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Study Guides > College Algebra

Finding a New Representation of the Given Equation after Rotating through a Given Angle

Until now, we have looked at equations of conic sections without an xyxy term, which aligns the graphs with the x- and y-axes. When we add an xyxy term, we are rotating the conic about the origin. If the x- and y-axes are rotated through an angle, say θ\theta , then every point on the plane may be thought of as having two representations: (x,y)\left(x,y\right) on the Cartesian plane with the original x-axis and y-axis, and (x,y)\left({x}^{\prime },{y}^{\prime }\right) on the new plane defined by the new, rotated axes, called the x'-axis and y'-axis.

Figure 3. The graph of the rotated ellipse x2+y2xy15=0{x}^{2}+{y}^{2}-xy - 15=0
We will find the relationships between xx and yy on the Cartesian plane with x{x}^{\prime } and y{y}^{\prime } on the new rotated plane.
Figure 4. The Cartesian plane with x- and y-axes and the resulting x′− and y′−axes formed by a rotation by an angle  θ\text{ }\theta .
The original coordinate x- and y-axes have unit vectors ii and jj. The rotated coordinate axes have unit vectors i{i}^{\prime } and j{j}^{\prime }. The angle θ\theta is known as the angle of rotation. We may write the new unit vectors in terms of the original ones.

i=cos θi+sin θjj=sin θi+cos θj\begin{array}{l}{i}^{\prime }=\cos \text{ }\theta i+\sin \text{ }\theta j\hfill \\ {j}^{\prime }=-\sin \text{ }\theta i+\cos \text{ }\theta j\hfill \end{array}
Figure 5. Relationship between the old and new coordinate planes.
Consider a vector uu in the new coordinate plane. It may be represented in terms of its coordinate axes.
u=xi+yju=x(i cos θ+j sin θ)+y(i sin θ+j cos θ)Substitute.u=ix’ cos θ+jx’ sin θiy’ sin θ+jy’ cos θDistribute.u=ix’ cos θiy’ sin θ+jx’ sin θ+jy’ cos θApply commutative property.u=(x’ cos θy’ sin θ)i+(x’ sin θ+y’ cos θ)jFactor by grouping.\begin{array}{ll}u={x}^{\prime }{i}^{\prime }+{y}^{\prime }{j}^{\prime }\hfill & \hfill \\ u={x}^{\prime }\left(i\text{ }\cos \text{ }\theta +j\text{ }\sin \text{ }\theta \right)+{y}^{\prime }\left(-i\text{ }\sin \text{ }\theta +j\text{ }\cos \text{ }\theta \right)\hfill & \begin{array}{cccc}& & & \end{array}\text{Substitute}.\hfill \\ u=ix\text{'}\text{ }\cos \text{ }\theta +jx\text{'}\text{ }\sin \text{ }\theta -iy\text{'}\text{ }\sin \text{ }\theta +jy\text{'}\text{ }\cos \text{ }\theta \hfill & \begin{array}{cccc}& & & \end{array}\text{Distribute}.\hfill \\ u=ix\text{'}\text{ }\cos \text{ }\theta -iy\text{'}\text{ }\sin \text{ }\theta +jx\text{'}\text{ }\sin \text{ }\theta +jy\text{'}\text{ }\cos \text{ }\theta \hfill & \begin{array}{cccc}& & & \end{array}\text{Apply commutative property}.\hfill \\ u=\left(x\text{'}\text{ }\cos \text{ }\theta -y\text{'}\text{ }\sin \text{ }\theta \right)i+\left(x\text{'}\text{ }\sin \text{ }\theta +y\text{'}\text{ }\cos \text{ }\theta \right)j\hfill & \begin{array}{cccc}& & & \end{array}\text{Factor by grouping}.\hfill \end{array}
Because u=xi+yju={x}^{\prime }{i}^{\prime }+{y}^{\prime }{j}^{\prime }, we have representations of xx and yy in terms of the new coordinate system.
x=xcos θysin θandy=xsin θ+ycos θ\begin{array}{c}x={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta \\ \text{and}\\ y={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta \end{array}

A General Note: Equations of Rotation

If a point (x,y)\left(x,y\right) on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an angle θ\theta from the positive x-axis, then the coordinates of the point with respect to the new axes are (x,y)\left({x}^{\prime },{y}^{\prime }\right). We can use the following equations of rotation to define the relationship between (x,y)\left(x,y\right) and (x,y):\left({x}^{\prime },{y}^{\prime }\right):
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

How To: Given the equation of a conic, find a new representation after rotating through an angle.

