Which Figure Shows How a Shape Can Be Rotated About an Axis to Form the Given Object?

arning Objectives

In this section, you will:

• Identify nondegenerate conic sections given their general course equations.
• Use rotation of axes formulas.
• Write equations of rotated conics in standard grade.
• Identify conics without rotating axes.

Figure 1. The nondegenerate conic sections

Equally 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 nosotros as well call a cone. The style in which nosotros piece the cone will make up one's mind 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 past slicing a single cone with a slanted airplane 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 height and bottom of the cone. Run across (Effigy).

Figure 2. Degenerate 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). 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.

Identifying Nondegenerate Conics in Full general Form

In previous sections of this chapter, we have focused on the standard class 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 club, as shown below.

[latex]A{x}^{2}+Bxy+C{y}^{ii}+Dx+Ey+F=0[/latex]

where[latex]\,A,B,[/latex] and [latex]\,C\,[/latex]are not all nil. We can use the values of the coefficients to identify which blazon conic is represented past a given equation.

Yous may detect that the general form equation has an[latex]\,xy\,[/latex]term that we accept non seen in whatever of the standard form equations. As nosotros will discuss later, the[latex]\,xy\,[/latex]term rotates the conic whenever[latex]\text{ }B\text{ }[/latex]is non equal to zero.

Conic Sections	Case
ellipse	[latex]4{x}^{2}+9{y}^{2}=1[/latex]
circle	[latex]4{ten}^{ii}+four{y}^{2}=i[/latex]
hyperbola	[latex]4{x}^{2}-9{y}^{ii}=1[/latex]
parabola	[latex]4{x}^{2}=9y\text{ or }4{y}^{two}=9x[/latex]
one line	[latex]4x+9y=one[/latex]
intersecting lines	[latex]\left(x-iv\correct)\left(y+iv\right)=0[/latex]
parallel lines	[latex]\left(x-4\correct)\left(x-ix\correct)=0[/latex]
a signal	[latex]4{x}^{2}+four{y}^{ii}=0[/latex]
no graph	[latex]4{x}^{2}+four{y}^{2}=\,-\,1[/latex]

Full general Form of Conic Sections

A conic section has the full general form

[latex]A{10}^{2}+Bxy+C{y}^{ii}+Dx+Ey+F=0[/latex]

where[latex]\,A,B,[/latex] and[latex]\,C\,[/latex]are non all zero.

(Figure) summarizes the dissimilar conic sections where[latex]\,B=0,[/latex] and[latex]\,A\,[/latex]and[latex]\,C\,[/latex]are nonzero real numbers. This indicates that the conic has not been rotated.

ellipse	[latex]A{x}^{ii}+C{y}^{two}+Dx+Ey+F=0,\text{ }A\ne C\text{ and }Air-conditioning>0[/latex]
circle	[latex]A{10}^{2}+C{y}^{2}+Dx+Ey+F=0,\text{ }A=C[/latex]
hyperbola	[latex]A{10}^{ii}-C{y}^{2}+Dx+Ey+F=0\text{ or }-A{ten}^{two}+C{y}^{ii}+Dx+Ey+F=0,[/latex]where[latex]\,A\,[/latex]and[latex]\,C\,[/latex]are positive
parabola	[latex]A{10}^{2}+Dx+Ey+F=0\text{ or }C{y}^{2}+Dx+Ey+F=0[/latex]

How To

Given the equation of a conic, identify the type of conic.

1. Rewrite the equation in the general form, [latex]A{10}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0.[/latex]
2. Identify the values of[latex]\,A\,[/latex]and[latex]\,C\,[/latex]from the general form.
   1. If[latex]\,A\,[/latex]and[latex]\,C\,[/latex] are nonzero, have the same sign, and are not equal to each other, then the graph may exist an ellipse.
   2. If[latex]\,A\,[/latex]and[latex]\,C\,[/latex]are equal and nonzero and have the same sign, then the graph may be a circle.
   3. If[latex]\,A\,[/latex]and[latex]\,C\,[/latex]are nonzero and have opposite signs, then the graph may be a hyperbola.
   4. If either[latex]\,A\,[/latex]or[latex]\,C\,[/latex] is zero, then the graph may exist a parabola.

