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

  1. x 2 9 x y + 3 y 2 12 = 0
  2. 10 x 2 9 x y + 4 y 2 4 = 0
  1. hyperbola
  2. ellipse
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Access this online resource for additional instruction and practice with conic sections and rotation of axes.

Key equations

General Form equation of a conic section A x 2 + B x y + C y 2 + D x + E y + F = 0
Rotation of a conic section x = x cos   θ y sin   θ y = x sin   θ + y cos   θ
Angle of rotation θ , where  cot ( 2 θ ) = 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 A x 2 + B x y + C y 2 + D x + E y + F = 0 where A , B and C are not all zero. The values of A , B , and C determine the type of conic. See [link] .
  • Equations of conic sections with an x y term have been rotated about the origin. See [link] .
  • The general form can be transformed into an equation in the x and y coordinate system without the x y term. See [link] and [link] .
  • 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. See [link] .

Section exercises

Verbal

What effect does the x y term have on the graph of a conic section?

The x y term causes a rotation of the graph to occur.

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If the equation of a conic section is written in the form A x 2 + B y 2 + C x + D y + E = 0 and A B = 0 , what can we conclude?

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If the equation of a conic section is written in the form A x 2 + B x y + C y 2 + D x + E y + F = 0 , and B 2 4 A C > 0 , what can we conclude?

The conic section is a hyperbola.

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Given the equation a x 2 + 4 x + 3 y 2 12 = 0 , what can we conclude if a > 0 ?

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For the equation A x 2 + B x y + C y 2 + D x + E y + F = 0 , the value of θ that satisfies cot ( 2 θ ) = A C B gives us what information?

It gives the angle of rotation of the axes in order to eliminate the x y term.

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Algebraic

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

9 x 2 + 4 y 2 + 72 x + 36 y 500 = 0

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x 2 10 x + 4 y 10 = 0

A B = 0 , parabola

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2 x 2 2 y 2 + 4 x 6 y 2 = 0

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4 x 2 y 2 + 8 x 1 = 0

A B = 4 < 0 , hyperbola

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4 y 2 5 x + 9 y + 1 = 0

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2 x 2 + 3 y 2 8 x 12 y + 2 = 0

A B = 6 > 0 , ellipse

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4 x 2 + 9 x y + 4 y 2 36 y 125 = 0

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3 x 2 + 6 x y + 3 y 2 36 y 125 = 0

B 2 4 A C = 0 , parabola

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3 x 2 + 3 3 x y 4 y 2 + 9 = 0

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2 x 2 + 4 3 x y + 6 y 2 6 x 3 = 0

B 2 4 A C = 0 , parabola

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x 2 + 4 2 x y + 2 y 2 2 y + 1 = 0

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8 x 2 + 4 2 x y + 4 y 2 10 x + 1 = 0

B 2 4 A C = 96 < 0 , ellipse

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

3 x 2 + x y + 3 y 2 5 = 0 , θ = 45°

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4 x 2 x y + 4 y 2 2 = 0 , θ = 45°

7 x 2 + 9 y 2 4 = 0

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2 x 2 + 8 x y 1 = 0 , θ = 30°

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2 x 2 + 8 x y + 1 = 0 , θ = 45°

3 x 2 + 2 x y 5 y 2 + 1 = 0

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4 x 2 + 2 x y + 4 y 2 + y + 2 = 0 , θ = 45°

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For the following exercises, determine the angle θ that will eliminate the x y term and write the corresponding equation without the x y term.

x 2 + 3 3 x y + 4 y 2 + y 2 = 0

θ = 60 , 11 x 2 y 2 + 3 x + y 4 = 0

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4 x 2 + 2 3 x y + 6 y 2 + y 2 = 0

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9 x 2 3 3 x y + 6 y 2 + 4 y 3 = 0

θ = 150 , 21 x 2 + 9 y 2 + 4 x 4 3 y 6 = 0

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−3 x 2 3 x y 2 y 2 x = 0

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16 x 2 + 24 x y + 9 y 2 + 6 x 6 y + 2 = 0

θ 36.9 , 125 x 2 + 6 x 42 y + 10 = 0

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x 2 + 4 x y + 4 y 2 + 3 x 2 = 0

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x 2 + 4 x y + y 2 2 x + 1 = 0

θ = 45 , 3 x 2 y 2 2 x + 2 y + 1 = 0

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4 x 2 2 3 x y + 6 y 2 1 = 0

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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

2 2 ( x + y ) = 1 2 ( x y ) 2

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x 2 4 + y 2 1 = 1 , θ = 45

( x y ) 2 8 + ( x + y ) 2 2 = 1

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y 2 16 + x 2 9 = 1 , θ = 45

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y 2 x 2 = 1 , θ = 45

( x + y ) 2 2 ( x y ) 2 2 = 1

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x = ( y 1 ) 2 , θ = 30

3 2 x 1 2 y = ( 1 2 x + 3 2 y 1 ) 2

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x 2 9 + y 2 4 = 1 , θ = 30

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For the following exercises, graph the equation relative to the x y system in which the equation has no x y term.

x 2 + 10 x y + y 2 6 = 0

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x 2 10 x y + y 2 24 = 0

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4 x 2 3 3 x y + y 2 22 = 0

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6 x 2 + 2 3 x y + 4 y 2 21 = 0

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11 x 2 + 10 3 x y + y 2 64 = 0

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21 x 2 + 2 3 x y + 19 y 2 18 = 0

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16 x 2 + 24 x y + 9 y 2 130 x + 90 y = 0

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16 x 2 + 24 x y + 9 y 2 60 x + 80 y = 0

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13 x 2 6 3 x y + 7 y 2 16 = 0

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4 x 2 4 x y + y 2 8 5 x 16 5 y = 0

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For the following exercises, determine the angle of rotation in order to eliminate the x y term. Then graph the new set of axes.

6 x 2 5 3 x y + y 2 + 10 x 12 y = 0

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6 x 2 5 x y + 6 y 2 + 20 x y = 0

θ = 45

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6 x 2 8 3 x y + 14 y 2 + 10 x 3 y = 0

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4 x 2 + 6 3 x y + 10 y 2 + 20 x 40 y = 0

θ = 60

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8 x 2 + 3 x y + 4 y 2 + 2 x 4 = 0

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16 x 2 + 24 x y + 9 y 2 + 20 x 44 y = 0

θ 36.9

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For the following exercises, determine the value of k based on the given equation.

Given 4 x 2 + k x y + 16 y 2 + 8 x + 24 y 48 = 0 , find k for the graph to be a parabola.

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Given 2 x 2 + k x y + 12 y 2 + 10 x 16 y + 28 = 0 , find k for the graph to be an ellipse.

4 6 < k < 4 6

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Given 3 x 2 + k x y + 4 y 2 6 x + 20 y + 128 = 0 , find k for the graph to be a hyperbola.

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Given k x 2 + 8 x y + 8 y 2 12 x + 16 y + 18 = 0 , find k for the graph to be a parabola.

k = 2

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Given 6 x 2 + 12 x y + k y 2 + 16 x + 10 y + 4 = 0 , find k for the graph to be an ellipse.

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Questions & Answers

x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
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Practice Key Terms 3

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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