Learning Objectives
In this section, you will:
- 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. See Figure 1.
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=0𝐴𝑥2+𝐵𝑥𝑦+𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0
where A,B,𝐴,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 B𝐵 is not equal to zero.
| Conic Sections | Example |
|---|---|
| ellipse | 4×2+9y2=14𝑥2+9𝑦2=1 |
| circle | 4×2+4y2=14𝑥2+4𝑦2=1 |
| hyperbola | 4×2−9y2=14𝑥2−9𝑦2=1 |
| parabola | 4×2=9yor 4y2=9×4𝑥2=9𝑦or 4𝑦2=9𝑥 |
| one line | 4x+9y=14𝑥+9𝑦=1 |
| intersecting lines | (x−4)(y+4)=0(𝑥−4)(𝑦+4)=0 |
| parallel lines | (x−4)(x−9)=0(𝑥−4)(𝑥−9)=0 |
| a point | 4×2+4y2=04𝑥2+4𝑦2=0 |
| no graph | 4×2+4y2=−14𝑥2+4𝑦2=−1 |
GENERAL FORM OF CONIC SECTIONS
A conic section has the general form
Ax2+Bxy+Cy2+Dx+Ey+F=0𝐴𝑥2+𝐵𝑥𝑦+𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0
where A,B,𝐴,𝐵, and C𝐶 are not all zero.
Table 2 summarizes the different conic sections where B=0,𝐵=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≠Cand AC>0𝐴𝑥2+𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0,𝐴≠𝐶and 𝐴𝐶>0 |
| circle | Ax2+Cy2+Dx+Ey+F=0,A=C𝐴𝑥2+𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0,𝐴=𝐶 |
| hyperbola | Ax2−Cy2+Dx+Ey+F=0or −Ax2+Cy2+Dx+Ey+F=0,𝐴𝑥2−𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0or −𝐴𝑥2+𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0, where A𝐴 and C𝐶 are positive |
| parabola | Ax2+Dx+Ey+F=0or Cy2+Dx+Ey+F=0𝐴𝑥2+𝐷𝑥+𝐸𝑦+𝐹=0or 𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0 |
HOW TO
Given the equation of a conic, identify the type of conic.
- Rewrite the equation in the general form, Ax2+Bxy+Cy2+Dx+Ey+F=0.𝐴𝑥2+𝐵𝑥𝑦+𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0.
- Identify the values of A𝐴 and C𝐶 from the general form.
- If A𝐴 and C𝐶 are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse.
- If A𝐴 and C𝐶 are equal and nonzero and have the same sign, then the graph may be a circle.
- If A𝐴 and C𝐶 are nonzero and have opposite signs, then the graph may be a hyperbola.
- If either A𝐴 or C𝐶 is zero, then the graph may be a parabola.
Ax2+By2=0,𝐴𝑥2+𝐵𝑦2=0, when A and B have the same sign.
The degenerate case of a hyperbola is two intersecting straight lines: Ax2+By2=0,𝐴𝑥2+𝐵𝑦2=0, when A and B have opposite signs.
On the other hand, the equation, Ax2+By2+1=0,𝐴𝑥2+𝐵𝑦2+1=0, when A and B are positive does not represent a graph at all, since there are no real ordered pairs which satisfy it.
EXAMPLE 1
Identifying a Conic from Its General Form
Identify the graph of each of the following nondegenerate conic sections.
- ⓐ4×2−9y2+36x+36y−125=04𝑥2−9𝑦2+36𝑥+36𝑦−125=0
- ⓑ 9y2+16x+36y−10=09𝑦2+16𝑥+36𝑦−10=0
- ⓒ 3×2+3y2−2x−6y−4=03𝑥2+3𝑦2−2𝑥−6𝑦−4=0
- ⓓ −25×2−4y2+100x+16y+20=0−25𝑥2−4𝑦2+100𝑥+16𝑦+20=0
Solution
- ⓐ Rewriting the general form, we have
A=4𝐴=4 and C=−9,𝐶=−9, so we observe that A𝐴 and C𝐶 have opposite signs. The graph of this equation is a hyperbola.
- ⓑ Rewriting the general form, we have
A=0𝐴=0 and C=9.𝐶=9. We can determine that the equation is a parabola, since A𝐴 is zero.
- ⓒ Rewriting the general form, we have
A=3𝐴=3 and C=3.𝐶=3. Because A=C,𝐴=𝐶, the graph of this equation is a circle.