   If B = 0, the conic department will take a vertical and/or horizontal axes. If B does not equal 0, every bit shown beneath, the conic section is rotated.  Notice the phrase "may be" in the definitions. That is because the equation may not represent a conic department at all, depending on the values of A, B, C, D, E, and F. For example, the degenerate case of a circle or an ellipse is a bespeak:

   [latex]A{x}^{2}+B{y}^{2}=0\text{,}[/latex]when A and B have the same sign.

   The degenerate case of a hyperbola is 2 intersecting direct lines:[latex]A{x}^{two}+B{y}^{ii}=0\text{,}[/latex]when A and B have reverse signs.

   On the other hand, the equation, [latex]A{x}^{2}+B{y}^{ii}+i=0\text{,}[/latex] when A and B are positive does not represent a graph at all, since there are no existent ordered pairs which satisfy it.

Identifying a Conic from Its Full general Form

Place the graph of each of the post-obit nondegenerate conic sections.

1. [latex]4{x}^{2}-9{y}^{2}+36x+36y-125=0[/latex]
2. [latex]ix{y}^{2}+16x+36y-10=0[/latex]
3. [latex]3{x}^{2}+3{y}^{2}-2x-6y-four=0[/latex]
4. [latex]-25{x}^{ii}-four{y}^{2}+100x+16y+20=0[/latex]

Try It

Identify the graph of each of the following nondegenerate conic sections.

1. [latex]16{y}^{2}-{x}^{two}+x-4y-nine=0[/latex]
2. [latex]16{x}^{2}+4{y}^{2}+16x+49y-81=0[/latex]

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

Until now, we take looked at equations of conic sections without an[latex]\,xy\,[/latex]term, which aligns the graphs with the ten– and y-axes. When nosotros add together an[latex]\,xy\,[/latex]term, nosotros are rotating the conic nearly the origin. If the ten– and y-axes are rotated through an angle, say[latex]\,\theta ,[/latex]so every point on the plane may be thought of as having two representations:[latex]\,\left(x,y\correct)\,[/latex]on the Cartesian plane with the original x-axis and y-centrality, and[latex]\,\left({ten}^{\prime },{y}^{\prime }\correct)\,[/latex]on the new plane defined by the new, rotated axes, called the x'-axis and y'-centrality. See (Figure).

Figure three. The graph of the rotated ellipse[latex]\,{ten}^{two}+{y}^{2}–xy–15=0[/latex]

We volition find the relationships between[latex]\,10\,[/latex]and[latex]\,y\,[/latex]on the Cartesian plane with[latex]\,{x}^{\prime }\,[/latex]and[latex]\,{y}^{\prime }\,[/latex]on the new rotated plane. Run into (Effigy).

Figure 4. The Cartesian plane with x- and y-axes and the resulting x′− and y′−axes formed by a rotation past an angle[latex]\text{ }\theta .[/latex]

The original coordinate x– and y-axes have unit vectors[latex]\,i\,[/latex]and[latex]\,j\,.[/latex]The rotated coordinate axes have unit of measurement vectors[latex]\,{i}^{\prime }\,[/latex]and[latex]\,{j}^{\prime }.[/latex]The angle[latex]\,\theta \,[/latex]is known equally the angle of rotation. See (Figure). We may write the new unit vectors in terms of the original ones.

[latex]\begin{assortment}{50}{i}^{\prime }=\mathrm{cos}\text{ }\theta i+\mathrm{sin}\text{ }\theta j\hfill \\ {j}^{\prime }=-\mathrm{sin}\text{ }\theta i+\mathrm{cos}\text{ }\theta j\hfill \finish{array}[/latex]

Figure 5. Human relationship betwixt the quondam and new coordinate planes.

Consider a vector[latex]\,u\,[/latex]in the new coordinate airplane. It may be represented in terms of its coordinate axes.

