Section 10.5 Rotation of Axes; General Form of a Conic 8 Objective 1: Identifying a Non-rotated Conic. The graph of the equation Ax + Bxy + Cy + Dx + Ey + F = 0 where A, B, and C cannot all be zero is a conic. The conics we have study thus far are non-rotated which correspond to the cases where B = 0. In some instances, the graph of the conic will be a pair of lines, or a single point, or no points. In these instances, the graph is called a degenerate conic. In the non-generative cases, we can determine the type of conic by examining the signs of A and C. General Equation of a Conic (B = 0) Theorem The graph of the equation: Ax + Cy + Dx + Ey + F = 0 where A and C cannot be both zero is a conic or a degenerative conic. If the conic is nondegenerative, then the graph is a) a parabola if A or C = 0 b) an ellipse if A and C have the same sign (circle if A = C) c) a hyperbola if A and C have opposite signs. Identify each equation without completing the squares: 1a) y 6x x + y = 0 1b) 11x + 5.5y x + 11y = 0 1c) x + y + 7y + = 0 Solution: a) Since A = 6 and C = 1, they have opposite signs so the equation is a hyperbola. b) Since A = 11 and C = 5.5, they have same signs so the equation is an ellipse. c) Since A = 0 and C = 1, the equation is a parabola. Objective : Use a Rotation of Axes to transform equations. To work with a conic equation that includes an xy-term (B 0), we will need to rotate the x- and y-axis by an angle θ about the origin to transform the equation to get rid of the xy-term. We can then manipulate and graph the equation in relationship to the rotated axes. To see how the rotation works, we will fix the origin and rotate the axes by an angle θ. We will label the
8 new axes x' and y' respectively. A point (x, y) in the xy-plane will have coordinates (x', y') in the new x'y'- plane. y' θ y θ x x' r y α (x, y) θ x y' r (x', y') α x' The axes rotated by The point (x, y) is The point (x', y') is an angle θ to form equal to the point equal to the point the x'y' plane. (rcos(θ + α), rsin(θ + α)). (rcos(α), rsin(α)). From the picture on the right above, x' = rcos(α) and y' = rsin(α) and from the picture in the middle, x = rcos(θ + α) and y = rsin(θ + α). But, using the sum of angles formula and regrouping, we get: x = rcos(θ + α) and y = rsin(θ + α) = rcos(θ)cos(α) rsin(θ)sin(α) = rsin(θ)cos(α) + rcos(θ)sin(α) = rcos(α)cos(θ) rsin(α)sin(θ) = rcos(α)sin(θ) + rsin(α)cos(θ) Now, replace rcos(α) by x' and rsin(α) by y': = x'cos(θ) y'sin(θ) = x'sin(θ) + y'cos(θ) Thus, x = x'cos(θ) y'sin(θ) and y = x'sin(θ) + y'cos(θ) Rotation of Axes Theorem: Given that x-axis and y-axis are rotated by an angle θ to form the x'-axis and the y'-axis. The coordinates of a point (x, y) in the xy-plane and the coordinates of same point (x', y') in the x'y'-plane have the following relationship: x = x'cos(θ) y'sin(θ) and y = x'sin(θ) + y'cos(θ) Express the following equation in the xy-plane in terms of x' and y' by rotating the axes through the given angle θ: Ex. x xy + y = 10 ; θ = 60
Solution: Since θ = 60, then from the Rotation of Axes Theorem, x = x'cos(60 ) y'sin(60 ) = 1 x' y' and y = x'sin(60 ) + y'cos(60 ) = x' + 1 y' Replace x by 1 x' y' and y by x' + 1 y' in the equation: x + xy y = 10 ( 1 x' y') ( 1 x' y')( x' + 1 y') + ( x' + 1 y') = 10 (expand and simplify) [ 1 (x') x'y' + (y') ] [ + [ (x') + (x') x'y' + + (combine like terms) (x') + (x') + 5 ( x ) 0 + ( y ) (x') 1 x'y' (y') ] x'y' + 1 (y') ] = 10 (y') (x') + x'y' + (x') + x'y' + (y') = 10 (x') + (y') + (x') x'y' + x'y' + x'y' (y') + (y') = 10 (y') (y') = 10 (divide both sides by 10 ) = 1 This is a graph of an ellipse centered at (0, 0) with the major axis parallel to the x'- axis with vertices (± 5, 0) on the x'-axis. The ellipse intersects the y'-axis at (0, ± ). Since c = a b = 0 = 16, then c =. The foci are on the x'-axis at (±, 0). The graph is given on the right. Notice that the rotation allows us to get rid of the xy-term from the original equation. We 8