- ⓓ Rewriting the general form, we have
A=−25𝐴=−25 and C=−4.𝐶=−4. Because AC>0𝐴𝐶>0 and A≠C,𝐴≠𝐶, the graph of this equation is an ellipse.
TRY IT #1
Identify the graph of each of the following nondegenerate conic sections.
- ⓐ 16y2−x2+x−4y−9=016𝑦2−𝑥2+𝑥−4𝑦−9=0
- ⓑ 16×2+4y2+16x+49y−81=016𝑥2+4𝑦2+16𝑥+49𝑦−81=0
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,𝜃, 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. See Figure 3.
Figure 3 The graph of the rotated ellipse x2+y2–xy–15=0𝑥2+𝑦2–𝑥𝑦–15=0
We will find the relationships between x𝑥 and y𝑦 on the Cartesian plane with x′𝑥′ and y′𝑦′ on the new rotated plane. See Figure 4.
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 j.𝑗. The rotated coordinate axes have unit vectors i′𝑖′ and j′.𝑗′. The angle θ𝜃 is known as the angle of rotation. See Figure 5. We may write the new unit vectors in terms of the original ones.
i′=cosθi+sinθjj′=−sinθi+cosθj𝑖′=cos𝜃𝑖+sin𝜃𝑗𝑗′=−sin𝜃𝑖+cos𝜃𝑗
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′(icosθ+jsinθ)+y′(−isinθ+jcosθ)u=ix’cosθ+jx’sinθ−iy’sinθ+jy’cosθu=ix’cosθ−iy’sinθ+jx’sinθ+jy’cosθu=(x’cosθ−y’sinθ)i+(x’sinθ+y’cosθ)jSubstitute.Distribute.Apply commutative property.Factor by grouping.𝑢=𝑥′𝑖′+𝑦′𝑗′𝑢=𝑥′(𝑖cos𝜃+𝑗sin𝜃)+𝑦′(−𝑖sin𝜃+𝑗cos𝜃)Substitute.𝑢=𝑖𝑥’cos𝜃+𝑗𝑥’sin𝜃−𝑖𝑦’sin𝜃+𝑗𝑦’cos𝜃Distribute.𝑢=𝑖𝑥’cos𝜃−𝑖𝑦’sin𝜃+𝑗𝑥’sin𝜃+𝑗𝑦’cos𝜃Apply commutative property.𝑢=(𝑥’cos𝜃−𝑦’sin𝜃)𝑖+(𝑥’sin𝜃+𝑦’cos𝜃)𝑗Factor by grouping.
Because u=x′i′+y′j′,𝑢=𝑥′𝑖′+𝑦′𝑗′, we have representations of x𝑥 and y𝑦 in terms of the new coordinate system.
x=x′cosθ−y′sinθandy=x′sinθ+y′cosθ𝑥=𝑥′cos𝜃−𝑦′sin𝜃and𝑦=𝑥′sin𝜃+𝑦′cos𝜃
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=x′cosθ−y′sinθ𝑥=𝑥′cos𝜃−𝑦′sin𝜃
and
y=x′sinθ+y′cosθ𝑦=𝑥′sin𝜃+𝑦′cos𝜃
HOW TO
Given the equation of a conic, find a new representation after rotating through an angle.
- Find x𝑥 and y𝑦 where x=x′cosθ−y′sinθ𝑥=𝑥′cos𝜃−𝑦′sin𝜃 and y=x′sinθ+y′cosθ.𝑦=𝑥′sin𝜃+𝑦′cos𝜃.
- Substitute the expression for x𝑥 and y𝑦 into in the given equation, then simplify.
- Write the equations with x′𝑥′ and y′𝑦′ 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 2×2−xy+2y2−30=02𝑥2−𝑥𝑦+2𝑦2−30=0 after rotating through an angle of θ=45°.𝜃=45°.
Solution
Find x𝑥 and y,𝑦, where x=x′cosθ−y′sinθ𝑥=𝑥′cos𝜃−𝑦′sin𝜃 and y=x′sinθ+y′cosθ.𝑦=𝑥′sin𝜃+𝑦′cos𝜃.