[latex]\begin{array}{ll}u={x}^{\prime }{i}^{\prime }+{y}^{\prime }{j}^{\prime number }\hfill & \hfill \\ u={x}^{\prime }\left(i\text{ }\mathrm{cos}\text{ }\theta +j\text{ }\mathrm{sin}\text{ }\theta \right)+{y}^{\prime }\left(-i\text{ }\mathrm{sin}\text{ }\theta +j\text{ }\mathrm{cos}\text{ }\theta \correct)\hfill & \begin{array}{cccc}& & & \end{array}\text{Substitute}.\hfill \\ u=ix\text{'}\text{ }\mathrm{cos}\text{ }\theta +jx\text{'}\text{ }\mathrm{sin}\text{ }\theta -iy\text{'}\text{ }\mathrm{sin}\text{ }\theta +jy\text{'}\text{ }\mathrm{cos}\text{ }\theta \hfill & \begin{array}{cccc}& & & \end{array}\text{Distribute}.\hfill \\ u=ix\text{'}\text{ }\mathrm{cos}\text{ }\theta -iy\text{'}\text{ }\mathrm{sin}\text{ }\theta +jx\text{'}\text{ }\mathrm{sin}\text{ }\theta +jy\text{'}\text{ }\mathrm{cos}\text{ }\theta \hfill & \begin{assortment}{cccc}& & & \end{array}\text{Utilise commutative holding}.\hfill \\ u=\left(x\text{'}\text{ }\mathrm{cos}\text{ }\theta -y\text{'}\text{ }\mathrm{sin}\text{ }\theta \right)i+\left(x\text{'}\text{ }\mathrm{sin}\text{ }\theta +y\text{'}\text{ }\mathrm{cos}\text{ }\theta \correct)j\hfill & \begin{array}{cccc}& & & \end{array}\text{Factor by grouping}.\hfill \finish{array}[/latex]

Because[latex]\,u={x}^{\prime }{i}^{\prime }+{y}^{\prime }{j}^{\prime },[/latex] we have representations of[latex]\,ten\,[/latex]and[latex]\,y\,[/latex]in terms of the new coordinate system.

[latex]\begin{array}{c}x={x}^{\prime }\mathrm{cos}\text{ }\theta -{y}^{\prime }\mathrm{sin}\text{ }\theta \\ \text{and}\\ y={10}^{\prime number }\mathrm{sin}\text{ }\theta +{y}^{\prime number }\mathrm{cos}\text{ }\theta \end{assortment}[/latex]

Equations of Rotation

If a point[latex]\,\left(x,y\right)\,[/latex]on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an bending[latex]\,\theta \,[/latex]from the positive x-axis, then the coordinates of the bespeak with respect to the new axes are[latex]\,\left({ten}^{\prime },{y}^{\prime }\right).\,[/latex]We can employ the post-obit equations of rotation to define the relationship between[latex]\,\left(x,y\right)\,[/latex]and[latex]\,\left({x}^{\prime number },{y}^{\prime }\correct):[/latex]

[latex]x={x}^{\prime }\mathrm{cos}\text{ }\theta -{y}^{\prime }\mathrm{sin}\text{ }\theta [/latex]

and

[latex]y={ten}^{\prime }\mathrm{sin}\text{ }\theta +{y}^{\prime }\mathrm{cos}\text{ }\theta [/latex]

How To

Given the equation of a conic, notice a new representation after rotating through an angle.

1. Find[latex]\,x\,[/latex]and[latex]\,y\,[/latex]where[latex]\,10={10}^{\prime }\mathrm{cos}\text{ }\theta -{y}^{\prime number }\mathrm{sin}\text{ }\theta \,[/latex]and[latex]\,y={10}^{\prime }\mathrm{sin}\text{ }\theta +{y}^{\prime number }\mathrm{cos}\text{ }\theta .[/latex]
2. Substitute the expression for[latex]\,x\,[/latex]and[latex]\,y\,[/latex]into in the given equation, and then simplify.
3. Write the equations with[latex]\,{x}^{\prime }\,[/latex]and[latex]\,{y}^{\prime number }\,[/latex]in standard form.

