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Guide allo studio > Precalculus II

Rotation of Axes

Learning Objectives

By the end of this section, you will be able to:
  • Identify nondegenerate conic sections given their general form equations.
  • Use rotation of axes formulas.
  • Write equations of rotated conics in standard form.
  • Identify conics without rotating axes.
As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which we slice the cone will determine the type of conic section formed at the intersection. A circle is formed by slicing a cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices both the top and bottom of the cone.
Figure 1. The nondegenerate conic sections
Ellipses, circles, hyperbolas, and parabolas are sometimes called the nondegenerate conic sections, in contrast to the degenerate conic sections, which are shown in Figure 2. A degenerate conic results when a plane intersects the double cone and passes through the apex. Depending on the angle of the plane, three types of degenerate conic sections are possible: a point, a line, or two intersecting lines.
Figure 2. Degenerate conic sections

Identifying Nondegenerate Conics in General Form

In previous sections of this chapter, we have focused on the standard form equations for nondegenerate conic sections. In this section, we will shift our focus to the general form equation, which can be used for any conic. The general form is set equal to zero, and the terms and coefficients are given in a particular order, as shown below.
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. We can use the values of the coefficients to identify which type conic is represented by a given equation. You may notice that the general form equation has an xyxy term that we have not seen in any of the standard form equations. As we will discuss later, the xyxy term rotates the conic whenever  B \text{ }B\text{ } is not equal to zero.
Conic Sections Example
ellipse 4x2+9y2=14{x}^{2}+9{y}^{2}=1
circle 4x2+4y2=14{x}^{2}+4{y}^{2}=1
hyperbola 4x2−9y2=14{x}^{2}-9{y}^{2}=1
parabola 4x2=9y or 4y2=9x4{x}^{2}=9y\text{ or }4{y}^{2}=9x
one line 4x+9y=14x+9y=1
intersecting lines (x−4)(y+4)=0\left(x - 4\right)\left(y+4\right)=0
parallel lines (x−4)(x−9)=0\left(x - 4\right)\left(x - 9\right)=0
a point 4x2+4y2=04{x}^{2}+4{y}^{2}=0
no graph 4x2+4y2=−14{x}^{2}+4{y}^{2}=-1

A General Note: General Form of Conic Sections

A nondegenerate conic section has the general form
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. The table below summarizes the different conic sections where B=0B=0, and AA and CC are nonzero real numbers. This indicates that the conic has not been rotated.
ellipse Ax2+Cy2+Dx+Ey+F=0, A≠C and AC>0A{x}^{2}+C{y}^{2}+Dx+Ey+F=0,\text{ }A\ne C\text{ and }AC>0
circle Ax2+Cy2+Dx+Ey+F=0, A=CA{x}^{2}+C{y}^{2}+Dx+Ey+F=0,\text{ }A=C
hyperbola Ax2−Cy2+Dx+Ey+F=0 or −Ax2+Cy2+Dx+Ey+F=0A{x}^{2}-C{y}^{2}+Dx+Ey+F=0\text{ or }-A{x}^{2}+C{y}^{2}+Dx+Ey+F=0, where AA and CC are positive
parabola Ax2+Dx+Ey+F=0 or Cy2+Dx+Ey+F=0A{x}^{2}+Dx+Ey+F=0\text{ or }C{y}^{2}+Dx+Ey+F=0

How To: Given the equation of a conic, identify the type of conic.

  1. Rewrite the equation in the general form, Ax2+Bxy+Cy2+Dx+Ey+F=0A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0.
  2. Identify the values of AA and CC from the general form.
    1. If AA and CC are nonzero, have the same sign, and are not equal to each other, then the graph is an ellipse.
    2. If AA and CC are equal and nonzero and have the same sign, then the graph is a circle.
    3. If AA and CC are nonzero and have opposite signs, then the graph is a hyperbola.
    4. If either AA or CC is zero, then the graph is a parabola.