Because θ=45°,𝜃=45°,
x=x′cos(45°)−y′sin(45°)x=x′(12√)−y′(12√)x=x′−y′2√𝑥=𝑥′cos(45°)−𝑦′sin(45°)𝑥=𝑥′(12)−𝑦′(12)𝑥=𝑥′−𝑦′2
and
y=x′sin(45°)+y′cos(45°)y=x′(12√)+y′(12√)y=x′+y′2√𝑦=𝑥′sin(45°)+𝑦′cos(45°)𝑦=𝑥′(12)+𝑦′(12)𝑦=𝑥′+𝑦′2
Substitute x=x′cosθ−y′sinθ𝑥=𝑥′cos𝜃−𝑦′sin𝜃 and y=x′sinθ+y′cosθ𝑦=𝑥′sin𝜃+𝑦′cos𝜃 into 2×2−xy+2y2−30=0.2𝑥2−𝑥𝑦+2𝑦2−30=0.
2(x′−y′2–√)2−(x′−y′2–√)(x′+y′2–√)+2(x′+y′2–√)2−30=02(𝑥′−𝑦′2)2−(𝑥′−𝑦′2)(𝑥′+𝑦′2)+2(𝑥′+𝑦′2)2−30=0
Simplify.
2(x′−y′)(x′−y′)2−(x′−y′)(x′+y′)2+2(x′+y′)(x′+y′)2−30=0 x′2−2x′y′+y′2−(x′2−y′2)2+x′2+2x′y′+y′2−30=0 2x′2+2y′2−(x′2−y′2)2=30 2(2x′2+2y′2−(x′2−y′2)2)=2(30) 4x′2+4y′2−(x′2−y′2)=60 4x′2+4y′2−x′2+y′2=60 3x′260+5y′260=6060FOIL methodCombine like terms.Combine like terms.Multiply both sides by 2.Simplify.Distribute.Set equal to 1.2(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.
Write the equations with x′𝑥′ and y′𝑦′ in the standard form.
x′220+y′212=1𝑥′220+𝑦′212=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=0𝐴𝑥2+𝐵𝑥𝑦+𝐶𝑦2+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′ and y′𝑦′ coordinate system without the x′y′𝑥′𝑦′ term, by rotating the axes by a measure of θ𝜃 that satisfies
cot(2θ)=A−CBcot(2𝜃)=𝐴−𝐶𝐵
We have learned already that any conic may be represented by the second degree equation
Ax2+Bxy+Cy2+Dx+Ey+F=0𝐴𝑥2+𝐵𝑥𝑦+𝐶𝑦2+𝐷𝑥+𝐸𝑦+𝐹=0
where A,B,𝐴,𝐵, and C𝐶 are not all zero. However, if B≠0,𝐵≠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 θ𝜃 where cot(2θ)=A−CB.cot(2𝜃)=𝐴−𝐶𝐵.
- If cot(2θ)>0,cot(2𝜃)>0, then 2θ2𝜃 is in the first quadrant, and θ𝜃 is between (0°,45°).(0°,45°).
- If cot(2θ)<0,cot(2𝜃)<0, then 2θ2𝜃 is in the second quadrant, and θ𝜃 is between (45°,90°).(45°,90°).
- If A=C,𝐴=𝐶, then θ=45°.𝜃=45°.
HOW TO
Given an equation for a conic in the x′y′𝑥′𝑦′ system, rewrite the equation without the x′y′𝑥′𝑦′ term in terms of x′𝑥′ and y′,𝑦′, where the x′𝑥′ and y′𝑦′ axes are rotations of the standard axes by θ𝜃 degrees.
- Find cot(2θ).cot(2𝜃).
- Find sinθsin𝜃 and cosθ.cos𝜃.
- Substitute sinθsin𝜃 and cosθcos𝜃 into x=x′cosθ−y′sinθ𝑥=𝑥′cos𝜃−𝑦′sin𝜃 and y=x′sinθ+y′cosθ.𝑦=𝑥′sin𝜃+𝑦′cos𝜃.
- Substitute the expression for x𝑥 and y𝑦 into in the given equation, and then simplify.
- Write the equations with x′𝑥′ and y′𝑦′ 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 8×2−12xy+17y2=208𝑥2−12𝑥𝑦+17𝑦2=20 in the x′y′𝑥′𝑦′ system without an x′y′𝑥′𝑦′ term.
Solution
First, we find cot(2θ).cot(2𝜃). See Figure 7.
