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Study Guides > College Algebra

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 Ax2+Bxy+Cy2+Dx+Ey+F=0A{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{x}^{\prime } and y{y}^{\prime } coordinate system without the xy{x}^{\prime }{y}^{\prime } term, by rotating the axes by a measure of θ\theta that satisfies

cot(2θ)=ACB\cot \left(2\theta \right)=\frac{A-C}{B}
We have learned already that any conic may be represented by the second degree equation
Ax2+Bxy+Cy2+Dx+Ey+F=0A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0
where A,BA,B, and CC are not all zero. However, if B0B\ne 0, then we have an xyxy term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle θ\theta where cot(2θ)=ACB\cot \left(2\theta \right)=\frac{A-C}{B}.
  • If cot(2θ)>0\cot \left(2\theta \right)>0, then 2θ2\theta is in the first quadrant, and θ\theta is between (0,45)\left(0^\circ ,45^\circ \right).
  • If cot(2θ)<0\cot \left(2\theta \right)<0, then 2θ2\theta is in the second quadrant, and θ\theta is between (45,90)\left(45^\circ ,90^\circ \right).
  • If A=CA=C, then θ=45\theta =45^\circ .

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

  1. Find cot(2θ)\cot \left(2\theta \right).
  2. Find sin θ\sin \text{ }\theta and cos θ\cos \text{ }\theta .
  3. Substitute sin θ\sin \text{ }\theta and cos θ\cos \text{ }\theta into x=xcos θysin θx={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta and y=xsin θ+ycos θy={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta .
  4. Substitute the expression for xx and yy into in the given equation, and then simplify.
  5. Write the equations with x{x}^{\prime } and y{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 8x212xy+17y2=208{x}^{2}-12xy+17{y}^{2}=20 in the xy{x}^{\prime }{y}^{\prime } system without an xy{x}^{\prime }{y}^{\prime } term.

Solution

First, we find cot(2θ)\cot \left(2\theta \right).
8x212xy+17y2=20A=8,B=12andC=17 cot(2θ)=ACB=81712 cot(2θ)=912=34\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(2θ)=34=adjacentopposite\cot \left(2\theta \right)=\frac{3}{4}=\frac{\text{adjacent}}{\text{opposite}}
So the hypotenuse is
32+42=h29+16=h225=h2h=5\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 θ\sin \text{ }\theta and cos θ\cos \text{ }\theta .
sin θ=1cos(2θ)2=1352=55352=53512=210=15sin θ=15cos θ=1+cos(2θ)2=1+352=55+352=5+3512=810=45cos θ=25\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 θ\sin \text{ }\theta and cos θ\cos \text{ }\theta into x=xcos θysin θx={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta and y=xsin θ+ycos θy={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta .
x=xcos θysin θx=x(25)y(15)x=2xy5\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
y=xsin θ+ycos θy=x(15)+y(25)y=x+2y5\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 xx and yy into in the given equation, and then simplify.
 8(2xy5)212(2xy5)(x+2y5)+17(x+2y5)2=20  8((2xy)(2xy)5)12((2xy)(x+2y)5)+17((x+2y)(x+2y)5)=20  8(4x24xy+y2)12(2x2+3xy2y2)+17(x2+4xy+4y2)=10032x232xy+8y224x236xy+24y2+17x2+68xy+68y2=100 25x2+100y2=100  25100x2+100100y2=100100\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{x}^{\prime } and y{y}^{\prime } in the standard form with respect to the new coordinate system.
x24+y21=1\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 13x263xy+7y2=1613{x}^{2}-6\sqrt{3}xy+7{y}^{2}=16 in the xy{x}^{\prime }{y}^{\prime } system without the xy{x}^{\prime }{y}^{\prime } term. Solution

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

Graph the following equation relative to the xy{x}^{\prime }{y}^{\prime } system:
x2+12xy4y2=30{x}^{2}+12xy - 4{y}^{2}=30

Solution

First, we find cot(2θ)\cot \left(2\theta \right).
x2+12xy4y2=20A=1, B=12,and C=4{x}^{2}+12xy - 4{y}^{2}=20\Rightarrow A=1,\text{ }B=12,\text{and }C=-4
cot(2θ)=ACBcot(2θ)=1(4)12cot(2θ)=512\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(2θ)=512\cot \left(2\theta \right)=\frac{5}{12}, we can draw a reference triangle as in Figure 9.
Figure 9
cot(2θ)=512=adjacentopposite\cot \left(2\theta \right)=\frac{5}{12}=\frac{\text{adjacent}}{\text{opposite}}
Thus, the hypotenuse is
52+122=h225+144=h2169=h2h=13\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 θ\sin \text{ }\theta and cos θ\cos \text{ }\theta . We will use half-angle identities.
sin θ=1cos(2θ)2=15132=13135132=81312=213cos θ=1+cos(2θ)2=1+5132=1313+5132=181312=313\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 xx and y.y\text{.\hspace{0.17em}}
x=xcos θysin θx=x(313)y(213)x=3x2y13\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
y=xsin θ+ycos θy=x(213)+y(313)y=2x+3y13\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=3x2y13x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}} and y=2x+3y13y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}} into x2+12xy4y2=30{x}^{2}+12xy - 4{y}^{2}=30.
 (3x2y13)2+12(3x2y13)(2x+3y13)4(2x+3y13)2=30 (113)[(3x2y)2+12(3x2y)(2x+3y)4(2x+3y)2]=30Factor.(113)[9x212xy+4y2+12(6x2+5xy6y2)4(4x2+12xy+9y2)]=30Multiply. (113)[9x212xy+4y2+72x2+60xy72y216x248xy36y2]=30Distribute.  (113)[65x2104y2]=30Combine like terms. 65x2104y2=390Multiply.  x264y215=1Divide by 390.\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 x264y215=1. \frac{{{x}^{\prime }}^{2}}{6}-\frac{4{{y}^{\prime }}^{2}}{15}=1.\text{ }
Figure 10

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