Example 1: Identifying a Conic from Its General Form

Identify the graph of each of the following nondegenerate conic sections.
  1. 4x2−9y2+36x+36y−125=04{x}^{2}-9{y}^{2}+36x+36y - 125=0
  2. 9y2+16x+36y−10=09{y}^{2}+16x+36y - 10=0
  3. 3x2+3y2−2x−6y−4=03{x}^{2}+3{y}^{2}-2x - 6y - 4=0
  4. −25x2−4y2+100x+16y+20=0-25{x}^{2}-4{y}^{2}+100x+16y+20=0

Solution

  1. Rewriting the general form, we have A=4A=4 and C=−9C=-9, so we observe that AA and CC have opposite signs. The graph of this equation is a hyperbola.
  2. Rewriting the general form, we have A=0A=0 and C=9C=9. We can determine that the equation is a parabola, since AA is zero.
  3. Rewriting the general form, we have A=3A=3 and C=3C=3. Because A=CA=C, the graph of this equation is a circle.
  4. Rewriting the general form, we have A=−25A=-25 and C=−4C=-4. Because AC>0AC>0 and A≠CA\ne C, the graph of this equation is an ellipse.

Try It 1

Identify the graph of each of the following nondegenerate conic sections.
  1. 16y2−x2+x−4y−9=016{y}^{2}-{x}^{2}+x - 4y - 9=0
  2. 16x2+4y2+16x+49y−81=016{x}^{2}+4{y}^{2}+16x+49y - 81=0
Solution

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+y2−xy−15=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=x′i′+y′j′u=x′(i cos⁡ θ+j sin⁡ θ)+y′(−i sin⁡ θ+j cos⁡ θ)Substitute.u=ix’ cos⁡ θ+jx’ sin⁡ θ−iy’ sin⁡ θ+jy’ cos⁡ θDistribute.u=ix’ cos⁡ θ−iy’ sin⁡ θ+jx’ sin⁡ θ+jy’ cos⁡ θApply commutative property.u=(x’ cos⁡ θ−y’ 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=x′i′+y′j′u={x}^{\prime }{i}^{\prime }+{y}^{\prime }{j}^{\prime }, we have representations of xx and yy in terms of the new coordinate system.
x=x′cos⁡ θ−y′sin⁡ θandy=x′sin⁡ θ+y′cos⁡ 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=x′cos⁡ θ−y′sin⁡ θx={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta
and
y=x′sin⁡ θ+y′cos⁡ 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=x′cos⁡ θ−y′sin⁡ θx={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta and y=x′sin⁡ θ+y′cos⁡ 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 2x2−xy+2y2−30=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=x′cos⁡ θ−y′sin⁡ θx={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta and y=x′sin⁡ θ+y′cos⁡ θy={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta . Because θ=45∘\theta =45^\circ ,
x=x′cos⁡(45∘)−y′sin⁡(45∘)x=x′(12)−y′(12)x=x′−y′2\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=x′sin⁡(45∘)+y′cos⁡(45∘)y=x′(12)+y′(12)y=x′+y′2\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=x′cos⁡θ−y′sin⁡θx={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta and y=x′sin⁡ θ+y′cos⁡ θy={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta into 2x2−xy+2y2−30=02{x}^{2}-xy+2{y}^{2}-30=0.
2(x′−y′2)2−(x′−y′2)(x′+y′2)+2(x′+y′2)2−30=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‾(x′−y′)(x′−y′))2‾−(x′−y′)(x′+y′)2+)2‾(x′+y′)(x′+y′))2‾−30=0FOIL method x′2)−2x′y‾′+y′2−(x′2−y′2)2+x′2)+2x′y′‾+y′2−30=0Combine like terms. 2x′2+2y′2−(x′2−y′2)2=30Combine like terms. 2(2x′2+2y′2−(x′2−y′2)2)=2(30)Multiply both sides by 2. 4x′2+4y′2−(x′2−y′2)=60Simplify. 4x′2+4y′2−x′2+y′2=60Distribute. 3x′260+5y′260=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.
x′220+y′212=1\frac{{{x}^{\prime }}^{2}}{20}+\frac{{{y}^{\prime }}^{2}}{12}=1
This equation is an ellipse. Figure 6 shows the graph.
Figure 6

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 x′y′{x}^{\prime }{y}^{\prime } term, by rotating the axes by a measure of θ\theta that satisfies
cot⁡(2θ)=A−CB\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 B≠0B\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)=A−CB\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 x′y′{x}^{\prime }{y}^{\prime } system, rewrite the equation without the x′y′{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=x′cos⁡ θ−y′sin⁡ θx={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta and y=x′sin⁡ θ+y′cos⁡ 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 8x2−12xy+17y2=208{x}^{2}-12xy+17{y}^{2}=20 in the x′y′{x}^{\prime }{y}^{\prime } system without an x′y′{x}^{\prime }{y}^{\prime } term.