8×2−12xy+17y2=20⇒A=8,B=−12andC=17 cot(2θ)=A−CB=8−17−12 cot(2θ)=−9−12=348𝑥2−12𝑥𝑦+17𝑦2=20⇒𝐴=8,𝐵=−12and𝐶=17 cot(2𝜃)=𝐴−𝐶𝐵=8−17−12 cot(2𝜃)=−9−12=34
Figure 7
cot(2θ)=34=adjacentoppositecot(2𝜃)=34=adjacentopposite
So the hypotenuse is
32+42=h29+16=h225=h2h=532+42=ℎ29+16=ℎ225=ℎ2ℎ=5
Next, we find sinθsin𝜃 and cosθ.cos𝜃.
sinθ=1−cos(2θ)2−−−−−−−√=1−352−−−−√=55−352−−−−√=5−35⋅12−−−−−−√=210−−√=15−−√sinθ=15√cosθ=1+cos(2θ)2−−−−−−−√=1+352−−−−√=55+352−−−−√=5+35⋅12−−−−−−√=810−−√=45−−√cosθ=25√sin𝜃=1−cos(2𝜃)2=1−352=55−352=5−35⋅12=210=15sin𝜃=15cos𝜃=1+cos(2𝜃)2=1+352=55+352=5+35⋅12=810=45cos𝜃=25
Substitute the values of sinθsin𝜃 and cosθcos𝜃 into x=x′cosθ−y′sinθ𝑥=𝑥′cos𝜃−𝑦′sin𝜃 and y=x′sinθ+y′cosθ.𝑦=𝑥′sin𝜃+𝑦′cos𝜃.
x=x′cosθ−y′sinθx=x′(25√)−y′(15√)x=2x′−y′5√𝑥=𝑥′cos𝜃−𝑦′sin𝜃𝑥=𝑥′(25)−𝑦′(15)𝑥=2𝑥′−𝑦′5
and
y=x′sinθ+y′cosθy=x′(15√)+y′(25√)y=x′+2y′5√𝑦=𝑥′sin𝜃+𝑦′cos𝜃𝑦=𝑥′(15)+𝑦′(25)𝑦=𝑥′+2𝑦′5
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 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
Write the equations with x′𝑥′ and y′𝑦′ in the standard form with respect to the new coordinate system.
x′24+y′21=1𝑥′24+𝑦′21=1
Figure 8 shows the graph of the ellipse.
Figure 8
TRY IT #2
Rewrite the 13×2−63–√xy+7y2=1613𝑥2−63𝑥𝑦+7𝑦2=16 in the x′y′𝑥′𝑦′ system without the x′y′𝑥′𝑦′ term.
EXAMPLE 4
Graphing an Equation That Has No x′y′ Terms
Graph the following equation relative to the x′y′𝑥′𝑦′ system:
x2+12xy−4y2=30𝑥2+12𝑥𝑦−4𝑦2=30
Solution
First, we find cot(2θ).cot(2𝜃).
x2+12xy−4y2=20⇒A=1,B=12,and C=−4𝑥2+12𝑥𝑦−4𝑦2=20⇒𝐴=1,𝐵=12,and 𝐶=−4
cot(2θ)=A−CBcot(2θ)=1−(−4)12cot(2θ)=512cot(2𝜃)=𝐴−𝐶𝐵cot(2𝜃)=1−(−4)12cot(2𝜃)=512
Because cot(2θ)=512,cot(2𝜃)=512, we can draw a reference triangle as in Figure 9.
Figure 9
cot(2θ)=512=adjacentoppositecot(2𝜃)=512=adjacentopposite
Thus, the hypotenuse is
52+122=h225+144=h2169=h2h=1352+122=ℎ225+144=ℎ2169=ℎ2ℎ=13
Next, we find sinθsin𝜃 and cosθ.cos𝜃. We will use half-angle identities.
sinθ=1−cos(2θ)2−−−−−−−√=1−5132−−−−√=1313−5132−−−−−√=813⋅12−−−−−√=213√cosθ=1+cos(2θ)2−−−−−−−√=1+5132−−−−√=1313+5132−−−−−√=1813⋅12−−−−−√=313√sin𝜃=1−cos(2𝜃)2=1−5132=1313−5132=813⋅12=213cos𝜃=1+cos(2𝜃)2=1+5132=1313+5132=1813⋅12=313
Now we find x𝑥 and y. 𝑦.
