Rotation of Axes: Learn It 3

Lumen Learning, OpenStax

Rotation of Axes: Learn It 3

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

[latex]\cot \left(2\theta \right)=\frac{A-C}{B}[/latex]

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

[latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex]

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

If [latex]\cot \left(2\theta \right)>0[/latex], then [latex]2\theta[/latex] is in the first quadrant, and [latex]\theta[/latex] is between [latex]\left(0^\circ ,45^\circ \right)[/latex].
If [latex]\cot \left(2\theta \right)<0[/latex], then [latex]2\theta[/latex] is in the second quadrant, and [latex]\theta[/latex] is between [latex]\left(45^\circ ,90^\circ \right)[/latex].
If [latex]A=C[/latex], then [latex]\theta =45^\circ[/latex].

How To: Given an equation for a conic in the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] system, rewrite the equation without the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] term in terms of [latex]\begin{align}{x}^{\prime }\end{align}[/latex] and [latex]\begin{align}{y}^{\prime }\end{align}[/latex], where the [latex]\begin{align}{x}^{\prime }\end{align}[/latex] and [latex]\begin{align}{y}^{\prime }\end{align}[/latex] axes are rotations of the standard axes by [latex]\theta[/latex] degrees.

Find [latex]\cot \left(2\theta \right)[/latex].
Find [latex]\sin \theta[/latex] and [latex]\cos \theta[/latex].
Substitute [latex]\sin \theta[/latex] and [latex]\cos \theta[/latex] into [latex]\begin{align}x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \end{align}[/latex] and [latex]\begin{align} y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \end{align}[/latex].
Substitute the expression for [latex]x[/latex] and [latex]y[/latex] into in the given equation, and then simplify.
Write the equations with [latex]\begin{align}{x}^{\prime }\end{align}[/latex] and [latex]\begin{align}{y}^{\prime }\end{align}[/latex] in the standard form with respect to the rotated axes.

Rewrite the equation [latex]8{x}^{2}-12xy+17{y}^{2}=20[/latex] in the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] system without an [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] term.

First, we find [latex]\cot \left(2\theta \right)[/latex].

[latex]\begin{align}8{x}^{2}-12xy+17{y}^{2}=20\Rightarrow A=8,B=-12\text{ and }C=17\end{align}[/latex]

[latex]\begin{align} &\cot \left(2\theta \right)=\frac{A-C}{B}=\frac{8 - 17}{-12} \\ &\cot \left(2\theta \right)=\frac{-9}{-12}=\frac{3}{4} \end{align}[/latex]

[latex]\cot \left(2\theta \right)=\frac{3}{4}=\frac{\text{adjacent}}{\text{opposite}}[/latex]

So the hypotenuse is

[latex]\begin{gathered} {3}^{2}+{4}^{2}={h}^{2}\\ 9+16={h}^{2}\\ 25={h}^{2}\\ h=5\end{gathered}[/latex]

Next, we find [latex]\sin \text{ }\theta[/latex] and [latex]\cos \text{ }\theta[/latex].

[latex]\begin{align} \sin \theta &=\sqrt{\frac{1-\cos \left(2\theta \right)}{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}} \\ &=\frac{1}{\sqrt{5}} \\[2mm] \cos \theta &=\sqrt{\frac{1+\cos \left(2\theta \right)}{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}} \\ &=\frac{2}{\sqrt{5}} \end{align}[/latex]

Substitute the values of [latex]\sin \text{ }\theta[/latex] and [latex]\cos \text{ }\theta[/latex] into [latex]\begin{align}x={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta \end{align}[/latex] and [latex]\begin{align}y={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta \end{align}[/latex].

