ਸਿੱਖਣ ਦੇ ਉਦੇਸ਼
ਇਸ ਭਾਗ ਵਿੱਚ, ਤੁਸੀਂ:
ਆਮ ਰੂਪ ਸਮੀਕਰਨਾਂ ਦੁਆਰਾ ਦਿੱਤੇ ਗਏ ਗੈਰ-ਅਪਭ੍ਰਸ਼ਟ ਕੋਨਿਕ ਭਾਗਾਂ ਦੀ ਪਛਾਣ ਕਰੋ।
ਧੁਰਿਆਂ ਦੇ ਘੁੰਮਾਓ ਦੇ ਸੂਤਰਾਂ ਦੀ ਵਰਤੋਂ ਕਰੋ।
ਘੁੰਮਾਏ ਗਏ ਕੋਨਿਕਸ ਦੇ ਸਮੀਕਰਨਾਂ ਨੂੰ ਮਿਆਰੀ ਰੂਪ ਵਿੱਚ ਲਿਖੋ।
ਧੁਰਿਆਂ ਨੂੰ ਘੁੰਮਾਏ ਬਿਨਾਂ ਕੋਨਿਕਸ ਦੀ ਪਛਾਣ ਕਰੋ।
ਜਿਵੇਂ ਕਿ ਅਸੀਂ ਦੇਖਿਆ ਹੈ, ਕੋਨਿਕ ਭਾਗ ਉਦੋਂ ਬਣਦੇ ਹਨ ਜਦੋਂ ਇੱਕ ਸਮਤਲ ਦੋ ਸੱਜੇ ਚੱਕਰੀ ਕੋਨਾਂ ਨੂੰ ਸਿਰੇ-ਤੋਂ-ਸਿਰੇ ਜੋੜ ਕੇ ਕੱਟਦਾ ਹੈ ਅਤੇ ਵਿਰੋਧੀ ਦਿਸ਼ਾਵਾਂ ਵਿੱਚ ਅਨੰਤ ਤੱਕ ਫੈਲਦਾ ਹੈ, ਜਿਸਨੂੰ ਅਸੀਂ ਕੋਨ ਵੀ ਕਹਿੰਦੇ ਹਾਂ। ਜਿਸ ਤਰੀਕੇ ਨਾਲ ਅਸੀਂ ਕੋਨ ਨੂੰ ਕੱਟਦੇ ਹਾਂ, ਉਹ ਬਣਨ ਵਾਲੇ ਕੋਨਿਕ ਭਾਗ ਦੀ ਕਿਸਮ ਨਿਰਧਾਰਤ ਕਰੇਗਾ। ਇੱਕ ਚੱਕਰ ਕੋਨ ਦੇ ਸਮਰੂਪਤਾ ਧੁਰੇ ਦੇ ਲੰਬਵਤ ਸਮਤਲ ਦੁਆਰਾ ਕੋਨ ਨੂੰ ਕੱਟਣ ਨਾਲ ਬਣਦਾ ਹੈ। ਇੱਕ ਅੰਡਾਕਾਰ (ellipse) ਇੱਕ ਸਿੰਗਲ ਕੋਨ ਨੂੰ ਇੱਕ ਤਿਰਛੇ ਸਮਤਲ ਦੁਆਰਾ ਕੱਟਣ ਨਾਲ ਬਣਦਾ ਹੈ ਜੋ ਸਮਰੂਪਤਾ ਧੁਰੇ ਦੇ ਲੰਬਵਤ ਨਹੀਂ ਹੈ। ਇੱਕ ਪਰਵਲਯ (parabola) ਡਬਲ-ਕੋਨ ਦੇ ਉੱਪਰਲੇ ਜਾਂ ਹੇਠਲੇ ਹਿੱਸੇ ਵਿੱਚੋਂ ਸਮਤਲ ਨੂੰ ਲੰਘਾ ਕੇ ਬਣਦਾ ਹੈ, ਜਦੋਂ ਕਿ ਇੱਕ ਅਤਿ-ਪਰਵਲਯ (hyperbola) ਉਦੋਂ ਬਣਦਾ ਹੈ ਜਦੋਂ ਸਮਤਲ ਕੋਨ ਦੇ ਉੱਪਰਲੇ ਅਤੇ ਹੇਠਲੇ ਦੋਵੇਂ ਹਿੱਸਿਆਂ ਨੂੰ ਕੱਟਦਾ ਹੈ। ਚਿੱਤਰ 1 ਦੇਖੋ।
