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Topics in
Analytic Geometry
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10.5
ROTATION OF CONICS
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3. What You Should Learn
Rotate the coordinate axes to eliminate the xy-term in equations of conics.
Use the discriminant to classify conics.

4. Rotation

5. Rotation
We 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 form
Ax2 + 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 = 0
Horizontal or vertical axis
Equation in xy-plane

6. Rotation
To 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 Figure
Figure 10.44

7. Rotation
After the rotation, the equation of the conic in the new xy-plane will have the form
A(x)2 + C(y)2 + Dx + Ey + F  = 0.
Because this equation has no xy-term, you can obtain a standard form by completing the square.
Equation in xy-plane

8. Rotation
The 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.

9. Example 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 have
which implies that

10. Example 1 – Solution
cont’d
and

11. Example 1 – Solution
cont’d
The equation in the xy-system is obtained by substituting these expressions in the equation xy – 1 = 0.
Write in standard form.

12. Example 1 – Solution
cont’d
In the xy-system, this is a hyperbola centered at the origin with vertices at , as shown in Figure
Figure 10.45

13. Example 1 – Solution
cont’d
To find the coordinates of the vertices in the xy-system,
substitute the coordinates in the equations
This 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.

14. Invariants Under Rotation

15. Invariants 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.

16. Invariants 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 to
B2 – 4AC = – 4AC.
This quantity is called the discriminant of the equation
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
Discriminant

17. Invariants Under Rotation
Now, from the classification procedure, you know that the sign of AC determines the type of graph for the equation
A(x)2 + C(y)2 + Dx + Ey + F  = 0.

18. Invariants Under Rotation
Consequently, the sign of B2 – 4AC will determine the type of graph for the original equation, as given in the following classification.

19. Invariants Under Rotation
For example, in the general equation
3x2 + 7xy + 5y2 – 6x – 7y + 15 = 0
you have A = 3, B = 7, and C = 5. So the discriminant is
B2 – 4AC = 72 – 4(3)(5) = 49 – 60 = –11.
Because –11 < 0, the graph of the equation is an ellipse or a circle.

20. Example 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 = 0
b. x2 – 6xy + 9y2 – 2y + 1 = 0
c. 3x2 + 8xy + 4y2 – 7 = 0

21. Example 4(a) – Solution
Because 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

22. Example 4(a) – Solution
cont’d
Graph both of the equations to obtain the ellipse shown in Figure
Figure 10.49
Top half of ellipse
Bottom half of ellipse

23. Example 4(b) – Solution
cont’d
Because B2 – 4AC = 36 – 36 = 0, the graph is a parabola.
x2 – 6xy + 9y2 – 2y + 1 = 0
9y2 – (6x + 2)y + (x2 + 1) = 0
Write original equation.
Quadratic form ay2 + by + c = 0

24. Example 4(b) – Solution
cont’d
Graphing both of the equations to obtain the parabola shown in Figure
Figure 10.50

25. Example 4(c) – Solution
cont’d
Because B2 – 4AC = 64 – 48 > 0, the graph is a hyperbola.
3x2 + 8xy + 4y2 – 7 = 0
4y2 + 8xy + (3x2 – 7) = 0
Write original equation.
Quadratic form ay2 + by + c = 0

26. Example 4(c) – Solution
cont’d
The graphs of these two equations yield the hyperbola shown in Figure
Figure 10.51