need to now derive a formula for finding the angle. We will begin with the general form of the conic equation where B 0 and rotate axes through an angle θ. We want the x'y'-term to disappear, thus we will set the coefficient of the x'y'-term equal to zero and solve for θ. Ax + Bxy + Cy + Dx + Ey + F = 0 Now, rotate the axes through an angle θ by replacing x by [x'cos(θ) y'sin(θ)] and y by [x'sin(θ) + y'cos(θ)]: A[x'cos(θ) y'sin(θ)] + B[x'cos(θ) y'sin(θ)][x'sin(θ) + y'cos(θ)] + C[x'sin(θ) + y'cos(θ)] + D[x'cos(θ) y'sin(θ)] + E[x'sin(θ) + y'cos(θ)] + F = 0 (expand) 85 A(x') cos (θ) A(x')(y')cos(θ)sin(θ) + A(y') sin (θ) + B(x') cos(θ)sin(θ) + Bx'y'cos (θ) Bx'y'sin (θ) B(y') cos(θ)sin(θ) + C(x') sin (θ) + C(x')(y')cos(θ)sin(θ) + C(y') cos (θ) + Ex'sin(θ) + Ey'cos(θ) + F = 0 Now, we will examine only the coefficient of the x'y'-terms (in bold) since we are looking for the angle θ such that the coefficient of the xy-term is 0. Acos(θ)sin(θ) + Bcos (θ) Bsin (θ) + Ccos(θ)sin(θ) = 0 (regroup) Bcos (θ) Bsin (θ) Acos(θ)sin(θ) + Ccos(θ)sin(θ) = 0 (factor out B from the first two terms and cos(θ)sin(θ) from the last two) B[cos (θ) sin (θ)] (A C)cos(θ)sin(θ) = 0 (use the double angle formulas for sine and cosine) Bcos(θ) (A C)sin(θ) = 0 (add (A C)sin(θ) to both sides) Bcos(θ) = (A C)sin(θ) (divide both sides by Bsin(θ)) cos(θ) = cot(θ) = A C, where B 0. sin(θ) B Angle of Rotation Theorem The angle of rotation θ of the axes used to transform the equation: Ax + Bxy + Cy + Dx + Ey + F = 0, B 0, into an equation in x' and y' without an x'y'-term is given by cot(θ) = A C, where 0 < θ < 180. B Objective : Transform an Equation and Graph. Analyze the following equation Ex. x + 1xy + 9y Solution: x = 0
86 Since there is a xy-term, we need to rotate the axes to get rid of that term. Since A =, B = 1, and C = 9, then by the Angle of Rotation Theorem, cot(θ) = A C 9 = = 5 B 1 1. Since the cot(θ) is negative, then 90 < θ < 180 or θ in quadrant II. By definition, cot(θ) = x y = 5, which implies x = 5 and y = 1. 1 Thus, r = ( 5) +(1) = 169 =. Hence, cos(θ) = x r Using the half-angle formulas, we get: cos(θ) = 1+cos(θ) 1 cos(θ) = 1 5 1+ 5 sin(θ) = = = From the Rotation of Axes Theorem, x = x'cos(θ) y'sin(θ) = y = x'sin(θ) + y'cos(θ) = Replace x by x' equation: x + 1xy + 9y [ x' + 9[ = y' ] + 1[ x' + 5 6 +5 6 x' x' + = = y' y' = 5. 8 6 = = 18 6 = 9 = and y' and y by x = 0 y'] x' [ x' + y'][ x' y' in the x' + y'] y'] = 0 Expand: 16 (x') 8 6 x'y' + (y') + 7 (x') 60 7 x'y' (y') + 81 (x') + 108 6 x'y' + (y') 6x' 9y' = 0 Combine like terms: 16 (x') + 7 (x') + 81 (x') 8 60 108 x'y' x'y' + x'y' + 6 (y') 7 (y') + 6 (y') 6x' 9y' = 0 169 ( x') + 0x'y' + 0(y') 6x' 9y' = 0 (simplify) (x') 6x' 9y' = (divide both sides by ) (x') x' y' = 1 (x') x' + 1 = y' (add y' + 1 to both sides) (factor)