ਚਿੱਤਰ 1: ਗੈਰ-ਅਪਭ੍ਰਸ਼ਟ ਕੋਨਿਕ ਭਾਗ
ਅੰਡਾਕਾਰ, ਚੱਕਰ, ਅਤਿ-ਪਰਵਲਯ, ਅਤੇ ਪਰਵਲਯ ਨੂੰ ਕਈ ਵਾਰ ਗੈਰ-ਅਪਭ੍ਰਸ਼ਟ ਕੋਨਿਕ ਭਾਗ ਕਿਹਾ ਜਾਂਦਾ ਹੈ, ਜਦੋਂ ਕਿ ਅਪਭ੍ਰਸ਼ਟ ਕੋਨਿਕ ਭਾਗਾਂ ਦੇ ਉਲਟ, ਜੋ ਚਿੱਤਰ 2 ਵਿੱਚ ਦਿਖਾਏ ਗਏ ਹਨ। ਇੱਕ ਅਪਭ੍ਰਸ਼ਟ ਕੋਨ ਉਦੋਂ ਪੈਦਾ ਹੁੰਦਾ ਹੈ ਜਦੋਂ ਇੱਕ ਸਮਤਲ ਡਬਲ ਕੋਨ ਨੂੰ ਕੱਟਦਾ ਹੈ ਅਤੇ ਸਿਖਰ ਵਿੱਚੋਂ ਲੰਘਦਾ ਹੈ। ਸਮਤਲ ਦੇ ਕੋਣ ਦੇ ਆਧਾਰ 'ਤੇ, ਤਿੰਨ ਕਿਸਮਾਂ ਦੇ ਅਪਭ੍ਰਸ਼ਟ ਕੋਨਿਕ ਭਾਗ ਸੰਭਵ ਹਨ: ਇੱਕ ਬਿੰਦੂ, ਇੱਕ ਰੇਖਾ, ਜਾਂ ਦੋ ਕੱਟਣ ਵਾਲੀਆਂ ਰੇਖਾਵਾਂ।
ਚਿੱਤਰ 2: ਅਪਭ੍ਰਸ਼ਟ ਕੋਨਿਕ ਭਾਗ
ਆਮ ਰੂਪ ਵਿੱਚ ਗੈਰ-ਅਪਭ੍ਰਸ਼ਟ ਕੋਨਿਕਸ ਦੀ ਪਛਾਣ
ਇਸ ਅਧਿਆਇ ਦੇ ਪਿਛਲੇ ਭਾਗਾਂ ਵਿੱਚ, ਅਸੀਂ ਗੈਰ-ਅਪਭ੍ਰਸ਼ਟ ਕੋਨਿਕ ਭਾਗਾਂ ਲਈ ਮਿਆਰੀ ਰੂਪ ਸਮੀਕਰਨਾਂ 'ਤੇ ਧਿਆਨ ਕੇਂਦਰਿਤ ਕੀਤਾ ਹੈ। ਇਸ ਭਾਗ ਵਿੱਚ, ਅਸੀਂ ਆਮ ਰੂਪ ਸਮੀਕਰਨ ਵੱਲ ਆਪਣਾ ਧਿਆਨ ਬਦਲਾਂਗੇ, ਜਿਸਦੀ ਵਰਤੋਂ ਕਿਸੇ ਵੀ ਕੋਨਿਕ ਲਈ ਕੀਤੀ ਜਾ ਸਕਦੀ ਹੈ। ਆਮ ਰੂਪ ਨੂੰ ਸਿਫ਼ਰ ਦੇ ਬਰਾਬਰ ਸੈੱਟ ਕੀਤਾ ਜਾਂਦਾ ਹੈ, ਅਤੇ ਪਦ ਅਤੇ ਗੁਣਾਂਕ ਇੱਕ ਖਾਸ ਕ੍ਰਮ ਵਿੱਚ ਦਿੱਤੇ ਜਾਂਦੇ ਹਨ, ਜਿਵੇਂ ਕਿ ਹੇਠਾਂ ਦਿਖਾਇਆ ਗਿਆ ਹੈ।