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

Find a new representation of the equation[latex]\,two{x}^{two}-xy+2{y}^{ii}-30=0\,[/latex]later rotating through an angle of[latex]\,\theta =45°.[/latex]

Writing Equations of Rotated Conics in Standard Form

Now that nosotros 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 class[latex]\,A{x}^{ii}+Bxy+C{y}^{two}+Dx+Ey+F=0\,[/latex]into standard course past rotating the axes. To do and then, nosotros volition rewrite the general form as an equation in the[latex]\,{ten}^{\prime }\,[/latex]and[latex]\,{y}^{\prime }\,[/latex]coordinate organization without the[latex]\,{x}^{\prime }{y}^{\prime }\,[/latex]term, by rotating the axes past a measure out of[latex]\,\theta \,[/latex]that satisfies

[latex]\mathrm{cot}\left(2\theta \right)=\frac{A-C}{B}[/latex]

We take learned already that any conic may be represented by the second caste equation

[latex]A{10}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex]

where[latex]\,A,B,[/latex]and[latex]\,C\,[/latex]are non all naught. However, if[latex]\,B\ne 0,[/latex] and then we have an[latex]\,xy\,[/latex]term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle[latex]\,\theta \,[/latex]where[latex]\,\mathrm{cot}\left(2\theta \correct)=\frac{A-C}{B}.[/latex]

• If[latex]\,\mathrm{cot}\left(2\theta \right)>0,[/latex] then[latex]\,2\theta \,[/latex] is in the start quadrant, and[latex]\,\theta \,[/latex] is between[latex]\,\left(0°,45°\right).[/latex]
• If[latex]\,\mathrm{cot}\left(ii\theta \right)<0,[/latex] then[latex]\,two\theta \,[/latex] is in the 2nd quadrant, and[latex]\,\theta \,[/latex] is between[latex]\,\left(45°,90°\right).[/latex]
• If[latex]\,A=C,[/latex] and so[latex]\,\theta =45°.[/latex]

How To

Given an equation for a conic in the[latex]\,{x}^{\prime }{y}^{\prime }\,[/latex]organization, rewrite the equation without the[latex]\,{x}^{\prime }{y}^{\prime number }\,[/latex]term in terms of[latex]\,{x}^{\prime }\,[/latex]and[latex]\,{y}^{\prime number },[/latex]where the[latex]\,{ten}^{\prime }\,[/latex]and[latex]\,{y}^{\prime }\,[/latex]axes are rotations of the standard axes by[latex]\,\theta \,[/latex]degrees.

1. Find[latex]\,\mathrm{cot}\left(2\theta \right).[/latex]
2. Find[latex]\,\mathrm{sin}\text{ }\theta \,[/latex]and[latex]\,\mathrm{cos}\text{ }\theta .[/latex]
3. Substitute[latex]\,\mathrm{sin}\text{ }\theta \,[/latex]and[latex]\,\mathrm{cos}\text{ }\theta \,[/latex]into[latex]\,ten={ten}^{\prime }\mathrm{cos}\text{ }\theta -{y}^{\prime }\mathrm{sin}\text{ }\theta \,[/latex]and[latex]\,y={10}^{\prime }\mathrm{sin}\text{ }\theta +{y}^{\prime }\mathrm{cos}\text{ }\theta .[/latex]
4. Substitute the expression for[latex]\,x\,[/latex]and[latex]\,y\,[/latex]into in the given equation, and then simplify.
5. Write the equations with[latex]\,{ten}^{\prime number }\,[/latex]and[latex]\,{y}^{\prime }\,[/latex]in the standard class with respect to the rotated axes.

Rewriting an Equation with respect to the x′ and y′ axes without the x′y′ Term

Rewrite the equation[latex]\,eight{ten}^{2}-12xy+17{y}^{two}=twenty\,[/latex]in the[latex]\,{x}^{\prime number }{y}^{\prime }\,[/latex]system without an[latex]\,{x}^{\prime }{y}^{\prime }\,[/latex]term.

