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Have a look at the following theorem:

THEOREM 1

Rotation of Axes

The general equation of the conic

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

where B ≠ 0, can be rewritten as

A'(x')2 + B'x'y' + C'(y')2 + D'x' + E'y' + F' = 0

by rotating the coordinate axes through an angle θ, where Note that the constant term is the same in both equations. Because of this, F is said to be invariant under rotation. The following theorem lists some other rotation invariants. The proof of this theorem is left as an exercise.

Theorem 2

Rotation Invariants

The rotation of coordinate axes through an angle θ that transforms the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 into the form

A'(x')2 + B'x'y' + C'(y')2 + D'x' + E'y' + F' = 0

has the following rotation invariants.

1. F = F'

2. A + C = A' + C'

3. B2 - 4AC = (B')2 - 4A'C'

You can use 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' Discriminant

which is called the discriminant of the equation

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

Because the sign of A'C' determines the type of graph for the equation

A'(x')2 + B'x'y' + C'(y')2 + D'x' + E'y' + F' = 0

the sign of B2 - 4AC must determine the type of graph for the original equation. This result is stated in the following theorem.

Theorem 3

Classification of Conics by the Discriminant

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

is, except in degenerate cases, determined by its discriminant as follows.

 1. Ellipse or circle B2 - 4AC < 0 2. Parabola B2 - 4AC = 0 3. Hyperbola B2 - 4AC > 0

Example

Using the Discriminant

Classify the graph of each of the following equations.

a. 4xy - 9 = 0

b. 2x2 - 3xy + 2y2 - 2x = 0

c. x2 - 6xy + 9y2 - 2y + 1 = 0

d. 3x2 + 8xy + 4y2 - 7 = 0

Solution

a. The graph is a hyperbola because

B2 - 4AC = 16 - 0 > 0.

b. The graph is a circle or an ellipse because

B2 - 4AC = 9 - 16 < 0.

c. The graph is a parabola because

B2 - 4AC = 36 - 36 = 0

d. The graph is a hyperbola because

B2 - 4AC = 64 - 48 > 0 