ਜਿੱਥੇ A, B, ਅਤੇ C ਸਾਰੇ ਸਿਫ਼ਰ ਨਹੀਂ ਹਨ। ਅਸੀਂ ਗੁਣਾਂਕਾਂ ਦੇ ਮੁੱਲਾਂ ਦੀ ਵਰਤੋਂ ਇਹ ਪਛਾਣਨ ਲਈ ਕਰ ਸਕਦੇ ਹਾਂ ਕਿ ਦਿੱਤਾ ਗਿਆ ਸਮੀਕਰਨ ਕਿਸ ਕਿਸਮ ਦੇ ਕੋਨਿਕ ਨੂੰ ਦਰਸਾਉਂਦਾ ਹੈ।
ਤੁਸੀਂ ਨੋਟ ਕਰ ਸਕਦੇ ਹੋ ਕਿ ਆਮ ਰੂਪ ਸਮੀਕਰਨ ਵਿੱਚ ਇੱਕ xy ਪਦ ਹੈ ਜੋ ਅਸੀਂ ਕਿਸੇ ਵੀ ਮਿਆਰੀ ਰੂਪ ਸਮੀਕਰਨ ਵਿੱਚ ਨਹੀਂ ਦੇਖਿਆ ਹੈ। ਜਿਵੇਂ ਕਿ ਅਸੀਂ ਬਾਅਦ ਵਿੱਚ ਚਰਚਾ ਕਰਾਂਗੇ, xy ਪਦ ਕੋਨਿਕ ਨੂੰ ਘੁਮਾਉਂਦਾ ਹੈ ਜਦੋਂ ਵੀ B ਸਿਫ਼ਰ ਦੇ ਬਰਾਬਰ ਨਹੀਂ ਹੁੰਦਾ।
ਕਤਾਰ: ਕੋਨਿਕ ਭਾਗ | ਉਦਾਹਰਨ
ਕਤਾਰ: ਅੰਡਾਕਾਰ | 4x² +9y² =1
ਕਤਾਰ: ਚੱਕਰ | 4x² +4y² =1
ਕਤਾਰ: ਅਤਿ-ਪਰਵਲਯ | 4x² −9y² =1
ਕਤਾਰ: ਪਰਵਲਯ | 4x² =9y ਜਾਂ 4y² =9x
ਕਤਾਰ: ਇੱਕ ਰੇਖਾ | 4x+9y=1
ਕਤਾਰ: ਕੱਟਣ ਵਾਲੀਆਂ ਰੇਖਾਵਾਂ | (x−4)(y+4)=0
ਕਤਾਰ: ਸਮਾਂਤਰ ਰੇਖਾਵਾਂ | (x−4)(x−9)=0
ਕਤਾਰ: ਇੱਕ ਬਿੰਦੂ | 4x² +4y² =0
ਕਤਾਰ: ਕੋਈ ਗ੍ਰਾਫ ਨਹੀਂ | 4x² +4y² =−1
General Form of Conic Sections
A conic section has the general form
where A,B, A,B, and C C are not all zero.
