## 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

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

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

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

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

### Example 3: Rewriting an Equation with respect to the x′ and y′ axes without the x′y′ Term

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

### Solution

First, we find $\cot \left(2\theta \right)$.

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

Figure 7

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

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 $\sin \text{ }\theta$ and $\cos \text{ }\theta$.

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

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

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

and

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

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

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

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

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

Figure 8 shows the graph of the ellipse.

Figure 8

### Try It 2

Rewrite the $13{x}^{2}-6\sqrt{3}xy+7{y}^{2}=16$ in the ${x}^{\prime }{y}^{\prime }$ system without the ${x}^{\prime }{y}^{\prime }$ term.

Solution

### Example 4: Graphing an Equation That Has No x′y′ Terms

Graph the following equation relative to the ${x}^{\prime }{y}^{\prime }$ system:

${x}^{2}+12xy - 4{y}^{2}=30$

### Solution

First, we find $\cot \left(2\theta \right)$.

${x}^{2}+12xy - 4{y}^{2}=20\Rightarrow A=1,\text{ }B=12,\text{and }C=-4$
$\begin{array}{l}\cot \left(2\theta \right)=\frac{A-C}{B}\hfill \\ \cot \left(2\theta \right)=\frac{1-\left(-4\right)}{12}\hfill \\ \cot \left(2\theta \right)=\frac{5}{12}\hfill \end{array}$

Because $\cot \left(2\theta \right)=\frac{5}{12}$, we can draw a reference triangle as in Figure 9.

Figure 9

$\cot \left(2\theta \right)=\frac{5}{12}=\frac{\text{adjacent}}{\text{opposite}}$

Thus, the hypotenuse is

$\begin{array}{r}\hfill {5}^{2}+{12}^{2}={h}^{2}\\ \hfill 25+144={h}^{2}\\ \hfill 169={h}^{2}\\ \hfill h=13\end{array}$

Next, we find $\sin \text{ }\theta$ and $\cos \text{ }\theta$. We will use half-angle identities.

$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ \sin \text{ }\theta =\sqrt{\frac{1-\cos \left(2\theta \right)}{2}}=\sqrt{\frac{1-\frac{5}{13}}{2}}=\sqrt{\frac{\frac{13}{13}-\frac{5}{13}}{2}}=\sqrt{\frac{8}{13}\cdot \frac{1}{2}}=\frac{2}{\sqrt{13}}\hfill \end{array}\hfill \\ \cos \text{ }\theta =\sqrt{\frac{1+\cos \left(2\theta \right)}{2}}=\sqrt{\frac{1+\frac{5}{13}}{2}}=\sqrt{\frac{\frac{13}{13}+\frac{5}{13}}{2}}=\sqrt{\frac{18}{13}\cdot \frac{1}{2}}=\frac{3}{\sqrt{13}}\hfill \end{array}$

Now we find $x$ and $y\text{.\hspace{0.17em}}$

$\begin{array}{l}\hfill \\ x={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta \hfill \\ x={x}^{\prime }\left(\frac{3}{\sqrt{13}}\right)-{y}^{\prime }\left(\frac{2}{\sqrt{13}}\right)\hfill \\ x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\hfill \end{array}$

and

$\begin{array}{l}\hfill \\ y={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta \hfill \\ y={x}^{\prime }\left(\frac{2}{\sqrt{13}}\right)+{y}^{\prime }\left(\frac{3}{\sqrt{13}}\right)\hfill \\ y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\hfill \end{array}$

Now we substitute $x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}$ and $y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}$ into ${x}^{2}+12xy - 4{y}^{2}=30$.

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

Figure 10 shows the graph of the hyperbola $\frac{{{x}^{\prime }}^{2}}{6}-\frac{4{{y}^{\prime }}^{2}}{15}=1.\text{ }$

Figure 10