Solution

First, we find cot⁡(2θ)\cot \left(2\theta \right).
8x2−12xy+17y2=20⇒A=8,B=−12andC=17 cot⁡(2θ)=A−CB=8−17−12 cot⁡(2θ)=−9−12=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=1−cos⁡(2θ)2=1−352=55−352=5−35⋅12=210=15sin⁡ θ=15cos⁡ θ=1+cos⁡(2θ)2=1+352=55+352=5+35⋅12=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=x′cos⁡ θ−y′sin⁡ θx={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta and y=x′sin⁡ θ+y′cos⁡ θy={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta .
x=x′cos⁡ θ−y′sin⁡ θx=x′(25)−y′(15)x=2x′−y′5\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=x′sin⁡ θ+y′cos⁡ θy=x′(15)+y′(25)y=x′+2y′5\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(2x′−y′5)2−12(2x′−y′5)(x′+2y′5)+17(x′+2y′5)2=20  8((2x′−y′)(2x′−y′)5)−12((2x′−y′)(x′+2y′)5)+17((x′+2y′)(x′+2y′)5)=20  8(4x′2−4x′y′+y′2)−12(2x′2+3x′y′−2y′2)+17(x′2+4x′y′+4y′2)=10032x′2−32x′y′+8y′2−24x′2−36x′y′+24y′2+17x′2+68x′y′+68y′2=100 25x′2+100y′2=100  25100x′2+100100y′2=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.
x′24+y′21=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 13x2−63xy+7y2=1613{x}^{2}-6\sqrt{3}xy+7{y}^{2}=16 in the x′y′{x}^{\prime }{y}^{\prime } system without the x′y′{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′y′{x}^{\prime }{y}^{\prime } system:
x2+12xy−4y2=30{x}^{2}+12xy - 4{y}^{2}=30

Solution

First, we find cot⁡(2θ)\cot \left(2\theta \right).
x2+12xy−4y2=20⇒A=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)=A−CBcot⁡(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=1−cos⁡(2θ)2=1−5132=1313−5132=813⋅12=213cos⁡ θ=1+cos⁡(2θ)2=1+5132=1313+5132=1813⋅12=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=x′cos⁡ θ−y′sin⁡ θx=x′(313)−y′(213)x=3x′−2y′13\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=x′sin⁡ θ+y′cos⁡ θy=x′(213)+y′(313)y=2x′+3y′13\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=3x′−2y′13x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}} and y=2x′+3y′13y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}} into x2+12xy−4y2=30{x}^{2}+12xy - 4{y}^{2}=30.
 (3x′−2y′13)2+12(3x′−2y′13)(2x′+3y′13)−4(2x′+3y′13)2=30 (113)[(3x′−2y′)2+12(3x′−2y′)(2x′+3y′)−4(2x′+3y′)2]=30Factor.(113)[9x′2−12x′y′+4y′2+12(6x′2+5x′y′−6y′2)−4(4x′2+12x′y′+9y′2)]=30Multiply. (113)[9x′2−12x′y′+4y′2+72x′2+60x′y′−72y′2−16x′2−48x′y′−36y′2]=30Distribute.  (113)[65x′2−104y′2]=30Combine like terms. 65x′2−104y′2=390Multiply.  x′26−4y′215=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 x′26−4y′215=1. \frac{{{x}^{\prime }}^{2}}{6}-\frac{4{{y}^{\prime }}^{2}}{15}=1.\text{ }
Figure 10

Identifying Conics without Rotating Axes

Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is
Ax2+Bxy+Cy2+Dx+Ey+F=0A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0
If we apply the rotation formulas to this equation we get the form
A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0{A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime }{x}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime }{x}^{\prime }+{E}^{\prime }{y}^{\prime }+{F}^{\prime }=0
It may be shown that B2−4AC=B′2−4A′C′{B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }. The expression does not vary after rotation, so we call the expression invariant. The discriminant, B2−4AC{B}^{2}-4AC, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section.