Table 2 summarizes the different conic sections where B=0, B=0, and A A and C C are nonzero real numbers. This indicates that the conic has not been rotated.
row: ellipse | A x 2 +C y 2 +Dx+Ey+F=0,A≠Cand AC>0 A x 2 +C y 2 +Dx+Ey+F=0,A≠Cand AC>0
row: circle | A x 2 +C y 2 +Dx+Ey+F=0,A=C A x 2 +C y 2 +Dx+Ey+F=0,A=C
row: hyperbola | A x 2 −C y 2 +Dx+Ey+F=0or −A x 2 +C y 2 +Dx+Ey+F=0, A x 2 −C y 2 +Dx+Ey+F=0or −A x 2 +C y 2 +Dx+Ey+F=0, where A A and C C are positive
row: parabola | A x 2 +Dx+Ey+F=0or C y 2 +Dx+Ey+F=0 A x 2 +Dx+Ey+F=0or C y 2 +Dx+Ey+F=0
How To
Given the equation of a conic, identify the type of conic.
Rewrite the equation in the general form, A x 2 +Bxy+C y 2 +Dx+Ey+F=0. A x 2 +Bxy+C y 2 +Dx+Ey+F=0.
Identify the values of A A and C C from the general form. If A A and C C are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse. If A A and C C are equal and nonzero and have the same sign, then the graph may be a circle. If A A and C C are nonzero and have opposite signs, then the graph may be a hyperbola. If either A A or C C is zero, then the graph may be a parabola.If B = 0, the conic section will have a vertical and/or horizontal axes. If B does not equal 0, as shown below, the conic section is rotated. Notice the phrase “may be” in the definitions. That is because the equation may not represent a conic section 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 point: A x 2 +By2=0, A x 2 +By2=0, when A and B have the same sign. The degenerate case of a hyperbola is two intersecting straight lines: A x 2 +By2=0, A x 2 +By2=0, when A and B have opposite signs. On the other hand, the equation, A x 2 +By2+1=0, A x 2 +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 x 2 −9 y 2 +36x+36y−125=0 4 x 2 −9 y 2 +36x+36y−125=0
ⓑ 9 y 2 +16x+36y−10=0 9 y 2 +16x+36y−10=0
ⓒ 3 x 2 +3 y 2 −2x−6y−4=0 3 x 2 +3 y 2 −2x−6y−4=0
ⓓ −25 x 2 −4 y 2 +100x+16y+20=0 −25 x 2 −4 y 2 +100x+16y+20=0
Solution
ⓐ Rewriting the general form, we have A=4 A=4 and C=−9, C=−9, so we observe that A A and C C have opposite signs. The graph of this equation is a hyperbola.
ⓑ Rewriting the general form, we have A=0 A=0 and C=9. C=9. We can determine that the equation is a parabola, since A A is zero.
ⓒ Rewriting the general form, we have A=3 A=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=−25 A=−25 and C=−4. C=−4. Because AC>0 AC>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.
ⓐ 16 y 2 − x 2 +x−4y−9=0 16 y 2 − x 2 +x−4y−9=0
ⓑ 16 x 2 +4 y 2 +16x+49y−81=0 16 x 2 +4 y 2 +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 xy xy term, which aligns the graphs with the x- and y-axes. When we add an xy xy 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 x 2 + y 2 –xy–15=0 x 2 + y 2 –xy–15=0
We will find the relationships between x x and y y 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 i i and j . j . The rotated coordinate axes have unit vectors i ′ i ′ and j ′ . j ′ . The angle θ θ is known as the angle of rotation. See Figure 5. We may write the new unit vectors in terms of the original ones.
Figure 5: Relationship between the old and new coordinate planes.
Consider a vector u u in the new coordinate plane. It may be represented in terms of its coordinate axes.
Because u= x ′ i ′ + y ′ j ′ , u= x ′ i ′ + y ′ j ′ , we have representations of x x and y y in terms of the new coordinate system.