  1. Find xx and yy where 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 .
  2. Substitute the expression for xx and yy into in the given equation, then simplify.
  3. Write the equations with x{x}^{\prime } and y{y}^{\prime } in standard form.

Example 2: Finding a New Representation of an Equation after Rotating through a Given Angle

Find a new representation of the equation 2x2xy+2y230=02{x}^{2}-xy+2{y}^{2}-30=0 after rotating through an angle of θ=45\theta =45^\circ .

Solution

Find xx and yy, where 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 . Because θ=45\theta =45^\circ ,
x=xcos(45)ysin(45)x=x(12)y(12)x=xy2\begin{array}{l}\hfill \\ x={x}^{\prime }\cos \left(45^\circ \right)-{y}^{\prime }\sin \left(45^\circ \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
y=xsin(45)+ycos(45)y=x(12)+y(12)y=x+y2\begin{array}{l}\\ \begin{array}{l}y={x}^{\prime }\sin \left(45^\circ \right)+{y}^{\prime }\cos \left(45^\circ \right)\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=xcosθysinθx={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta and y=xsin θ+ycos θy={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta into 2x2xy+2y230=02{x}^{2}-xy+2{y}^{2}-30=0.
2(xy2)2(xy2)(x+y2)+2(x+y2)230=02{\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.
)2(xy)(xy))2(xy)(x+y)2+)2(x+y)(x+y))230=0FOIL method x2)2xy+y2(x2y2)2+x2)+2xy+y230=0Combine like terms. 2x2+2y2(x2y2)2=30Combine like terms. 2(2x2+2y2(x2y2)2)=2(30)Multiply both sides by 2. 4x2+4y2(x2y2)=60Simplify. 4x2+4y2x2+y2=60Distribute. 3x260+5y260=6060Set equal to 1.\begin{array}{ll}\overline{)2}\frac{\left({x}^{\prime }-{y}^{\prime }\right)\left({x}^{\prime }-{y}^{\prime }\right)}{\overline{)2}}-\frac{\left({x}^{\prime }-{y}^{\prime }\right)\left({x}^{\prime }+{y}^{\prime }\right)}{2}+\overline{)2}\frac{\left({x}^{\prime }+{y}^{\prime }\right)\left({x}^{\prime }+{y}^{\prime }\right)}{\overline{)2}}-30=0\hfill & \begin{array}{cccc}& & & \end{array}\text{FOIL method}\hfill \\ \text{ }{x}^{\prime }{}^{2}{\overline{)-2{x}^{\prime }y}}^{\prime }+{y}^{\prime }{}^{2}-\frac{\left({x}^{\prime }{}^{2}-{y}^{\prime }{}^{2}\right)}{2}+{x}^{\prime }{}^{2}\overline{)+2{x}^{\prime }{y}^{\prime }}+{y}^{\prime }{}^{2}-30=0\hfill & \begin{array}{cccc}& & & \end{array}\text{Combine like terms}.\hfill \\ \text{ }2{x}^{\prime }{}^{2}+2{y}^{\prime }{}^{2}-\frac{\left({x}^{\prime }{}^{2}-{y}^{\prime }{}^{2}\right)}{2}=30\hfill & \begin{array}{cccc}& & & \end{array}\text{Combine like terms}.\hfill \\ \text{ }2\left(2{x}^{\prime }{}^{2}+2{y}^{\prime }{}^{2}-\frac{\left({x}^{\prime }{}^{2}-{y}^{\prime }{}^{2}\right)}{2}\right)=2\left(30\right)\hfill & \begin{array}{cccc}& & & \end{array}\text{Multiply both sides by 2}.\hfill \\ \text{ }4{x}^{\prime }{}^{2}+4{y}^{\prime }{}^{2}-\left({x}^{\prime }{}^{2}-{y}^{\prime }{}^{2}\right)=60\hfill & \begin{array}{cccc}& & & \end{array}\text{Simplify}.\hfill \\ \text{ }4{x}^{\prime }{}^{2}+4{y}^{\prime }{}^{2}-{x}^{\prime }{}^{2}+{y}^{\prime }{}^{2}=60\hfill & \begin{array}{cccc}& & & \end{array}\text{Distribute}.\hfill \\ \text{ }\frac{3{x}^{\prime }{}^{2}}{60}+\frac{5{y}^{\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{x}^{\prime } and y{y}^{\prime } in the standard form.
x220+y212=1\frac{{{x}^{\prime }}^{2}}{20}+\frac{{{y}^{\prime }}^{2}}{12}=1
This equation is an ellipse. Figure 6 shows the graph.
Figure 6

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