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 θθ, then every point on the plane may be thought of as having two representations: (x,y)(x,y) on the Cartesian plane with the original x-axis and y-axis, and (x,y)(x,y) 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=0x2+y2xy15=0

We will find the relationships between xx and yy on the Cartesian plane with xx and yy 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 θ.

The original coordinate x– and y-axes have unit vectors ii and jj. The rotated coordinate axes have unit vectors ii and jj. The angle θθ 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 θji=cos θi+sin θjj=sin θi+cos θj

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.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.

Because u=xi+yju=xi+yj, we have representations of xx and yy in terms of the new coordinate system.

x=xcos θysin θandy=xsin θ+ycos θx=xcos θysin θandy=xsin θ+ycos θ

A General Note: Equations of Rotation

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

x=xcos θysin θx=xcos θysin θ

and

y=xsin θ+ycos θy=xsin θ+ycos θ

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=xcos θysin θ and y=xsin θ+ycos θy=xsin θ+ycos θ.
  2. Substitute the expression for xx and yy into in the given equation, then simplify.
  3. Write the equations with xx and yy 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=02x2xy+2y230=0 after rotating through an angle of θ=45θ=45.

Solution

Find xx and yy, where x=xcos θysin θx=xcos θysin θ and y=xsin θ+ycos θy=xsin θ+ycos θ.

Because θ=45θ=45,

x=xcos(45)ysin(45)x=x(12)y(12)x=xy2x=xcos(45)ysin(45)x=x(12)y(12)x=xy2

and

y=xsin(45)+ycos(45)y=x(12)+y(12)y=x+y2y=xsin(45)+ycos(45)y=x(12)+y(12)y=x+y2

Substitute x=xcosθysinθx=xcosθysinθ and y=xsin θ+ycos θy=xsin θ+ycos θ into 2x2xy+2y230=02x2xy+2y230=0.

2(xy2)2(xy2)(x+y2)+2(x+y2)230=02(xy2)2(xy2)(x+y2)+2(x+y2)230=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.¯¯¯¯¯)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.

Write the equations with xx and yy in the standard form.

x220+y212=1x220+y212=1

This equation is an ellipse. Figure 6 shows the graph.

Figure 6