5RotationWe have learned that the equation of a conic with axes parallel to one of the coordinate axes has a standard form that can be written in the general formAx2 + Cy2 + Dx + Ey + F = 0.In this section, you will study the equations of conics whose axes are rotated so that they are not parallel to either the x-axis or the y-axis.The general equation for such conics contains an xy-term.Ax2 + Bxy + Cy2 + Dx + Ey + F = 0Horizontal or vertical axisEquation in xy-plane
6RotationTo eliminate this xy-term, you can use a procedure called rotation of axes. The objective is to rotate the x- and y-axes until they are parallel to the axes of the conic.The rotated axes are denoted as the x-axis and the y-axis, as shown in FigureFigure 10.44
7RotationAfter the rotation, the equation of the conic in the new xy-plane will have the formA(x)2 + C(y)2 + Dx + Ey + F = 0.Because this equation has no xy-term, you can obtain a standard form by completing the square.Equation in xy-plane
8RotationThe following theorem identifies how much to rotate the axes to eliminate the xy-term and also the equations for determining the new coefficients A, C, D, E, and F.
9Example 1 – Rotation of Axes for a Hyperbola Write the equation xy – 1 = 0 in standard form.Solution:Because A = 0, B = 1, and C = 0, you havewhich implies that
11Example 1 – Solutioncont’dThe equation in the xy-system is obtained by substituting these expressions in the equation xy – 1 = 0.Write in standard form.
12Example 1 – Solutioncont’dIn the xy-system, this is a hyperbola centered at the origin with vertices at , as shown in FigureFigure 10.45
13Example 1 – Solutioncont’dTo find the coordinates of the vertices in the xy-system,substitute the coordinates in the equationsThis substitution yields the vertices (1, 1) and (–1, –1) in the xy-system.Note also that the asymptotes of the hyperbola have equations y = x, which correspond to the original x- and y-axes.
15Invariants Under Rotation In the rotation of axes theorem listed at the beginning of this section, note that the constant term is the same in both equations, F= F. Such quantities are invariant under rotation.The next theorem lists some other rotation invariants.
16Invariants Under Rotation You can use the results of this theorem to classify the graph of a second-degree equation with an xy-term in much the same way you do for a second-degree equation without an xy-term.Note that because B = 0, the invariant B2 – 4AC reduces toB2 – 4AC = – 4AC.This quantity is called the discriminant of the equationAx2 + Bxy + Cy2 + Dx + Ey + F = 0.Discriminant
17Invariants Under Rotation Now, from the classification procedure, you know that the sign of AC determines the type of graph for the equationA(x)2 + C(y)2 + Dx + Ey + F = 0.
18Invariants Under Rotation Consequently, the sign of B2 – 4AC will determine the type of graph for the original equation, as given in the following classification.
19Invariants Under Rotation For example, in the general equation3x2 + 7xy + 5y2 – 6x – 7y + 15 = 0you have A = 3, B = 7, and C = 5. So the discriminant isB2 – 4AC = 72 – 4(3)(5) = 49 – 60 = –11.Because –11 < 0, the graph of the equation is an ellipse or a circle.
20Example 4 – Rotation and Graphing Utilities For each equation, classify the graph of the equation, use the Quadratic Formula to solve for y, and then use a graphing utility to graph the equation.a. 2x2 – 3xy + 2y2 – 2x = 0b. x2 – 6xy + 9y2 – 2y + 1 = 0c. 3x2 + 8xy + 4y2 – 7 = 0
21Example 4(a) – SolutionBecause B2 – 4AC = 9 – 16 < 0, the graph is a circle or an ellipse. Solve for y as follows.Write original equation.Quadratic form ay2 + by + c = 0
22Example 4(a) – Solutioncont’dGraph both of the equations to obtain the ellipse shown in FigureFigure 10.49Top half of ellipseBottom half of ellipse
23Example 4(b) – Solutioncont’dBecause B2 – 4AC = 36 – 36 = 0, the graph is a parabola.x2 – 6xy + 9y2 – 2y + 1 = 09y2 – (6x + 2)y + (x2 + 1) = 0Write original equation.Quadratic form ay2 + by + c = 0
24Example 4(b) – Solutioncont’dGraphing both of the equations to obtain the parabola shown in FigureFigure 10.50
25Example 4(c) – Solutioncont’dBecause B2 – 4AC = 64 – 48 > 0, the graph is a hyperbola.3x2 + 8xy + 4y2 – 7 = 04y2 + 8xy + (3x2 – 7) = 0Write original equation.Quadratic form ay2 + by + c = 0
26Example 4(c) – Solutioncont’dThe graphs of these two equations yield the hyperbola shown in FigureFigure 10.51