Effort It

Rewrite the[latex]\,13{10}^{ii}-6\sqrt{3}xy+7{y}^{2}=16\,[/latex]in the[latex]\,{x}^{\prime }{y}^{\prime }\,[/latex]system without the[latex]\,{ten}^{\prime number }{y}^{\prime }\,[/latex]term.

Show Solution

[latex]\frac{{{x}^{\prime }}^{2}}{4}+\frac{{{y}^{\prime }}^{ii}}{i}=1[/latex]

Graphing an Equation That Has No x′y′ Terms

Graph the following equation relative to the[latex]\,{x}^{\prime }{y}^{\prime }\,[/latex]system:

[latex]{10}^{ii}+12xy-4{y}^{two}=xxx[/latex]

Identifying Conics without Rotating Axes

Now we have come up full circumvolve. How practise we identify the blazon of conic described by an equation? What happens when the axes are rotated? Recall, the full general form of a conic is

[latex]A{x}^{2}+Bxy+C{y}^{ii}+Dx+Ey+F=0[/latex]

If we apply the rotation formulas to this equation we get the course

[latex]{A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime number }{ten}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime }{x}^{\prime number }+{E}^{\prime number }{y}^{\prime }+{F}^{\prime }=0[/latex]

It may exist shown that[latex]\,{B}^{ii}-4AC={{B}^{\prime number }}^{2}-4{A}^{\prime number }{C}^{\prime }.\,[/latex]The expression does not vary subsequently rotation, and so we phone call the expression invariant. The discriminant,[latex]\,{B}^{2}-4AC,[/latex] is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section.

Using the Discriminant to Place a Conic

If the equation[latex]\,A{ten}^{ii}+Bxy+C{y}^{ii}+Dx+Ey+F=0\,[/latex]is transformed by rotating axes into the equation[latex]\,{A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime number }{10}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime number }{x}^{\prime }+{Eastward}^{\prime }{y}^{\prime number }+{F}^{\prime }=0,[/latex] then[latex]\,{B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime number }.[/latex]

The equation[latex]\,A{ten}^{ii}+Bxy+C{y}^{2}+Dx+Ey+F=0\,[/latex]is an ellipse, a parabola, or a hyperbola, or a degenerate instance of one of these.

If the discriminant,[latex]\,{B}^{2}-4AC,[/latex]is

• [latex]<0,[/latex] the conic department is an ellipse
• [latex]=0,[/latex] the conic section is a parabola
• [latex]>0,[/latex] the conic section is a hyperbola

Identifying the Conic without Rotating Axes

Identify the conic for each of the post-obit without rotating axes.

1. [latex]five{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0[/latex]
2. [latex]5{ten}^{two}+ii\sqrt{3}xy+12{y}^{2}-5=0[/latex]

Effort It

Identify the conic for each of the following without rotating axes.

1. [latex]{ten}^{2}-9xy+3{y}^{2}-12=0[/latex]
2. [latex]10{x}^{2}-9xy+4{y}^{2}-4=0[/latex]

Key Equations

General Form equation of a conic section	[latex]A{ten}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex]
Rotation of a conic department	[latex]\begin{assortment}{l}ten={x}^{\prime }\mathrm{cos}\text{ }\theta -{y}^{\prime }\mathrm{sin}\text{ }\theta \hfill \\ y={ten}^{\prime }\mathrm{sin}\text{ }\theta +{y}^{\prime }\mathrm{cos}\text{ }\theta \hfill \stop{array}[/latex]
Angle of rotation	[latex]\theta ,\text{where }\mathrm{cot}\left(2\theta \right)=\frac{A-C}{B}[/latex]

