Week 11, Day 2

Today we learned about rotation and systems of equations. 

Horizontal or Vertices Axes: Ax^2 + Cy^2 + Dx +Ey + F = 0

Equation in an xy-Plane: Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

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. After the rotation, the equation of the conic in thw 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. The following theorem identifies how much to rotate the axes eliminate the xy-term and also the eauations for determining the new coefficients A', C', D', E', and F'. 

Rotation of Axes to Eliminate an xy-Term

The general second-degree equation Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 can be rewritten as:

A'(x')^2 + C'(y')^2 + D'x' + E'y' + F' = 0 by rotating the coordinate axes theough an angle theta, where x = x' cos(theta) - y'sin(theta) and y = x' sin(theta) + y' cos(theta). 

Example 1: Rotation of Axes for a Hyperbola

Write the equation xy -1 = 0 in standard form.

Because A = 0, B = 1, and C = 0, you have

cot2(theta) = A - C / B = 0

2(theta) = pi / 2

Theta = pi / 4

which implies that

x = x' cos (pi /4) - y' sin(pi / 4)

x = x' ((2)^1/2 / 2) - y' ((2)^1/2 / 2) 

x = x' - y' / (2)^1/2 and

y = x' sin (pi / 4) + y' cos (pi / 4)

y = x' ((2)^1/2 / 2) + y' ((2)^1/2 /2)

y = x' + y' / (2)^1/2

The equation in the x'y'-system is obtained by substituting these expressions into the equation xy -1 =  0. (x' - y' / (2)^1/2) x (x' + y' / (2)^1/2) - 1 = 0

(x')^2 - (y')^2 / 2 - 1 = 0

(x')^2 / ((2)^1/2)^2 - (y')^2 / ((2)^1/2)^2 = 1

In the x'y'-system, this is a hyperbola centered at the origin with vertices at (+/- (2)^2, 0). To find the coordinates of the vertices in the xy-system, substitute the coordinates (+/- (2)^1/2, 0) into the equations x = x' - y' /(2)^1/2 and y = x' + y' / (2)^1/2. 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.

Example 2: Rotation of Axes for an Ellipse

Sketch the graph of 7x^2 - 6(3)^1/2xy + 13y^2 - 16 = 0

Because A = 7, B = -6(3)^1/2, and C = 13, you have

cot2(theta) = A - C / B = 7 - 13 / 6(3)^1/2 = 1 / (3)^1/2

which implies that theta = pi / 6. The equation in the x'y'-system is obtained by making the substitutions x = x' cos(pi / 6) - y' sin (pi / 6)

x = x'((3)^1/2 / 2) - y'(1/2)

x = (3)^1/2x' - y' / 2 and

y = x' sin(pi / 6) + y' cos(pi / 6)

y = x'(1/2) + y'((3)^1/2 /2)

y = x' + (3)^1/2y' / 2

into the original equation. Thus, you have

7x^2 - 6(3)^1/2xy + 13y^2 - 16 = 0

7((3)^1/2x' - y' / 2)^2 - 6(3)^1/2((3)^1/2x' - y' / 2)(x' + (3)^1/2y' / 2) + 13(x' + (3)^1/2y'/ 2)^2 - 16= 0

which simplifies to: 4(x')^2 + 16(y')^2 - 16 = 0

4(x')^2 + 16(y')^2 = 16

(x')^2 / 4 + (y')^2 / 1 = 1.

This is the equation of an ellipse centered at the origin with vertices (+/- 2, 0) in the x'y'-system.

Example 3: Rotation of Axes for a Parabola

Sketch the graph of x^2 - 4xy + 4y^2 + 5(5)^1/2y + 1 = 0

Because A = 1, B = -4, and C = 4, you have cot2(theta) = A - C / B = 1 - 4 / -4 = 3/4

Using the identity cot2(theta) = (cot^2(theta) - 1) / (2cot(theta)) produces

cot2(theta) = 3/4 = cot^2 - 1 / 2cot(theta) from which you obtain the equation: 

4cot^2(theta) - 4 = 6cot(theta)

4cot^2(theta) - 6cot(theta) - 4 = 0

(2cot(theta) - 4)(2cot(theta) + 1) = 0

Considering 0 < theta < pi / 2, you have 2cot(theta) = 4.

Thus, cot(theta) = 2 ---> theta = 26.6 degrees.

From the triangle above, you obtain sin(theta) = 1 / (5)^1/2 and cos(theta) = 2 / (5)^1/2. 

Consequently, you use the substitutions: 

x = x' cos(theta) - y' sin(theta) = x'(2 / (5)^1/2) - y' (1 / (5)^1/2) = 2x' - y' / (5)^1/2

y = x' sin(theta) + y' cos(theta) = x' (1 / (5)^1/2) + y' (2 / (5)^1/2) = x' + 2y' / (5)^1/2.

Substituting these expressions into the original equation, you have x^2 - 4xy + 4y^2 + 5(5)^1/2 + 1=0

(2x' - y' / (5)^1/2) - 4(2x' - y' / (5)^1/2)(x' + 2y' / (5)^1/2) + 4(x' + 2y' / (5)^1/2)^2 + 5(5)^1/2(x' + 2y' / (5)^1/2) + 1 = 0, which simplifies as follows. 5(y')^2 + 5x' + 10y' + 1 = 0 ---> 5(y' + 1)^2 = -5x' + 4 

---> (y' + 1)^2 = (-1)(x' - 4/5). The graph of this equation is a parabola with vertex at (4/5, -1). Its axis is parallel to the x'- axis in the x'y'-system.

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---that is, F' = F. Such quantities are invariant under rotation. 

The next theorem lists other rotation invariants:

The rotation of the coordinate axes through an angle that transforms the equation:

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 has the following rotation invariants:

1. F = F'

2. A + C = A' + C'

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

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 B^2 - 4AC reduces to B^2 - 4AC = -4A'C'. This quantity is called the discriminant of the equation Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. Now, from the classification procedure given in the previous lesson, you know that the sign of A'C' determines the type of graph for the eauation A'(x')^2 + C'(y')^2 + D'x' + E'y' + F' = 0. Consequently, the sign of B^2 - 4AC will determine the type of graph for the original equation, as given in the following classification.

1. Ellipse or circle: B^2 - 4AC < 0

2. Parabola: B^2 - 4AC = 0

3. Hyperbola: B^2 - 4AC > 0