[latex]\begin{align}&x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \\ &x={x}^{\prime }\left(\frac{2}{\sqrt{5}}\right)-{y}^{\prime }\left(\frac{1}{\sqrt{5}}\right) \\ &x=\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}} \end{align}[/latex]

and

[latex]\begin{align} &y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \\ &y={x}^{\prime }\left(\frac{1}{\sqrt{5}}\right)+{y}^{\prime }\left(\frac{2}{\sqrt{5}}\right) \\ &y=\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}} \end{align}[/latex]

Substitute the expressions for [latex]x[/latex] and [latex]y[/latex] into in the given equation, and then simplify.

[latex]\begin{gathered}8{\left(\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}}\right)}^{2}-12\left(\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}}\right)\left(\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}}\right)+17{\left(\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}}\right)}^{2}=20 \\ 8\left(\frac{\left(2{x}^{\prime }-{y}^{\prime }\right)\left(2{x}^{\prime }-{y}^{\prime }\right)}{5}\right)-12\left(\frac{\left(2{x}^{\prime }-{y}^{\prime }\right)\left({x}^{\prime }+2{y}^{\prime }\right)}{5}\right)+17\left(\frac{\left({x}^{\prime }+2{y}^{\prime }\right)\left({x}^{\prime }+2{y}^{\prime }\right)}{5}\right)=20 \\ 8\left(4{x}^{\prime }{}^{2}-4{x}^{\prime }{y}^{\prime }+{y}^{\prime }{}^{2}\right)-12\left(2{x}^{\prime }{}^{2}+3{x}^{\prime }{y}^{\prime }-2{y}^{\prime }{}^{2}\right)+17\left({x}^{\prime }{}^{2}+4{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}\right)=100 \\ 32{x}^{\prime }{}^{2}-32{x}^{\prime }{y}^{\prime }+8{y}^{\prime }{}^{2}-24{x}^{\prime }{}^{2}-36{x}^{\prime }{y}^{\prime }+24{y}^{\prime }{}^{2}+17{x}^{\prime }{}^{2}+68{x}^{\prime }{y}^{\prime }+68{y}^{\prime }{}^{2}=100 \\ 25{x}^{\prime }{}^{2}+100{y}^{\prime }{}^{2}=100 \\ \frac{25}{100}{x}^{\prime }{}^{2}+\frac{100}{100}{y}^{\prime }{}^{2}=\frac{100}{100} \end{gathered}[/latex]

Write the equations with [latex]\begin{align}{x}^{\prime }\end{align}[/latex] and [latex]\begin{align}{y}^{\prime }\end{align}[/latex] in the standard form with respect to the new coordinate system.

[latex]\begin{align}\frac{{{x}^{\prime }}^{2}}{4}+\frac{{{y}^{\prime }}^{2}}{1}=1\end{align}[/latex]

Rewrite the [latex]13{x}^{2}-6\sqrt{3}xy+7{y}^{2}=16[/latex] in the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] system without the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] term.

Graph the following equation relative to the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] system:

[latex]{x}^{2}+12xy - 4{y}^{2}=30[/latex]

First, we find [latex]\cot \left(2\theta \right)[/latex].

[latex]{x}^{2}+12xy - 4{y}^{2}=20\Rightarrow A=1,B=12,\text{ and }C=-4[/latex]

[latex]\begin{align}&\cot \left(2\theta \right)=\frac{A-C}{B} \\ &\cot \left(2\theta \right)=\frac{1-\left(-4\right)}{12} \\ &\cot \left(2\theta \right)=\frac{5}{12} \end{align}[/latex]

Because [latex]\cot \left(2\theta \right)=\frac{5}{12}[/latex], we can draw a reference triangle.

[latex]\cot \left(2\theta \right)=\frac{5}{12}=\frac{\text{adjacent}}{\text{opposite}}[/latex]

Thus, the hypotenuse is

[latex]\begin{gathered} {5}^{2}+{12}^{2}={h}^{2}\\ 25+144={h}^{2}\\ 169={h}^{2}\\ h=13\end{gathered}[/latex]

Next, we find [latex]\sin \theta[/latex] and [latex]\cos \theta[/latex]. We will use half-angle identities.