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 θ θ 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 ′ ):
and
How To
Given the equation of a conic, find a new representation after rotating through an angle.
Find x x and y y where x= x ′ cosθ− y ′ sinθ x= x ′ cosθ− y ′ sinθ and y= x ′ sinθ+ y ′ cosθ. y= x ′ sinθ+ y ′ cosθ.
Substitute the expression for x x and y y 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 x 2 −xy+2 y 2 −30=0 2 x 2 −xy+2 y 2 −30=0 after rotating through an angle of θ=45°. θ=45°.
Solution
Find x x and y, y, where x= x ′ cosθ− y ′ sinθ x= x ′ cosθ− y ′ sinθ and y= x ′ sinθ+ y ′ cosθ. y= x ′ sinθ+ y ′ cosθ.
Because θ=45°, θ=45°,
and
Substitute x= x ′ cosθ− y ′ sinθ x= x ′ cosθ− y ′ sinθ and y= x ′ sinθ+ y ′ cosθ y= x ′ sinθ+ y ′ cosθ into 2 x 2 −xy+2 y 2 −30=0. 2 x 2 −xy+2 y 2 −30=0.
Simplify.
Write the equations with x ′ x ′ and y ′ y ′ in the standard form.
This equation is an ellipse. Figure 6 shows the graph.
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 A x 2 +Bxy+C y 2 +Dx+Ey+F=0 A 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 ′ and y ′ y ′ coordinate system without the x ′ y ′ x ′ y ′ term, by rotating the axes by a measure of θ θ that satisfies
We have learned already that any conic may be represented by the second degree equation
where A,B, A,B, and C C are not all zero. However, if B≠0, B≠0, then we have an xy xy term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle θ θ where cot( 2θ )= A−C B . cot( 2θ )= A−C B .
If cot(2θ)>0, cot(2θ)>0, then 2θ 2θ is in the first quadrant, and θ θ is between (0°,45°). (0°,45°).
If cot(2θ)<0, cot(2θ)<0, then 2θ 2θ is in the second quadrant, and θ θ 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 θ θ degrees.
Find cot(2θ). cot(2θ).
Find sinθ sinθ and cosθ. cosθ.
Substitute sinθ sinθ and cosθ cosθ into x= x ′ cosθ− y ′ sinθ x= x ′ cosθ− y ′ sinθ and y= x ′ sinθ+ y ′ cosθ. y= x ′ sinθ+ y ′ cosθ.
Substitute the expression for x x and y y 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 x 2 −12xy+17 y 2 =20 8 x 2 −12xy+17 y 2 =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.
So the hypotenuse is
Next, we find sinθ sinθ and cosθ. cosθ.
Substitute the values of sinθ sinθ and cosθ cosθ into x= x ′ cosθ− y ′ sinθ x= x ′ cosθ− y ′ sinθ and y= x ′ sinθ+ y ′ cosθ. y= x ′ sinθ+ y ′ cosθ.
and
Substitute the expressions for x x and y y 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 new coordinate system.
Figure 8 shows the graph of the ellipse.
Try It #2
Rewrite the 13 x 2 −6 3 xy+7 y 2 =16 13 x 2 −6 3 xy+7 y 2 =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:
Solution
First, we find cot( 2θ ). cot( 2θ ).
Because cot( 2θ )= 5 12 , cot( 2θ )= 5 12 , we can draw a reference triangle as in Figure 9.
Thus, the hypotenuse is
Next, we find sinθ sinθ and cosθ. cosθ. We will use half-angle identities.
Now we find x x and y. y.
and
Now we substitute x= 3 x ′ −2 y ′ 13 x= 3 x ′ −2 y ′ 13 and y= 2 x ′ +3 y ′ 13 y= 2 x ′ +3 y ′ 13 into x 2 +12xy−4 y 2 =30. x 2 +12xy−4 y 2 =30.