A General Note: Using the Discriminant to Identify a Conic

If the equation Ax2+Bxy+Cy2+Dx+Ey+F=0A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0 is transformed by rotating axes into the equation A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0{A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime }{x}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime }{x}^{\prime }+{E}^{\prime }{y}^{\prime }+{F}^{\prime }=0, then B2−4AC=B′2−4A′C′{B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }. The equation Ax2+Bxy+Cy2+Dx+Ey+F=0A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0 is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these. If the discriminant, B2−4AC{B}^{2}-4AC, is
  • <0<0, the conic section is an ellipse
  • =0=0, the conic section is a parabola
  • >0>0, the conic section is a hyperbola

Example 5: Identifying the Conic without Rotating Axes

Identify the conic for each of the following without rotating axes.
  1. 5x2+23xy+2y2−5=05{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0
  2. 5x2+23xy+12y2−5=05{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0

Solution

  1. Let’s begin by determining A,BA,B, and CC.
    5⏟Ax2+23⏟Bxy+2⏟Cy2−5=0\underset{A}{\underbrace{5}}{x}^{2}+\underset{B}{\underbrace{2\sqrt{3}}}xy+\underset{C}{\underbrace{2}}{y}^{2}-5=0
    Now, we find the discriminant.
    B2−4AC=(23)2−4(5)(2) =4(3)−40 =12−40 =−28<0\begin{array}{l}{B}^{2}-4AC={\left(2\sqrt{3}\right)}^{2}-4\left(5\right)\left(2\right)\hfill \\ \text{ }=4\left(3\right)-40\hfill \\ \text{ }=12 - 40\hfill \\ \text{ }=-28<0\hfill \end{array}
    Therefore, 5x2+23xy+2y2−5=05{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0 represents an ellipse.
  2. Again, let’s begin by determining A,BA,B, and CC.
    5⏟Ax2+23⏟Bxy+12⏟Cy2−5=0\underset{A}{\underbrace{5}}{x}^{2}+\underset{B}{\underbrace{2\sqrt{3}}}xy+\underset{C}{\underbrace{12}}{y}^{2}-5=0
    Now, we find the discriminant.
    B2−4AC=(23)2−4(5)(12) =4(3)−240 =12−240 =−228<0\begin{array}{l}{B}^{2}-4AC={\left(2\sqrt{3}\right)}^{2}-4\left(5\right)\left(12\right)\hfill \\ \text{ }=4\left(3\right)-240\hfill \\ \text{ }=12 - 240\hfill \\ \text{ }=-228<0\hfill \end{array}
    Therefore, 5x2+23xy+12y2−5=05{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0 represents an ellipse.

Try It 3

Identify the conic for each of the following without rotating axes.
  1. x2−9xy+3y2−12=0{x}^{2}-9xy+3{y}^{2}-12=0
  2. 10x2−9xy+4y2−4=010{x}^{2}-9xy+4{y}^{2}-4=0
Solution

Key Equations

General Form equation of a conic section Ax2+Bxy+Cy2+Dx+Ey+F=0A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0
Rotation of a conic section x=x′cos⁡ θ−y′sin⁡ θy=x′sin⁡ θ+y′cos⁡ θ\begin{array}{l}x={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta \hfill \\ y={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta \hfill \end{array}
Angle of rotation θ,where cot⁡(2θ)=A−CB\theta ,\text{where }\cot \left(2\theta \right)=\frac{A-C}{B}

Key Concepts

  • Four basic shapes can result from the intersection of a plane with a pair of right circular cones connected tail to tail. They include an ellipse, a circle, a hyperbola, and a parabola.
  • A nondegenerate conic section has the general form 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. The values of A,BA,B, and CC determine the type of conic.
  • Equations of conic sections with an xyxy term have been rotated about the origin.
  • The general form can be transformed into an equation in the x′{x}^{\prime } and y′{y}^{\prime } coordinate system without the x′y′{x}^{\prime }{y}^{\prime } term.
  • An expression is described as invariant if it remains unchanged after rotating. Because the discriminant is invariant, observing it enables us to identify the conic section.

Glossary

angle of rotation
an acute angle formed by a set of axes rotated from the Cartesian plane where, if cot⁡(2θ)>0\cot \left(2\theta \right)>0, then θ\theta is between (0∘,45∘)\left(0^\circ ,45^\circ \right); if cot⁡(2θ)<0\cot \left(2\theta \right)<0, then θ\theta is between (45∘,90∘)\left(45^\circ ,90^\circ \right); and if cot⁡(2θ)=0\cot \left(2\theta \right)=0, then θ=45∘\theta =45^\circ
degenerate conic sections
any of the possible shapes formed when a plane intersects a double cone through the apex. Types of degenerate conic sections include a point, a line, and intersecting lines.
nondegenerate conic section
a shape formed by the intersection of a plane with a double right cone such that the plane does not pass through the apex; nondegenerate conics include circles, ellipses, hyperbolas, and parabolas

