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Identify the conic for each of the following without rotating axes.
Access this online resource for additional instruction and practice with conic sections and rotation of axes.
General Form equation of a conic section | $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ |
Rotation of a conic section | $$\begin{array}{l}x={x}^{\prime}\mathrm{cos}\text{}\theta -{y}^{\prime}\mathrm{sin}\text{}\theta \hfill \\ y={x}^{\prime}\mathrm{sin}\text{}\theta +{y}^{\prime}\mathrm{cos}\text{}\theta \hfill \end{array}$$ |
Angle of rotation | $\theta ,\text{where}\mathrm{cot}\left(2\theta \right)=\frac{A-C}{B}$ |
What effect does the $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term have on the graph of a conic section?
The $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term causes a rotation of the graph to occur.
If the equation of a conic section is written in the form $\text{\hspace{0.17em}}A{x}^{2}+B{y}^{2}+Cx+Dy+E=0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}AB=0,$ what can we conclude?
If the equation of a conic section is written in the form $\text{\hspace{0.17em}}A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0,$ and $\text{\hspace{0.17em}}{B}^{2}-4AC>0,$ what can we conclude?
The conic section is a hyperbola.
Given the equation $\text{\hspace{0.17em}}a{x}^{2}+4x+3{y}^{2}-12=0,$ what can we conclude if $\text{\hspace{0.17em}}a>0?$
For the equation $\text{\hspace{0.17em}}A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0,$ the value of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ that satisfies $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right)=\frac{A-C}{B}\text{\hspace{0.17em}}$ gives us what information?
It gives the angle of rotation of the axes in order to eliminate the $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term.
For the following exercises, determine which conic section is represented based on the given equation.
$9{x}^{2}+4{y}^{2}+72x+36y-500=0$
$2{x}^{2}-2{y}^{2}+4x-6y-2=0$
$4{y}^{2}-5x+9y+1=0$
$4{x}^{2}+9xy+4{y}^{2}-36y-125=0$
$-3{x}^{2}+3\sqrt{3}xy-4{y}^{2}+9=0$
$2{x}^{2}+4\sqrt{3}xy+6{y}^{2}-6x-3=0$
${B}^{2}-4AC=0,$ parabola
$-{x}^{2}+4\sqrt{2}xy+2{y}^{2}-2y+1=0$
$8{x}^{2}+4\sqrt{2}xy+4{y}^{2}-10x+1=0$
${B}^{2}-4AC=-96<0,$ ellipse
For the following exercises, find a new representation of the given equation after rotating through the given angle.
$3{x}^{2}+xy+3{y}^{2}-5=0,\theta =\mathrm{45\xb0}$
$4{x}^{2}-xy+4{y}^{2}-2=0,\theta =\mathrm{45\xb0}$
$7{{x}^{\prime}}^{2}+9{{y}^{\prime}}^{2}-4=0$
$2{x}^{2}+8xy-1=0,\theta =\mathrm{30\xb0}$
$-2{x}^{2}+8xy+1=0,\theta =\mathrm{45\xb0}$
$3{{x}^{\prime}}^{2}+2{x}^{\prime}{y}^{\prime}-5{{y}^{\prime}}^{2}+1=0$
$4{x}^{2}+\sqrt{2}xy+4{y}^{2}+y+2=0,\theta =\mathrm{45\xb0}$
For the following exercises, determine the angle $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ that will eliminate the $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term and write the corresponding equation without the $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term.
${x}^{2}+3\sqrt{3}xy+4{y}^{2}+y-2=0$
$\theta ={60}^{\circ},11{{x}^{\prime}}^{2}-{{y}^{\prime}}^{2}+\sqrt{3}{x}^{\prime}+{y}^{\prime}-4=0$
$4{x}^{2}+2\sqrt{3}xy+6{y}^{2}+y-2=0$
$9{x}^{2}-3\sqrt{3}xy+6{y}^{2}+4y-3=0$
$\theta ={150}^{\circ},21{{x}^{\prime}}^{2}+9{{y}^{\prime}}^{2}+4{x}^{\prime}-4\sqrt{3}{y}^{\prime}-6=0$
$\mathrm{-3}{x}^{2}-\sqrt{3}xy-2{y}^{2}-x=0$
$16{x}^{2}+24xy+9{y}^{2}+6x-6y+2=0$
$\theta \approx {36.9}^{\circ},125{{x}^{\prime}}^{2}+6{x}^{\prime}-42{y}^{\prime}+10=0$
${x}^{2}+4xy+4{y}^{2}+3x-2=0$
${x}^{2}+4xy+{y}^{2}-2x+1=0$
$\theta ={45}^{\circ},3{{x}^{\prime}}^{2}-{{y}^{\prime}}^{2}-\sqrt{2}{x}^{\prime}+\sqrt{2}{y}^{\prime}+1=0$
$4{x}^{2}-2\sqrt{3}xy+6{y}^{2}-1=0$
For the following exercises, rotate through the given angle based on the given equation. Give the new equation and graph the original and rotated equation.