87 (x' 1) = y' This is the graph of a parabola with the vertex at (1, 0) on the x'y' plane that opens up in the positive y' direction and it's axis of symmetry is parallel to the y'-axis. To find the actual angle θ that the axes has been rotated, solve cos(θ) = for θ. This yields θ 56.. Ex. x + 6xy + y 8 = 0 Solution: Since there is a xy-term, we need to rotate the axes to get rid of that term. Since A = 1, B = 6, and C = 1, then by the Angle of Rotation Theorem, cot(θ) = A C B θ = 5. = 1 1 6 = 0. Thus, θ = cot 1 (0) = 90 or Since cos(5 ) = and sin(5 ) =, then from the Rotation of Axes Theorem: x = x'cos(5 ) y'sin(5 ) = x' y' and y = x'sin(5 ) + y'cos(5 ) = x' + y' Replace x by x' y' and y by x' + y' in the equation: x + 6xy + y 8 = 0 [ x' y'] + 6[ x' y'][ x' + y'] + [ x' + y'] 8 = 0 (expand) 1 (x') x'y' + 1 (y') + 6[ 1 (x') 1 (y') ] + 1 (x') + x'y' + 1 (y') 8 = 0 1 (x') x'y' + 1 (y') + (x') (y') + 1 (x') + x'y' + 1 (y') 8 = 0 (combine like terms)
88 1 (x') + (x') + 1 (x') x'y' + x'y' + 1 (y') (y') + 1 (y') 8 = 0 (x') (y') 8 = 0 (add 8 to both sides) (x') (y') = 8 (divide both sides by 8) ( x ) ( y ) = 1 This is a hyperbola centered at (0, 0) on the x'y'-plane. The transverse axis runs along the x' axis and the vertices are (±, 0). The oblique asymptotes are y' = ± b a x' = ± x' = ± x' Objective : Identifying a Conic with and xy-term without having to rotate the axes. Recall that b ac was the discriminant for a quadratic equation. With a general conic equation Ax + Bxy + Cy + Dx + Ey + F = 0 where A, B, and C cannot all be zero, we also have a discriminant B AC. The value of this expression does not change no matter what angle the axes is rotated. The discriminant for conics will allow us to determine what type of conic the equation is without having to perform the rotation of axes. General Equation of a Conic Theorem The graph of the equation: Ax + Bxy + Cy + Dx + Ey + F = 0 where A and C cannot be both zero is a conic or a degenerative conic. If the conic is non-degenerative, then the graph is a) a parabola if B AC = 0. b) an ellipse if B AC < 0. c) a hyperbola if B AC > 0. We will not prove it formerly, but merely outline the proof. Step 1: Rotate the equation: Ax + Bxy + Cy + Dx + Ey + F = 0 through and angle θ to get:
Step : Step : 89 A*(x') + B*x'y' + C*(y') + D*x' + E*y' + F* = 0 where A* = Acos (θ) + Bsin(θ)cos(θ) + Csin (θ), B* = B(cos (θ) sin (θ)) + (C A)(sin(θ)cos(θ)), C* = Asin (θ) Bsin(θ)cos(θ) + Ccos (θ), D* = Dcos(θ) + Esin(θ), E* = Dsin(θ) + Ecos(θ), & F* = F Using the results from step 1, show that: A* + C* = A + C (This shows that the sum of the coefficients of the squared terms does not change under the rotation). Using the results from steps 1 &, show that: B* A*C*= B AC (This shows that the discriminant does not change under the rotation). Step : Pick an angle θ to rotate the axes so that B* = 0. From step, B AC = 0 A*C* = A*C* Step 5: Use General Equation of a Conic (B = 0) Theorem for the transformed equation in step : Parabola If either A* or C* are zero, the conic is a parabola. But if either A* or C* are 0, then A*C* = 0 which implies B AC = 0 Ellipse If A* or C* have the same signs, the conic is an ellipse. But if A* or C* have the same signs, then A*C* < 0 which implies B AC < 0 Hyperbola If A* or C* have the opposite signs, the conic is a hyperbola. But if A* or C* have the opposite signs, then A*C* > 0 which implies B AC > 0 Identify each equation without rotating the axes: Ex. 5a x + xy + 5y x + y 7 = 0 Ex. 5b 7x + xy y + x + 9y 5 = 0 Ex. 5c 16x xy + 9y 7x 5y + 8 = 0 Solution: a) Since B AC = () ()(5) =, the equation is an ellipse. b) Since B AC = () (7)( ) = 88, the equation is a hyperbola. c) Since B AC = ( ) (16)(9) = 0, the equation is a parabola.