Figure 10 shows the graph of the hyperbola x ′ 2 6 − 4 y ′ 2 15 =1. x ′ 2 6 − 4 y ′ 2 15 =1.
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
If we apply the rotation formulas to this equation we get the form
It may be shown that B 2 −4AC= B ′ 2 −4 A ′ C ′ . B 2 −4AC= B ′ 2 −4 A ′ C ′ . The expression does not vary after rotation, so we call the expression invariant. The discriminant, B 2 −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.
Using the Discriminant to Identify a Conic
If the equation A x 2 +Bxy+C y 2 +Dx+Ey+F=0 A 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 ′ x ′ 2 + B ′ x ′ y ′ + C ′ y ′ 2 + D ′ x ′ + E ′ y ′ + F ′ =0, then B 2 −4AC= B ′ 2 −4 A ′ C ′ . B 2 −4AC= B ′ 2 −4 A ′ C ′ .
The equation A x 2 +Bxy+C y 2 +Dx+Ey+F=0 A 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, B 2 −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.
ⓐ 5 x 2 +2 3 xy+2 y 2 −5=0 5 x 2 +2 3 xy+2 y 2 −5=0
ⓑ 5 x 2 +2 3 xy+12 y 2 −5=0 5 x 2 +2 3 xy+12 y 2 −5=0
Solution
ⓐ Let’s begin by determining A,B, A,B, and C. C. 5 ︸ A x 2 + 2 3 ︸ B xy+ 2 ︸ C y 2 −5=0 5 ︸ A x 2 + 2 3 ︸ B xy+ 2 ︸ C y 2 −5=0 Now, we find the discriminant. B 2 −4AC= ( 2 3 ) 2 −4(5)(2) =4(3)−40 =12−40 =−28<0 B 2 −4AC= ( 2 3 ) 2 −4(5)(2) =4(3)−40 =12−40 =−28<0 Therefore, 5 x 2 +2 3 xy+2 y 2 −5=0 5 x 2 +2 3 xy+2 y 2 −5=0 represents an ellipse.
ⓑ Again, let’s begin by determining A,B, A,B, and C. C. 5 ︸ A x 2 + 2 3 ︸ B xy+ 12 ︸ C y 2 −5=0 5 ︸ A x 2 + 2 3 ︸ B xy+ 12 ︸ C y 2 −5=0 Now, we find the discriminant. B 2 −4AC= ( 2 3 ) 2 −4(5)(12) =4(3)−240 =12−240 =−228<0 B 2 −4AC= ( 2 3 ) 2 −4(5)(12) =4(3)−240 =12−240 =−228<0 Therefore, 5 x 2 +2 3 xy+12 y 2 −5=0 5 x 2 +2 3 xy+12 y 2 −5=0 represents an ellipse.
Try It #3
Identify the conic for each of the following without rotating axes.
ⓐ x 2 −9xy+3 y 2 −12=0 x 2 −9xy+3 y 2 −12=0
ⓑ 10 x 2 −9xy+4 y 2 −4=0 10 x 2 −9xy+4 y 2 −4=0
Media
Access this online resource for additional instruction and practice with conic sections and rotation of axes.
Introduction to Conic Sections
Verbal
What effect does the xy xy term have on the graph of a conic section?
If the equation of a conic section is written in the form A x 2 +B y 2 +Cx+Dy+E=0 A x 2 +B y 2 +Cx+Dy+E=0 and AB=0, AB=0, what can we conclude?
If the equation of a conic section is written in the form A x 2 +Bxy+C y 2 +Dx+Ey+F=0, A x 2 +Bxy+C y 2 +Dx+Ey+F=0, and B 2 −4AC>0, B 2 −4AC>0, what can we conclude?
Given the equation a x 2 +4x+3 y 2 −12=0, a x 2 +4x+3 y 2 −12=0, what can we conclude if a>0? a>0?