x=x′cosθ−y′sinθx=x′(313√)−y′(213√)x=3x′−2y′13√𝑥=𝑥′cos𝜃−𝑦′sin𝜃𝑥=𝑥′(313)−𝑦′(213)𝑥=3𝑥′−2𝑦′13
and
y=x′sinθ+y′cosθy=x′(213√)+y′(313√)y=2x′+3y′13√𝑦=𝑥′sin𝜃+𝑦′cos𝜃𝑦=𝑥′(213)+𝑦′(313)𝑦=2𝑥′+3𝑦′13
Now we substitute x=3x′−2y′13√𝑥=3𝑥′−2𝑦′13 and y=2x′+3y′13√𝑦=2𝑥′+3𝑦′13 into x2+12xy−4y2=30.𝑥2+12𝑥𝑦−4𝑦2=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]=30 (113)[9x′2−12x′y′+4y′2+12(6x′2+5x′y′−6y′2)−4(4x′2+12x′y′+9y′2)]=30 (113)[9x′2−12x′y′+4y′2+72x′2+60x′y′−72y′2−16x′2−48x′y′−36y′2]=30 (113)[65x′2−104y′2]=30 65x′2−104y′2=390 x′26−4y′215=1 Factor.Multiply.Distribute.Combine like terms.Multiply. Divide by 390. (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]=30 Factor.(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=1 Divide by 390.
Figure 10 shows the graph of the hyperbola x′26−4y′215=1. 𝑥′26−4𝑦′215=1.
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=0𝐴𝑥2+𝐵𝑥𝑦+𝐶𝑦2+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′=0𝐴′𝑥′2+𝐵′𝑥′𝑦′+𝐶′𝑦′2+𝐷′𝑥′+𝐸′𝑦′+𝐹′=0
It may be shown that B2−4AC=B′2−4A′C′.𝐵2−4𝐴𝐶=𝐵′2−4𝐴′𝐶′. The expression does not vary after rotation, so we call the expression invariant. The discriminant, B2−4AC,𝐵2−4𝐴𝐶, 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 IDENTIFY A CONIC
If the equation Ax2+Bxy+Cy2+Dx+Ey+F=0𝐴𝑥2+𝐵𝑥𝑦+𝐶𝑦2+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′𝑥′2+𝐵′𝑥′𝑦′+𝐶′𝑦′2+𝐷′𝑥′+𝐸′𝑦′+𝐹′=0, then B2−4AC=B′2−4A′C′.𝐵2−4𝐴𝐶=𝐵′2−4𝐴′𝐶′.
The equation Ax2+Bxy+Cy2+Dx+Ey+F=0𝐴𝑥2+𝐵𝑥𝑦+𝐶𝑦2+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,𝐵2−4𝐴𝐶, 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.
- ⓐ 5×2+23–√xy+2y2−5=05𝑥2+23𝑥𝑦+2𝑦2−5=0
- ⓑ 5×2+23–√xy+12y2−5=05𝑥2+23𝑥𝑦+12𝑦2−5=0
Solution
- ⓐ Let’s begin by determining A,B,𝐴,𝐵, and C.𝐶.5Ax2+23–√Bxy+2Cy2−5=05︸𝐴𝑥2+23︸𝐵𝑥𝑦+2︸𝐶𝑦2−5=0Now, we find the discriminant.B2−4AC=(23–√)2−4(5)(2) =4(3)−40 =12−40 =−28<0𝐵2−4𝐴𝐶=(23)2−4(5)(2) =4(3)−40 =12−40 =−28<0Therefore, 5×2+23–√xy+2y2−5=05𝑥2+23𝑥𝑦+2𝑦2−5=0 represents an ellipse.
- ⓑ Again, let’s begin by determining A,B,𝐴,𝐵, and C.𝐶.5Ax2+23–√Bxy+12Cy2−5=05︸𝐴𝑥2+23︸𝐵𝑥𝑦+12︸𝐶𝑦2−5=0Now, we find the discriminant.B2−4AC=(23–√)2−4(5)(12) =4(3)−240 =12−240 =−228<0𝐵2−4𝐴𝐶=(23)2−4(5)(12) =4(3)−240 =12−240 =−228<0Therefore, 5×2+23–√xy+12y2−5=05𝑥2+23𝑥𝑦+12𝑦2−5=0 represents an ellipse.
TRY IT #3
Identify the conic for each of the following without rotating axes.
- ⓐ x2−9xy+3y2−12=0𝑥2−9𝑥𝑦+3𝑦2−12=0
- ⓑ 10×2−9xy+4y2−4=0
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