Primal Concepts

• Four basic shapes can consequence from the intersection of a airplane with a pair of right round cones connected tail to tail. They include an ellipse, a circle, a hyperbola, and a parabola.
• A nondegenerate conic section has the general class[latex]\,A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0\,[/latex]where[latex]\,A,B\,[/latex]and[latex]\,C\,[/latex]are not all zero. The values of[latex]\,A,B,[/latex] and[latex]\,C\,[/latex]determine the type of conic. Meet (Figure).
• Equations of conic sections with an[latex]\,xy\,[/latex]term have been rotated almost the origin. Meet (Figure).
• The full general form tin can be transformed into an equation in the[latex]\,{x}^{\prime number }\,[/latex]and[latex]\,{y}^{\prime }\,[/latex]coordinate system without the[latex]\,{x}^{\prime }{y}^{\prime }\,[/latex]term. See (Figure) and (Figure).
• An expression is described as invariant if it remains unchanged later on rotating. Because the discriminant is invariant, observing it enables us to identify the conic section. See (Figure).

Section Exercises

Exact

What outcome does the[latex]\,xy\,[/latex]term have on the graph of a conic department?

Bear witness Solution

The[latex]\,xy\,[/latex]term causes a rotation of the graph to occur.

If the equation of a conic section is written in the form[latex]\,A{ten}^{2}+B{y}^{2}+Cx+Dy+E=0\,[/latex]and[latex]\,AB=0,[/latex] what tin can nosotros conclude?

If the equation of a conic section is written in the grade[latex]\,A{x}^{2}+Bxy+C{y}^{ii}+Dx+Ey+F=0,[/latex]and[latex]\,{B}^{2}-4AC>0,[/latex] what can we conclude?

Prove Solution

The conic department is a hyperbola.

Given the equation[latex]\,a{x}^{ii}+4x+3{y}^{2}-12=0,[/latex] what can we conclude if[latex]\,a>0?[/latex]

For the equation[latex]\,A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0,[/latex] the value of[latex]\,\theta \,[/latex]that satisfies[latex]\,\mathrm{cot}\left(ii\theta \correct)=\frac{A-C}{B}\,[/latex]gives the states what data?

Show Solution

It gives the angle of rotation of the axes in social club to eliminate the[latex]\,xy\,[/latex]term.

Algebraic

For the following exercises, determine which conic section is represented based on the given equation.

[latex]9{x}^{2}+4{y}^{two}+72x+36y-500=0[/latex]

[latex]{x}^{2}-10x+4y-ten=0[/latex]

Prove Solution

[latex]AB=0,[/latex] parabola

[latex]2{x}^{2}-two{y}^{2}+4x-6y-2=0[/latex]

[latex]4{x}^{2}-{y}^{2}+8x-ane=0[/latex]

Show Solution

[latex]AB=-four<0,[/latex] hyperbola

[latex]iv{y}^{ii}-5x+9y+1=0[/latex]

[latex]two{x}^{two}+iii{y}^{ii}-8x-12y+2=0[/latex]

Bear witness Solution

[latex]AB=6>0,[/latex] ellipse

[latex]4{x}^{ii}+9xy+four{y}^{2}-36y-125=0[/latex]

[latex]iii{x}^{2}+6xy+3{y}^{2}-36y-125=0[/latex]

Show Solution

[latex]{B}^{2}-4AC=0,[/latex] parabola

[latex]-3{x}^{two}+3\sqrt{3}xy-four{y}^{2}+9=0[/latex]

[latex]2{x}^{2}+4\sqrt{3}xy+half-dozen{y}^{two}-6x-3=0[/latex]

Show Solution

[latex]{B}^{2}-4AC=0,[/latex] parabola

[latex]-{x}^{2}+4\sqrt{2}xy+ii{y}^{2}-2y+one=0[/latex]

[latex]8{x}^{2}+4\sqrt{2}xy+4{y}^{2}-10x+1=0[/latex]

Show Solution

[latex]{B}^{2}-4AC=-96<0,[/latex] ellipse

For the following exercises, find a new representation of the given equation later rotating through the given angle.

