=\dfrac{X+Y}{\sqrt{2}}}$

Example 1: Rotate the axes to remove the ${xy}$ term from ${xy=2}$, and identify the graph.

Solution:

Here ${A=C=0}$ and ${B=1}$, so ${\varphi=45^\circ}$. Substitute:

$\begin{align*}\dfrac{X-Y}{\sqrt{2}}\cdot\dfrac{X+Y}{\sqrt{2}}&=2\\\dfrac{X^2-Y^2}{2}&=2\\\dfrac{X^2}{4}-\dfrac{Y^2}{4}&=1\end{align*}$

This is a hyperbola with ${a=b=2}$, opening along the ${X}$-axis, which is the line ${y=x}$. Its vertices are ${2}$ units from the origin along that line, at ${(\sqrt{2},\sqrt{2})}$ and ${(-\sqrt{2},-\sqrt{2})}$.

The graph of ${xy=2}$ is a hyperbola. In the turned ${XY}$-axes (dashed), its equation is ${\dfrac{X^2}{4}-\dfrac{Y^2}{4}=1}$.

Example 2: Rotate the axes to remove the ${xy}$ term from ${5x^2-6xy+5y^2=32}$, and identify the graph.

Solution:

Here ${A=C=5}$, so again ${\varphi=45^\circ}$. Work out each term:

${x^2=\dfrac{X^2-2XY+Y^2}{2}}$${y^2=\dfrac{X^2+2XY+Y^2}{2}}$

${xy=\dfrac{X^2-Y^2}{2}}$

So ${5x^2+5y^2=5(X^2+Y^2)}$ (the ${XY}$ terms cancel), and ${-6xy=-3X^2+3Y^2}$. The equation becomes

$\begin{align*}2X^2+8Y^2&=32\\\dfrac{X^2}{16}+\dfrac{Y^2}{4}&=1\end{align*}$

This is an ellipse with ${a=4}$ along the ${X}$-axis (the line ${y=x}$) and ${b=2}$.

The graph of ${5x^2-6xy+5y^2=32}$ is an ellipse tilted by ${45^\circ}$: ${\dfrac{X^2}{16}+\dfrac{Y^2}{4}=1}$.

Identifying a conic without rotating

The number ${B^2-4AC}$, called the discriminant, does not change when the axes are turned. Its sign tells you which conic you have:

${B^2-4AC}$	The graph is
negative	an ellipse (or a circle)
zero	a parabola
positive	a hyperbola

(In a few special cases, the graph is "degenerate": a pair of lines, one point, or nothing at all.)

Example 3: Identify each conic.

(a) ${x^2+4xy+4y^2-6x=0}$

(b) ${3x^2+2xy-y^2+7=0}$

Solution:

(a) ${A=1}$, ${B=4}$, ${C=4}$: ${B^2-4AC=16-16=0}$. A parabola.

(b) ${A=3}$, ${B=2}$, ${C=-1}$: ${B^2-4AC=4+12=16>0}$. A hyperbola.

For Example 2, ${B^2-4AC=36-100}$, which is ${-64}$. It is negative: an ellipse, as we found.

Summary

A conic whose equation has an ${xy}$ term (${B\ne 0}$) is tilted.
Rotation formulas: ${x=X\cos\varphi-Y\sin\varphi}$ and ${y=X\sin\varphi+Y\cos\varphi}$.
Choose ${\varphi}$ with ${\cot 2\varphi=\dfrac{A-C}{B}}$ to remove the ${xy}$ term. If ${A=C}$, use ${\varphi=45^\circ}$.
${B^2-4AC}$: negative means ellipse, zero means parabola, positive means hyperbola.

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