For the equation A x 2 +Bxy+C y 2 +Dx+Ey+F=0, A x 2 +Bxy+C y 2 +Dx+Ey+F=0, the value of θ θ that satisfies cot( 2θ )= A−C B cot( 2θ )= A−C B gives us what information?
Algebraic
For the following exercises, determine which conic section is represented based on the given equation.
9 x 2 +4 y 2 +72x+36y−500=0 9 x 2 +4 y 2 +72x+36y−500=0
x 2 −10x+4y−10=0 x 2 −10x+4y−10=0
2 x 2 −2 y 2 +4x−6y−2=0 2 x 2 −2 y 2 +4x−6y−2=0
4 x 2 − y 2 +8x−1=0 4 x 2 − y 2 +8x−1=0
4 y 2 −5x+9y+1=0 4 y 2 −5x+9y+1=0
2 x 2 +3 y 2 −8x−12y+2=0 2 x 2 +3 y 2 −8x−12y+2=0
4 x 2 +9xy+4 y 2 −36y−125=0 4 x 2 +9xy+4 y 2 −36y−125=0
3 x 2 +6xy+3 y 2 −36y−125=0 3 x 2 +6xy+3 y 2 −36y−125=0
−3 x 2 +3 3 xy−4 y 2 +9=0 −3 x 2 +3 3 xy−4 y 2 +9=0
2 x 2 +4 3 xy+6 y 2 −6x−3=0 2 x 2 +4 3 xy+6 y 2 −6x−3=0
− x 2 +4 2 xy+2 y 2 −2y+1=0 − x 2 +4 2 xy+2 y 2 −2y+1=0
8 x 2 +4 2 xy+4 y 2 −10x+1=0 8 x 2 +4 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.
3 x 2 +xy+3 y 2 −5=0,θ=45° 3 x 2 +xy+3 y 2 −5=0,θ=45°
4 x 2 −xy+4 y 2 −2=0,θ=45° 4 x 2 −xy+4 y 2 −2=0,θ=45°
2 x 2 +8xy−1=0,θ=30° 2 x 2 +8xy−1=0,θ=30°
−2 x 2 +8xy+1=0,θ=45° −2 x 2 +8xy+1=0,θ=45°
4 x 2 + 2 xy+4 y 2 +y+2=0,θ=45° 4 x 2 + 2 xy+4 y 2 +y+2=0,θ=45°
For the following exercises, determine the angle θ θ that will eliminate the xy xy term and write the corresponding equation without the xy xy term.
x 2 +3 3 xy+4 y 2 +y−2=0 x 2 +3 3 xy+4 y 2 +y−2=0
4 x 2 +2 3 xy+6 y 2 +y−2=0 4 x 2 +2 3 xy+6 y 2 +y−2=0
9 x 2 −3 3 xy+6 y 2 +4y−3=0 9 x 2 −3 3 xy+6 y 2 +4y−3=0
−3 x 2 − 3 xy−2 y 2 −x=0 −3 x 2 − 3 xy−2 y 2 −x=0
16 x 2 +24xy+9 y 2 +6x−6y+2=0 16 x 2 +24xy+9 y 2 +6x−6y+2=0
x 2 +4xy+4 y 2 +3x−2=0 x 2 +4xy+4 y 2 +3x−2=0
x 2 +4xy+ y 2 −2x+1=0 x 2 +4xy+ y 2 −2x+1=0
4 x 2 −2 3 xy+6 y 2 −1=0 4 x 2 −2 3 xy+6 y 2 −1=0
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.