Section Exercises

1. What effect does the xyxy term have on the graph of a conic section? 2. If the equation of a conic section is written in the form Ax2+By2+Cx+Dy+E=0A{x}^{2}+B{y}^{2}+Cx+Dy+E=0 and AB=0AB=0, what can we conclude? 3. If the equation of a conic section is written in the form Ax2+Bxy+Cy2+Dx+Ey+F=0A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0, and B2−4AC>0{B}^{2}-4AC>0, what can we conclude? 4. Given the equation ax2+4x+3y2−12=0a{x}^{2}+4x+3{y}^{2}-12=0, what can we conclude if a>0?a>0? 5. For the equation Ax2+Bxy+Cy2+Dx+Ey+F=0A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0, the value of θ\theta that satisfies cot⁡(2θ)=A−CB\cot \left(2\theta \right)=\frac{A-C}{B} gives us what information? For the following exercises, determine which conic section is represented based on the given equation. 6. 9x2+4y2+72x+36y−500=09{x}^{2}+4{y}^{2}+72x+36y - 500=0 7. x2−10x+4y−10=0{x}^{2}-10x+4y - 10=0 8. 2x2−2y2+4x−6y−2=02{x}^{2}-2{y}^{2}+4x - 6y - 2=0 9. 4x2−y2+8x−1=04{x}^{2}-{y}^{2}+8x - 1=0 10. 4y2−5x+9y+1=04{y}^{2}-5x+9y+1=0 11. 2x2+3y2−8x−12y+2=02{x}^{2}+3{y}^{2}-8x - 12y+2=0 12. 4x2+9xy+4y2−36y−125=04{x}^{2}+9xy+4{y}^{2}-36y - 125=0 13. 3x2+6xy+3y2−36y−125=03{x}^{2}+6xy+3{y}^{2}-36y - 125=0 14. −3x2+33xy−4y2+9=0-3{x}^{2}+3\sqrt{3}xy - 4{y}^{2}+9=0 15. 2x2+43xy+6y2−6x−3=02{x}^{2}+4\sqrt{3}xy+6{y}^{2}-6x - 3=0 16. −x2+42xy+2y2−2y+1=0-{x}^{2}+4\sqrt{2}xy+2{y}^{2}-2y+1=0 17. 8x2+42xy+4y2−10x+1=08{x}^{2}+4\sqrt{2}xy+4{y}^{2}-10x+1=0 For the following exercises, find a new representation of the given equation after rotating through the given angle. 18. 3x2+xy+3y2−5=0,θ=45∘3{x}^{2}+xy+3{y}^{2}-5=0,\theta =45^\circ 19. 4x2−xy+4y2−2=0,θ=45∘4{x}^{2}-xy+4{y}^{2}-2=0,\theta =45^\circ 20. 2x2+8xy−1=0,θ=30∘2{x}^{2}+8xy - 1=0,\theta =30^\circ 21. −2x2+8xy+1=0,θ=45∘-2{x}^{2}+8xy+1=0,\theta =45^\circ 22. 4x2+2xy+4y2+y+2=0,θ=45∘4{x}^{2}+\sqrt{2}xy+4{y}^{2}+y+2=0,\theta =45^\circ For the following exercises, determine the angle θ\theta that will eliminate the xyxy term and write the corresponding equation without the xyxy term. 23. x2+33xy+4y2+y−2=0{x}^{2}+3\sqrt{3}xy+4{y}^{2}+y - 2=0 24. 4x2+23xy+6y2+y−2=04{x}^{2}+2\sqrt{3}xy+6{y}^{2}+y - 2=0 25. 9x2−33xy+6y2+4y−3=09{x}^{2}-3\sqrt{3}xy+6{y}^{2}+4y - 3=0 26. −3x2−3xy−2y2−x=0-3{x}^{2}-\sqrt{3}xy - 2{y}^{2}-x=0 27. 16x2+24xy+9y2+6x−6y+2=016{x}^{2}+24xy+9{y}^{2}+6x - 6y+2=0 28. x2+4xy+4y2+3x−2=0{x}^{2}+4xy+4{y}^{2}+3x - 2=0 29. x2+4xy+y2−2x+1=0{x}^{2}+4xy+{y}^{2}-2x+1=0 30. 4x2−23xy+6y2−1=04{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. 