[latex]3{x}^{2}+xy+iii{y}^{2}-5=0,\theta =45°[/latex]

[latex]iv{x}^{2}-xy+4{y}^{2}-2=0,\theta =45°[/latex]

Show Solution

[latex]vii{{x}^{\prime number }}^{2}+nine{{y}^{\prime }}^{2}-4=0[/latex]

[latex]2{x}^{2}+8xy-ane=0,\theta =30°[/latex]

[latex]-2{x}^{2}+8xy+ane=0,\theta =45°[/latex]

Show Solution

[latex]3{{x}^{\prime }}^{2}+2{x}^{\prime number }{y}^{\prime }-5{{y}^{\prime }}^{2}+1=0[/latex]

[latex]4{x}^{2}+\sqrt{2}xy+iv{y}^{2}+y+2=0,\theta =45°[/latex]

For the post-obit exercises, make up one's mind the angle[latex]\,\theta \,[/latex]that will eliminate the[latex]\,xy\,[/latex]term and write the corresponding equation without the[latex]\,xy\,[/latex]term.

[latex]{x}^{two}+three\sqrt{iii}xy+iv{y}^{2}+y-2=0[/latex]

Evidence Solution

[latex]\theta ={60}^{\circ },11{{x}^{\prime }}^{2}-{{y}^{\prime }}^{2}+\sqrt{3}{10}^{\prime }+{y}^{\prime }-iv=0[/latex]

[latex]four{10}^{ii}+2\sqrt{iii}xy+6{y}^{ii}+y-2=0[/latex]

[latex]ix{x}^{2}-3\sqrt{3}xy+half-dozen{y}^{ii}+4y-3=0[/latex]

Show Solution

[latex]\theta ={150}^{\circ },21{{x}^{\prime }}^{2}+9{{y}^{\prime }}^{2}+4{x}^{\prime }-4\sqrt{three}{y}^{\prime }-6=0[/latex]

[latex]-three{x}^{two}-\sqrt{3}xy-2{y}^{2}-10=0[/latex]

[latex]16{x}^{2}+24xy+ix{y}^{2}+6x-6y+2=0[/latex]

Prove Solution

[latex]\theta \approx {36.9}^{\circ },125{{x}^{\prime number }}^{2}+half-dozen{x}^{\prime }-42{y}^{\prime number }+10=0[/latex]

[latex]{x}^{2}+4xy+4{y}^{2}+3x-2=0[/latex]

[latex]{10}^{2}+4xy+{y}^{2}-2x+1=0[/latex]

Show Solution

[latex]\theta ={45}^{\circ },3{{x}^{\prime number }}^{2}-{{y}^{\prime }}^{2}-\sqrt{ii}{x}^{\prime }+\sqrt{ii}{y}^{\prime }+1=0[/latex]

[latex]4{x}^{2}-two\sqrt{3}xy+half dozen{y}^{2}-1=0[/latex]

Graphical

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.

[latex]y=-{x}^{ii},\theta =-{45}^{\circ }[/latex]

[latex]x={y}^{two},\theta ={45}^{\circ }[/latex]

[latex]\frac{{x}^{2}}{4}+\frac{{y}^{two}}{one}=1,\theta ={45}^{\circ }[/latex]

[latex]\frac{{y}^{2}}{16}+\frac{{x}^{2}}{9}=1,\theta ={45}^{\circ }[/latex]

[latex]{y}^{2}-{10}^{2}=i,\theta ={45}^{\circ }[/latex]

[latex]y=\frac{{10}^{2}}{2},\theta ={xxx}^{\circ }[/latex]

[latex]x={\left(y-ane\right)}^{2},\theta ={30}^{\circ }[/latex]

[latex]\frac{{x}^{2}}{9}+\frac{{y}^{2}}{four}=one,\theta ={30}^{\circ }[/latex]

For the following exercises, graph the equation relative to the[latex]\,{x}^{\prime number }{y}^{\prime }\,[/latex]organization in which the equation has no[latex]\,{x}^{\prime }{y}^{\prime }\,[/latex]term.