$y=-{x}^{2},\theta =-{45}^{\circ}$
$\frac{\sqrt{2}}{2}\left({x}^{\prime}+{y}^{\prime}\right)=\frac{1}{2}{\left({x}^{\prime}-{y}^{\prime}\right)}^{2}$
$x={y}^{2},\theta ={45}^{\circ}$
$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{1}=1,\theta ={45}^{\circ}$
$\frac{{\left({x}^{\prime}-{y}^{\prime}\right)}^{2}}{8}+\frac{{\left({x}^{\prime}+{y}^{\prime}\right)}^{2}}{2}=1$
$\frac{{y}^{2}}{16}+\frac{{x}^{2}}{9}=1,\theta ={45}^{\circ}$
${y}^{2}-{x}^{2}=1,\theta ={45}^{\circ}$
$\frac{{\left({x}^{\prime}+{y}^{\prime}\right)}^{2}}{2}-\frac{{\left({x}^{\prime}-{y}^{\prime}\right)}^{2}}{2}=1$
$y=\frac{{x}^{2}}{2},\theta ={30}^{\circ}$
$x={\left(y-1\right)}^{2},\theta ={30}^{\circ}$
$\frac{\sqrt{3}}{2}{x}^{\prime}-\frac{1}{2}{y}^{\prime}={\left(\frac{1}{2}{x}^{\prime}+\frac{\sqrt{3}}{2}{y}^{\prime}-1\right)}^{2}$
$\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1,\theta ={30}^{\circ}$
For the following exercises, graph the equation relative to the $\text{\hspace{0.17em}}{x}^{\prime}{y}^{\prime}\text{\hspace{0.17em}}$ system in which the equation has no $\text{\hspace{0.17em}}{x}^{\prime}{y}^{\prime}\text{\hspace{0.17em}}$ term.
${x}^{2}+10xy+{y}^{2}-6=0$
$4{x}^{2}-3\sqrt{3}xy+{y}^{2}-22=0$
$11{x}^{2}+10\sqrt{3}xy+{y}^{2}-64=0$
$16{x}^{2}+24xy+9{y}^{2}-130x+90y=0$
$13{x}^{2}-6\sqrt{3}xy+7{y}^{2}-16=0$
For the following exercises, determine the angle of rotation in order to eliminate the $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term. Then graph the new set of axes.
$6{x}^{2}-5\sqrt{3}xy+{y}^{2}+10x-12y=0$
$6{x}^{2}-8\sqrt{3}xy+14{y}^{2}+10x-3y=0$
$4{x}^{2}+6\sqrt{3}xy+10{y}^{2}+20x-40y=0$
$\theta ={60}^{\circ}$
$8{x}^{2}+3xy+4{y}^{2}+2x-4=0$
$16{x}^{2}+24xy+9{y}^{2}+20x-44y=0$
$\theta \approx {36.9}^{\circ}$
For the following exercises, determine the value of $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ based on the given equation.
Given $\text{\hspace{0.17em}}4{x}^{2}+kxy+16{y}^{2}+8x+24y-48=0,$ find $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ for the graph to be a parabola.
Given $\text{\hspace{0.17em}}2{x}^{2}+kxy+12{y}^{2}+10x-16y+28=0,$ find $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ for the graph to be an ellipse.
$-4\sqrt{6}<k<4\sqrt{6}$
Given $\text{\hspace{0.17em}}3{x}^{2}+kxy+4{y}^{2}-6x+20y+128=0,$ find $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ for the graph to be a hyperbola.
Given $\text{\hspace{0.17em}}k{x}^{2}+8xy+8{y}^{2}-12x+16y+18=0,$ find $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ for the graph to be a parabola.
$k=2$
Given $\text{\hspace{0.17em}}6{x}^{2}+12xy+k{y}^{2}+16x+10y+4=0,$ find $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ for the graph to be an ellipse.
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