31. y=−x2,θ=−45∘y=-{x}^{2},\theta =-{45}^{\circ } 32. x=y2,θ=45∘x={y}^{2},\theta ={45}^{\circ } 33. x24+y21=1,θ=45∘\frac{{x}^{2}}{4}+\frac{{y}^{2}}{1}=1,\theta ={45}^{\circ } 34. y216+x29=1,θ=45∘\frac{{y}^{2}}{16}+\frac{{x}^{2}}{9}=1,\theta ={45}^{\circ } 35. y2−x2=1,θ=45∘{y}^{2}-{x}^{2}=1,\theta ={45}^{\circ } 36. y=x22,θ=30∘y=\frac{{x}^{2}}{2},\theta ={30}^{\circ } 37. x=(y−1)2,θ=30∘x={\left(y - 1\right)}^{2},\theta ={30}^{\circ } 38. x29+y24=1,θ=30∘\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1,\theta ={30}^{\circ } For the following exercises, graph the equation relative to the x′y′{x}^{\prime }{y}^{\prime } system in which the equation has no x′y′{x}^{\prime }{y}^{\prime } term. 39. xy=9xy=9 40. x2+10xy+y2−6=0{x}^{2}+10xy+{y}^{2}-6=0 41. x2−10xy+y2−24=0{x}^{2}-10xy+{y}^{2}-24=0 42. 4x2−33xy+y2−22=04{x}^{2}-3\sqrt{3}xy+{y}^{2}-22=0 43. 6x2+23xy+4y2−21=06{x}^{2}+2\sqrt{3}xy+4{y}^{2}-21=0 44. 11x2+103xy+y2−64=011{x}^{2}+10\sqrt{3}xy+{y}^{2}-64=0 45. 21x2+23xy+19y2−18=021{x}^{2}+2\sqrt{3}xy+19{y}^{2}-18=0 46. 16x2+24xy+9y2−130x+90y=016{x}^{2}+24xy+9{y}^{2}-130x+90y=0 47. 16x2+24xy+9y2−60x+80y=016{x}^{2}+24xy+9{y}^{2}-60x+80y=0 48. 13x2−63xy+7y2−16=013{x}^{2}-6\sqrt{3}xy+7{y}^{2}-16=0 49. 4x2−4xy+y2−85x−165y=04{x}^{2}-4xy+{y}^{2}-8\sqrt{5}x - 16\sqrt{5}y=0 For the following exercises, determine the angle of rotation in order to eliminate the xyxy term. Then graph the new set of axes. 50. 6x2−53xy+y2+10x−12y=06{x}^{2}-5\sqrt{3}xy+{y}^{2}+10x - 12y=0 51. 6x2−5xy+6y2+20x−y=06{x}^{2}-5xy+6{y}^{2}+20x-y=0 52. 6x2−83xy+14y2+10x−3y=06{x}^{2}-8\sqrt{3}xy+14{y}^{2}+10x - 3y=0 53. 4x2+63xy+10y2+20x−40y=04{x}^{2}+6\sqrt{3}xy+10{y}^{2}+20x - 40y=0 54. 8x2+3xy+4y2+2x−4=08{x}^{2}+3xy+4{y}^{2}+2x - 4=0 55. 16x2+24xy+9y2+20x−44y=016{x}^{2}+24xy+9{y}^{2}+20x - 44y=0 For the following exercises, determine the value of kk based on the given equation. 56. Given 4x2+kxy+16y2+8x+24y−48=04{x}^{2}+kxy+16{y}^{2}+8x+24y - 48=0, find kk for the graph to be a parabola. 57. Given 2x2+kxy+12y2+10x−16y+28=02{x}^{2}+kxy+12{y}^{2}+10x - 16y+28=0, find kk for the graph to be an ellipse. 58. Given 3x2+kxy+4y2−6x+20y+128=03{x}^{2}+kxy+4{y}^{2}-6x+20y+128=0, find kk for the graph to be a hyperbola. 59. Given kx2+8xy+8y2−12x+16y+18=0k{x}^{2}+8xy+8{y}^{2}-12x+16y+18=0, find kk for the graph to be a parabola. 60. Given 6x2+12xy+ky2+16x+10y+4=06{x}^{2}+12xy+k{y}^{2}+16x+10y+4=0, find kk for the graph to be an ellipse.

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