[latex]xy=9[/latex]

Show Solution

[latex]{x}^{2}+10xy+{y}^{ii}-half dozen=0[/latex]

[latex]{x}^{2}-10xy+{y}^{2}-24=0[/latex]

Prove Solution

[latex]4{10}^{two}-3\sqrt{3}xy+{y}^{ii}-22=0[/latex]

[latex]6{x}^{2}+2\sqrt{3}xy+four{y}^{2}-21=0[/latex]

Show Solution

[latex]11{x}^{2}+10\sqrt{iii}xy+{y}^{2}-64=0[/latex]

[latex]21{ten}^{2}+two\sqrt{3}xy+xix{y}^{2}-xviii=0[/latex]

Evidence Solution

[latex]16{ten}^{two}+24xy+9{y}^{2}-130x+90y=0[/latex]

[latex]16{x}^{2}+24xy+9{y}^{2}-60x+80y=0[/latex]

Show Solution

[latex]13{10}^{two}-half dozen\sqrt{iii}xy+7{y}^{two}-16=0[/latex]

[latex]4{10}^{ii}-4xy+{y}^{2}-8\sqrt{v}x-16\sqrt{5}y=0[/latex]

Show Solution

For the following exercises, make up one's mind the angle of rotation in club to eliminate the[latex]\,xy\,[/latex]term. And then graph the new set of axes.

[latex]vi{x}^{2}-5\sqrt{3}xy+{y}^{two}+10x-12y=0[/latex]

[latex]6{ten}^{two}-5xy+6{y}^{2}+20x-y=0[/latex]

[latex]6{x}^{2}-8\sqrt{3}xy+xiv{y}^{ii}+10x-3y=0[/latex]

[latex]4{x}^{ii}+6\sqrt{3}xy+10{y}^{2}+20x-40y=0[/latex]

[latex]8{x}^{2}+3xy+4{y}^{ii}+2x-iv=0[/latex]

[latex]16{x}^{2}+24xy+nine{y}^{2}+20x-44y=0[/latex]

For the following exercises, determine the value of[latex]\,k\,[/latex]based on the given equation.

Given[latex]\,four{x}^{2}+kxy+16{y}^{2}+8x+24y-48=0,[/latex] find[latex]\,k\,[/latex]for the graph to be a parabola.

Given[latex]\,2{10}^{2}+kxy+12{y}^{2}+10x-16y+28=0,[/latex] find[latex]\,k\,[/latex]for the graph to be an ellipse.

Prove Solution

[latex]-iv\sqrt{6}<1000<4\sqrt{six}[/latex]

Given[latex]\,three{x}^{two}+kxy+4{y}^{2}-6x+20y+128=0,[/latex] notice[latex]\,k\,[/latex]for the graph to exist a hyperbola.

Given[latex]\,k{x}^{2}+8xy+8{y}^{ii}-12x+16y+18=0,[/latex] find[latex]\,k\,[/latex]for the graph to be a parabola.

Show Solution

[latex]thousand=ii[/latex]

Given[latex]\,6{x}^{two}+12xy+grand{y}^{2}+16x+10y+4=0,[/latex] find[latex]\,yard\,[/latex]for the graph to be an ellipse.

Glossary

angle of rotation

an acute bending formed by a set of axes rotated from the Cartesian plane where, if[latex]\,\mathrm{cot}\left(2\theta \right)>0,[/latex]then[latex]\,\theta \,[/latex]is between[latex]\,\left(0°,45°\right);[/latex]if[latex]\,\mathrm{cot}\left(2\theta \right)<0,[/latex]then[latex]\,\theta \,[/latex]is between[latex]\,\left(45°,xc°\right);\,[/latex]and if[latex]\,\mathrm{cot}\left(2\theta \right)=0,[/latex]so[latex]\,\theta =45°[/latex]

degenerate conic sections

any of the possible shapes formed when a airplane intersects a double cone through the apex. Types of degenerate conic sections include a bespeak, a line, and intersecting lines.

nondegenerate conic section

a shape formed by the intersection of a plane with a double correct cone such that the plane does not pass through the noon; nondegenerate conics include circles, ellipses, hyperbolas, and parabolas

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Source: https://courses.lumenlearning.com/suny-osalgebratrig/chapter/rotation-of-axes/

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