y=− x 2 ,θ=− 45 ∘ y=− x 2 ,θ=− 45 ∘
x= y 2 ,θ= 45 ∘ x= y 2 ,θ= 45 ∘
x 2 4 + y 2 1 =1,θ= 45 ∘ x 2 4 + y 2 1 =1,θ= 45 ∘
y 2 16 + x 2 9 =1,θ= 45 ∘ y 2 16 + x 2 9 =1,θ= 45 ∘
y 2 − x 2 =1,θ= 45 ∘ y 2 − x 2 =1,θ= 45 ∘
y= x 2 2 ,θ= 30 ∘ y= x 2 2 ,θ= 30 ∘
x= ( y−1 ) 2 ,θ= 30 ∘ x= ( y−1 ) 2 ,θ= 30 ∘
x 2 9 + y 2 4 =1,θ= 30 ∘ x 2 9 + y 2 4 =1,θ= 30 ∘
For the following exercises, graph the equation relative to the x ′ y ′ x ′ y ′ system in which the equation has no x ′ y ′ x ′ y ′ term.
xy=9 xy=9
x 2 +10xy+ y 2 −6=0 x 2 +10xy+ y 2 −6=0
x 2 −10xy+ y 2 −24=0 x 2 −10xy+ y 2 −24=0
4 x 2 −3 3 xy+ y 2 −22=0 4 x 2 −3 3 xy+ y 2 −22=0
6 x 2 +2 3 xy+4 y 2 −21=0 6 x 2 +2 3 xy+4 y 2 −21=0
11 x 2 +10 3 xy+ y 2 −64=0 11 x 2 +10 3 xy+ y 2 −64=0
21 x 2 +2 3 xy+19 y 2 −18=0 21 x 2 +2 3 xy+19 y 2 −18=0
16 x 2 +24xy+9 y 2 −130x+90y=0 16 x 2 +24xy+9 y 2 −130x+90y=0
16 x 2 +24xy+9 y 2 −60x+80y=0 16 x 2 +24xy+9 y 2 −60x+80y=0
13 x 2 −6 3 xy+7 y 2 −16=0 13 x 2 −6 3 xy+7 y 2 −16=0
4 x 2 −4xy+ y 2 −8 5 x−16 5 y=0 4 x 2 −4xy+ y 2 −8 5 x−16 5 y=0
For the following exercises, determine the angle of rotation in order to eliminate the xy xy term. Then graph the new set of axes.
6 x 2 −5 3 xy+ y 2 +10x−12y=0 6 x 2 −5 3 xy+ y 2 +10x−12y=0
6 x 2 −5xy+6 y 2 +20x−y=0 6 x 2 −5xy+6 y 2 +20x−y=0
6 x 2 −8 3 xy+14 y 2 +10x−3y=0 6 x 2 −8 3 xy+14 y 2 +10x−3y=0
4 x 2 +6 3 xy+10 y 2 +20x−40y=0 4 x 2 +6 3 xy+10 y 2 +20x−40y=0
8 x 2 +3xy+4 y 2 +2x−4=0 8 x 2 +3xy+4 y 2 +2x−4=0
16 x 2 +24xy+9 y 2 +20x−44y=0 16 x 2 +24xy+9 y 2 +20x−44y=0
For the following exercises, determine the value of k k based on the given equation.
Given 4 x 2 +kxy+16 y 2 +8x+24y−48=0, 4 x 2 +kxy+16 y 2 +8x+24y−48=0, find k k for the graph to be a parabola.
Given 2 x 2 +kxy+12 y 2 +10x−16y+28=0, 2 x 2 +kxy+12 y 2 +10x−16y+28=0, find k k for the graph to be an ellipse.
Given 3 x 2 +kxy+4 y 2 −6x+20y+128=0, 3 x 2 +kxy+4 y 2 −6x+20y+128=0, find k k for the graph to be a hyperbola.
Given k x 2 +8xy+8 y 2 −12x+16y+18=0, k x 2 +8xy+8 y 2 −12x+16y+18=0, find k k for the graph to be a parabola.
Given 6 x 2 +12xy+k y 2 +16x+10y+4=0, 6 x 2 +12xy+k y 2 +16x+10y+4=0, find k k for the graph to be an ellipse.