# 8.4 Rotation of axes  (Page 3/8)

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## Finding a new representation of an equation after rotating through a given angle

Find a new representation of the equation $\text{\hspace{0.17em}}2{x}^{2}-xy+2{y}^{2}-30=0\text{\hspace{0.17em}}$ after rotating through an angle of $\text{\hspace{0.17em}}\theta =45°.$

Find $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y,$ where and

Because $\text{\hspace{0.17em}}\theta =45°,$

$\begin{array}{l}\hfill \\ x={x}^{\prime }\mathrm{cos}\left(45°\right)-{y}^{\prime }\mathrm{sin}\left(45°\right)\hfill \\ x={x}^{\prime }\left(\frac{1}{\sqrt{2}}\right)-{y}^{\prime }\left(\frac{1}{\sqrt{2}}\right)\hfill \\ x=\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}}\hfill \end{array}$

and

$\begin{array}{l}\\ \begin{array}{l}y={x}^{\prime }\mathrm{sin}\left(45°\right)+{y}^{\prime }\mathrm{cos}\left(45°\right)\hfill \\ y={x}^{\prime }\left(\frac{1}{\sqrt{2}}\right)+{y}^{\prime }\left(\frac{1}{\sqrt{2}}\right)\hfill \\ y=\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}}\hfill \end{array}\end{array}$

Substitute $\text{\hspace{0.17em}}x={x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta \text{\hspace{0.17em}}$ and into $\text{\hspace{0.17em}}2{x}^{2}-xy+2{y}^{2}-30=0.$

$2{\left(\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}}\right)}^{2}-\left(\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}}\right)\left(\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}}\right)+2{\left(\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}}\right)}^{2}-30=0$

Simplify.

Write the equations with $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ in the standard form.

$\frac{{{x}^{\prime }}^{2}}{20}+\frac{{{y}^{\prime }}^{2}}{12}=1$

This equation is an ellipse. [link] shows the graph.

## Writing equations of rotated conics in standard form

Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form $\text{\hspace{0.17em}}A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0\text{\hspace{0.17em}}$ into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ coordinate system without the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ term, by rotating the axes by a measure of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ that satisfies

$\mathrm{cot}\left(2\theta \right)=\frac{A-C}{B}$

We have learned already that any conic may be represented by the second degree equation

$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$

where $\text{\hspace{0.17em}}A,B,$ and $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ are not all zero. However, if $\text{\hspace{0.17em}}B\ne 0,$ then we have an $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right)=\frac{A-C}{B}.$

• If $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right)>0,$ then $\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ is in the first quadrant, and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is between $\text{\hspace{0.17em}}\left(0°,45°\right).$
• If $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right)<0,$ then $\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ is in the second quadrant, and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is between $\text{\hspace{0.17em}}\left(45°,90°\right).$
• If $\text{\hspace{0.17em}}A=C,$ then $\text{\hspace{0.17em}}\theta =45°.$

Given an equation for a conic in the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ system, rewrite the equation without the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ term in terms of $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime },$ where the $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ axes are rotations of the standard axes by $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ degrees.

1. Find $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right).$
2. Find and
3. Substitute and into and
4. Substitute the expression for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ into in the given equation, and then simplify.
5. Write the equations with $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ in the standard form with respect to the rotated axes.

## Rewriting an equation with respect to the x′ And y′ Axes without the x′y′ Term

Rewrite the equation $\text{\hspace{0.17em}}8{x}^{2}-12xy+17{y}^{2}=20\text{\hspace{0.17em}}$ in the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ system without an $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ term.

First, we find $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right).\text{\hspace{0.17em}}$ See [link] .

$\mathrm{cot}\left(2\theta \right)=\frac{3}{4}=\frac{\text{adjacent}}{\text{opposite}}$

So the hypotenuse is

$\begin{array}{r}\hfill {3}^{2}+{4}^{2}={h}^{2}\\ \hfill 9+16={h}^{2}\\ \hfill 25={h}^{2}\\ \hfill h=5\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$

Next, we find and

Substitute the values of and into and

and

Substitute the expressions for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ into in the given equation, and then simplify.

Write the equations with $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ in the standard form with respect to the new coordinate system.

$\frac{{{x}^{\prime }}^{2}}{4}+\frac{{{y}^{\prime }}^{2}}{1}=1$

[link] shows the graph of the ellipse.

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