8.4 Rotation of axes  (Page 3/8)

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

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

Find $x$ and $y,$ where $x=x^{\prime }\cos \theta -y^{\prime }\sin \theta$ and $y=x^{\prime }\sin \theta +y^{\prime }\cos \theta .$

Because $\theta =\mathrm{45°},$

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

and

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

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

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

Simplify.

$$\begin{array}{ll}\cancel{2}\frac{(x^{\prime }-y^{\prime })(x^{\prime }-y^{\prime })}{\cancel{2}}-\frac{(x^{\prime }-y^{\prime })(x^{\prime }+y^{\prime })}{2}+\cancel{2}\frac{(x^{\prime }+y^{\prime })(x^{\prime }+y^{\prime })}{\cancel{2}}-30=0\hfill & \begin{array}{cccc}& & & \end{array}\text{FOIL method}\hfill \\ x^{\prime }{}^2{\cancel{-2x^{\prime }y}}^{\prime }+y^{\prime }{}^2-\frac{(x^{\prime }{}^2-y^{\prime }{}^2)}{2}+x^{\prime }{}^2\cancel{+2x^{\prime }{y}^{\prime }}+y^{\prime }{}^2-30=0\hfill & \begin{array}{cccc}& & & \end{array}\text{Combine like terms}.\hfill \\ 2x^{\prime }{}^2+2y^{\prime }{}^2-\frac{(x^{\prime }{}^2-y^{\prime }{}^2)}{2}=30\hfill & \begin{array}{cccc}& & & \end{array}\text{Combine like terms}.\hfill \\ 2\left(2x^{\prime }{}^2+2y^{\prime }{}^2-\frac{(x^{\prime }{}^2-y^{\prime }{}^2)}{2}\right)=2(30)\hfill & \begin{array}{cccc}& & & \end{array}\text{Multiply both sides by 2}.\hfill \\ 4x^{\prime }{}^2+4y^{\prime }{}^2-(x^{\prime }{}^2-y^{\prime }{}^2)=60\hfill & \begin{array}{cccc}& & & \end{array}\text{Simplify}.\hfill \\ 4x^{\prime }{}^2+4y^{\prime }{}^2-x^{\prime }{}^2+y^{\prime }{}^2=60\hfill & \begin{array}{cccc}& & & \end{array}\text{Distribute}.\hfill \\ \frac{3x^{\prime }{}^2}{60}+\frac{5y^{\prime }{}^2}{60}=\frac{60}{60}\hfill & \begin{array}{cccc}& & & \end{array}\text{Set equal to 1}.\hfill \end{array}$$

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

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

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

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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 $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the $x^{\prime }$ and $y^{\prime }$ coordinate system without the $x^{\prime }{y}^{\prime }$ term, by rotating the axes by a measure of $\theta$ that satisfies

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

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

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

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

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

First, we find $\cot (2\theta ).$ See [link] .

$$\begin{array}{l}8x^2-12xy+17y^2=20\Rightarrow A=8,B=-12\text{and}C=17\hfill \\ \cot (2\theta )=\frac{A-C}{B}=\frac{8-17}{-12}\hfill \\ \cot (2\theta )=\frac{-9}{-12}=\frac{3}{4}\hfill \end{array}$$

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

So the hypotenuse is

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

Next, we find $\sin \theta$ and $\cos \theta .$

$$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ \sin \theta =\sqrt{\frac{1-\cos (2\theta )}{2}}=\sqrt{\frac{1-\frac{3}{5}}{2}}=\sqrt{\frac{\frac{5}{5}-\frac{3}{5}}{2}}=\sqrt{\frac{5-3}{5}\cdot \frac{1}{2}}=\sqrt{\frac{2}{10}}=\sqrt{\frac{1}{5}}\hfill \end{array}\hfill \\ \sin \theta =\frac{1}{\sqrt{5}}\hfill \\ \cos \theta =\sqrt{\frac{1+\cos (2\theta )}{2}}=\sqrt{\frac{1+\frac{3}{5}}{2}}=\sqrt{\frac{\frac{5}{5}+\frac{3}{5}}{2}}=\sqrt{\frac{5+3}{5}\cdot \frac{1}{2}}=\sqrt{\frac{8}{10}}=\sqrt{\frac{4}{5}}\hfill \\ \cos \theta =\frac{2}{\sqrt{5}}\hfill \end{array}$$

Substitute the values of $\sin \theta$ and $\cos \theta$ into $x=x^{\prime }\cos \theta -y^{\prime }\sin \theta$ and $y=x^{\prime }\sin \theta +y^{\prime }\cos \theta .$

$$\begin{array}{l}\hfill \\ \begin{array}{l}x=x^{\prime }\cos \theta -y^{\prime }\sin \theta \hfill \\ x=x^{\prime }\left(\frac{2}{\sqrt{5}}\right)-y^{\prime }\left(\frac{1}{\sqrt{5}}\right)\hfill \\ x=\frac{2x^{\prime }-y^{\prime }}{\sqrt{5}}\hfill \end{array}\hfill \end{array}$$

and

$$\begin{array}{l}\begin{array}{l}\hfill \\ y=x^{\prime }\sin \theta +y^{\prime }\cos \theta \hfill \end{array}\hfill \\ y=x^{\prime }\left(\frac{1}{\sqrt{5}}\right)+y^{\prime }\left(\frac{2}{\sqrt{5}}\right)\hfill \\ y=\frac{x^{\prime }+2y^{\prime }}{\sqrt{5}}\hfill \end{array}$$

Substitute the expressions for $x$ and $y$ into in the given equation, and then simplify.

$$\begin{array}{l}8{\left(\frac{2x^{\prime }-y^{\prime }}{\sqrt{5}}\right)}^2-12\left(\frac{2x^{\prime }-y^{\prime }}{\sqrt{5}}\right)\left(\frac{x^{\prime }+2y^{\prime }}{\sqrt{5}}\right)+17{\left(\frac{x^{\prime }+2y^{\prime }}{\sqrt{5}}\right)}^2=20\hfill \\ 8\left(\frac{(2x^{\prime }-y^{\prime })(2x^{\prime }-y^{\prime })}{5}\right)-12\left(\frac{(2x^{\prime }-y^{\prime })(x^{\prime }+2y^{\prime })}{5}\right)+17\left(\frac{(x^{\prime }+2y^{\prime })(x^{\prime }+2y^{\prime })}{5}\right)=20\hfill \\ 8\left(4x^{\prime }{}^2-4x^{\prime }{y}^{\prime }+y^{\prime }{}^2\right)-12\left(2x^{\prime }{}^2+3x^{\prime }{y}^{\prime }-2y^{\prime }{}^2\right)+17\left(x^{\prime }{}^2+4x^{\prime }{y}^{\prime }+4y^{\prime }{}^2\right)=100\hfill \\ 32x^{\prime }{}^2-32x^{\prime }{y}^{\prime }+8y^{\prime }{}^2-24x^{\prime }{}^2-36x^{\prime }{y}^{\prime }+24y^{\prime }{}^2+17x^{\prime }{}^2+68x^{\prime }{y}^{\prime }+68y^{\prime }{}^2=100\hfill \\ 25x^{\prime }{}^2+100y^{\prime }{}^2=100\hfill \\ \frac{25}{100}{x}^{\prime }{}^2+\frac{100}{100}{y}^{\prime }{}^2=\frac{100}{100} \hfill \end{array}$$

Write the equations with $x^{\prime }$ and $y^{\prime }$ 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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