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  1. 4 A C B 2 > 0 . If so, the graph is an ellipse.
  2. 4 A C B 2 = 0 . If so, the graph is a parabola.
  3. 4 A C B 2 < 0 . If so, the graph is a hyperbola.

The simplest example of a second-degree equation involving a cross term is x y = 1 . This equation can be solved for y to obtain y = 1 x . The graph of this function is called a rectangular hyperbola as shown.

Graph of xy = 1, which has asymptotes at the x and y axes. This hyperbola is relegated to the first and third quadrants, and the graph also has red dashed lines along y = x and y = −x.
Graph of the equation x y = 1 ; The red lines indicate the rotated axes.

The asymptotes of this hyperbola are the x and y coordinate axes. To determine the angle θ of rotation of the conic section, we use the formula cot 2 θ = A C B . In this case A = C = 0 and B = 1 , so cot 2 θ = ( 0 0 ) / 1 = 0 and θ = 45 ° . The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled A , B , C , D , E , and F , and are given by the formulas

A = A cos 2 θ + B cos θ sin θ + C sin 2 θ B = 0 C = A sin 2 θ B sin θ cos θ + C cos 2 θ D = D cos θ + E sin θ E = D sin θ + E cos θ F = F .

The procedure for graphing a rotated conic is the following:

  1. Identify the conic section using the discriminant 4 A C B 2 .
  2. Determine θ using the formula cot 2 θ = A C B .
  3. Calculate A , B , C , D , E , and F .
  4. Rewrite the original equation using A , B , C , D , E , and F .
  5. Draw a graph using the rotated equation.

Identifying a rotated conic

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

13 x 2 6 3 x y + 7 y 2 256 = 0 .

In this equation, A = 13 , B = −6 3 , C = 7 , D = 0 , E = 0 , and F = −256 . The discriminant of this equation is 4 A C B 2 = 4 ( 13 ) ( 7 ) ( −6 3 ) 2 = 364 108 = 256 . Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use cot 2 θ = A C B . This gives

cot 2 θ = A C B = 13 7 −6 3 = 3 3 .

Therefore 2 θ = 120 o and θ = 60 o , which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

A = A cos 2 θ + B cos θ sin θ + C sin 2 θ = 13 cos 2 60 + ( −6 3 ) cos 60 sin 60 + 7 sin 2 60 = 13 ( 1 2 ) 2 6 3 ( 1 2 ) ( 3 2 ) + 7 ( 3 2 ) 2 = 4, B = 0 , C = A sin 2 θ B sin θ cos θ + C cos 2 θ = 13 sin 2 60 + ( −6 3 ) sin 60 cos 60 = 7 cos 2 60 = ( 3 2 ) 2 + 6 3 ( 3 2 ) ( 1 2 ) + 7 ( 1 2 ) 2 = 16, D = D cos θ + E sin θ = ( 0 ) cos 60 + ( 0 ) sin 60 = 0, E = D sin θ + E cos θ = ( 0 ) sin 60 + ( 0 ) cos 60 = 0, F = F = −256.

The equation of the conic in the rotated coordinate system becomes

4 ( x ) 2 + 16 ( y ) 2 = 256 ( x ) 2 64 + ( y ) 2 16 = 1.

A graph of this conic section appears as follows.

Graph of an ellipse with equation 13x2 – 6 times the square root of 3 times xy + 7y2 – 256 = 0. The center is at the origin, and the ellipse appears to be skewed 60 degrees. There are dashed red lines along the major and minor axes.
Graph of the ellipse described by the equation 13 x 2 6 3 x y + 7 y 2 256 = 0 . The axes are rotated 60 ° . The red dashed lines indicate the rotated axes.
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Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

3 x 2 + 5 x y 2 y 2 125 = 0 .

The conic is a hyperbola and the angle of rotation of the axes is θ = 22.5 ° .

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Key concepts

  • The equation of a vertical parabola in standard form with given focus and directrix is y = 1 4 p ( x h ) 2 + k where p is the distance from the vertex to the focus and ( h , k ) are the coordinates of the vertex.
  • The equation of a horizontal ellipse in standard form is ( x h ) 2 a 2 + ( y k ) 2 b 2 = 1 where the center has coordinates ( h , k ) , the major axis has length 2 a, the minor axis has length 2 b , and the coordinates of the foci are ( h ± c , k ) , where c 2 = a 2 b 2 .
  • The equation of a horizontal hyperbola in standard form is ( x h ) 2 a 2 ( y k ) 2 b 2 = 1 where the center has coordinates ( h , k ) , the vertices are located at ( h ± a , k ) , and the coordinates of the foci are ( h ± c , k ) , where c 2 = a 2 + b 2 .
  • The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
  • The polar equation of a conic section with eccentricity e is r = e p 1 ± e cos θ or r = e p 1 ± e sin θ , where p represents the focal parameter.
  • To identify a conic generated by the equation A x 2 + B x y + C y 2 + D x + E y + F = 0 , first calculate the discriminant D = 4 A C B 2 . If D > 0 then the conic is an ellipse, if D = 0 then the conic is a parabola, and if D < 0 then the conic is a hyperbola.

Questions & Answers

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are you nano engineer ?
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fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
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That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
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Graphene has a hexagonal structure
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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