[latex]\begin{align} \sin \theta &=\sqrt{\frac{1-\cos \left(2\theta \right)}{2}} \\ &=\sqrt{\frac{1-\frac{5}{13}}{2}} \\ &=\sqrt{\frac{\frac{13}{13}-\frac{5}{13}}{2}} \\&=\sqrt{\frac{8}{13}\cdot \frac{1}{2}} \\ &=\frac{2}{\sqrt{13}} \\ \cos \theta &=\sqrt{\frac{1+\cos \left(2\theta \right)}{2}} \\ &=\sqrt{\frac{1+\frac{5}{13}}{2}} \\ &=\sqrt{\frac{\frac{13}{13}+\frac{5}{13}}{2}} \\ &=\sqrt{\frac{18}{13}\cdot \frac{1}{2}} \\ &=\frac{3}{\sqrt{13}} \end{align}[/latex]

Now we find [latex]x[/latex] and [latex]y[/latex].

[latex]\begin{align} &x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \\ &x={x}^{\prime }\left(\frac{3}{\sqrt{13}}\right)-{y}^{\prime }\left(\frac{2}{\sqrt{13}}\right) \\ &x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}} \end{align}[/latex]

and

[latex]\begin{align} &y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \\ &y={x}^{\prime }\left(\frac{2}{\sqrt{13}}\right)+{y}^{\prime }\left(\frac{3}{\sqrt{13}}\right) \\ &y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}} \end{align}[/latex]

Now we substitute [latex]\begin{align}x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\end{align}[/latex] and [latex]\begin{align}y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\end{align}[/latex] into [latex]\begin{align}{x}^{2}+12xy - 4{y}^{2}=30\end{align}[/latex].

[latex]\begin{align} &{\left(\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\right)}^{2}+12\left(\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\right)\left(\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\right)-4{\left(\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\right)}^{2}=30 \\ &\left(\frac{1}{13}\right)\left[{\left(3{x}^{\prime }-2{y}^{\prime }\right)}^{2}+12\left(3{x}^{\prime }-2{y}^{\prime }\right)\left(2{x}^{\prime }+3{y}^{\prime }\right)-4{\left(2{x}^{\prime }+3{y}^{\prime }\right)}^{2}\right]=30 && \text{Factor}. \\ &\left(\frac{1}{13}\right)\left[9{x}^{\prime }{}^{2}-12{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}+12\left(6{x}^{\prime }{}^{2}+5{x}^{\prime }{y}^{\prime }-6{y}^{\prime }{}^{2}\right)-4\left(4{x}^{\prime }{}^{2}+12{x}^{\prime }{y}^{\prime }+9{y}^{\prime }{}^{2}\right)\right]=30 && \text{Multiply}. \\ &\left(\frac{1}{13}\right)\left[9{x}^{\prime }{}^{2}-12{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}+72{x}^{\prime }{}^{2}+60{x}^{\prime }{y}^{\prime }-72{y}^{\prime }{}^{2}-16{x}^{\prime }{}^{2}-48{x}^{\prime }{y}^{\prime }-36{y}^{\prime }{}^{2}\right]=30 && \text{Distribute}. \\ &\left(\frac{1}{13}\right)\left[65{x}^{\prime }{}^{2}-104{y}^{\prime }{}^{2}\right]=30 && \text{Combine like terms}. \\ &65{x}^{\prime }{}^{2}-104{y}^{\prime }{}^{2}=390 && \text{Multiply}. \\ &\frac{{x}^{\prime }{}^{2}}{6}-\frac{4{y}^{\prime }{}^{2}}{15}=1 && \text{Divide by 390}. \end{align}[/latex]

The graph of the hyperbola [latex]\begin{align}\frac{{{x}^{\prime }}^{2}}{6}-\frac{4{{y}^{\prime }}^{2}}{15}=1\end{align}[/latex].