第5.7节 坐标轴旋转

5.7 ROTATION OF AXES

We have been how to graph any second degree equation with no xy-term. These graphs are parabolas, ellipses, or hyperbolas with vertical and horizontal axes. When the equation has a nonzero xy-term, the graph will have diagonal axes. By rotating the axes, one can get new coordinate axes in the proper direction. The method will give us a new equation that has no xy-term and can be graphed by our previous method.

Suppose the x and y axes rotated counterclockwise by an angle α, and the new coordinate axes are called X and Y, as in Figure 5.7.1. A point P in the plane will have a pair of coordinates ( x, y ) in the old coordinate system and ( X, Y ) in the new coordinate system. The old and new coordinates of P are related to each other by the equations for rotation of axes.

EQUATIONS FOR ROTATION OF AXES

x=X cos α-Y sin α, y = X sin α+ Y cos α

These equations can be seen directly from Figure 5.7.2. If we substitute the equations for rotation of axes into a second degree equation in x and y, we get a new second degree equation in the coordinates X and Y.

EXAMPLE 1 Find the equation of the curve

xy - 4 = 0

With respect to the new coordinate axes X and Y formed by a counterclockwise rotation of 30

degrees (Figure 5.7.3).

In this example

________________________

Thus _________________________

Substitute into the original equation and collect terms.

xy-4 = 0

___________________

_______________________

Given any second degree equation

(1) Ax² + Bxy + Cy²+Dx+Ey+ F = 0

and any angle of rotation α,one can substitute the equations of rotation and collect terms to get a new second

degree equation in the X and Y coordinates,

(2) A1X² + B1XY + C1Y²+D1X+E1Y+ F1 = 0.

It can be shown that the discriminant is unchanged by the rotation; that is,

B² - 4AC = ______ - 4A1C1

This gives a useful check on the computations.

In Example 1 above, the original discriminant is

B² - 4AC = 1²- 4·0·0 = 1.

The new equation has the same discriminant.

_____________________________

The trouble with Example 1 is that the new equation is more complicated than the original equation, and in particular there is still a nonzero XY-term. We would like to be able to choose the angle of rotation α so that the new equation has no XY- term, because we could then sketch the curve. The next theorem tells us which angle of rotation is needed.

THEOREM 1

Given a second degree equation

Ax² + Bxy + Cy² + Dx + Ey + F = 0

With B nonzero. Rotate the coordinate axes counterclockwise through an angle α for which

____________________________

Then the equation

A1X²+ B1XY+ C1Y²+D1X+ E1Y+F1 = 0

With respect to the new coordinate axes X and Y has XY - term B1 = 0.

This theorem can be proved as follows. When the rotation equations are substituted and terms collected, the XY coefficient B1 comes out to be

B1 = B (cos²α- sin²α) -2 (A-C) sin α cosα.

From trigonometry,

cos²α- sin²α = cos(2α), 2 sin α cos α = sin(2α).

Thus B1 = B cos(2α) - ( A-C ) sin (2α).

So B1 = 0 if and only if

B cos(2α) - ( A-C ) sin (2α) = 0,

__________________

or ____________________

As shown in Figure 5.7.4, α is the angle between the original coordinate axes and the axes of the parabola, ellipse, or hyperbola.

We are now ready to use rotation of axes to sketch a second degree curve. We illustrate the method for the curve introduced in Example 1.

EXAMPLE 2 Sketch the curve xy - 4 = 0.

Step 1 Apply the Discriminant Test to find the type of curve.

B² - 4AC =1² - 4·0·0 = 1.

The discriminant is positive, so the curve is a hyperbola.

Step 2 Find an angle α with

___________________________

__________________________

2α=90°, α=45°.

Step 3 Change coordinate axes using the rotation equations.

Figure 5.7.4

Substituting, we get

As a check, the discriminant is still ______________________.

Step 4 Draw the X and Y axes as dotted lines and sketch the curve.

The new axes are found by rotating the old axes by α=45°.

The curve is shown in Figure 5.7.5.

METHOD OF ROTATION OF AXES

When to Use To graph an equation of the form Ax²+ Bxy+ Cy²+ Dx + Ey + F = 0 where B is nonzero.

Step 1 Use the Discriminant Test to determine the type of curve.

Figure 5.7.5 Example 2

Step 2 Find an angle α with

________________________

Step 3 Change coordinate axes using the Rotation Equations. The new equation has the form

A1X² + C1Y² + D1X+E1Y+F1 = 0,

Where x=X cos α - Y sin α, y= X sin α + Y cos α.

Step 4 Draw the X - and Y-axes by rotating the old axes through the angle α. The curve can now be sketched by our previous method, using Translation of Axes if necessary.

Here is an overall summary of the use of rotations and translations of axes. The problem is to graph an equation of the form

Ax² + Bxy + Cy² + Dx + Ey + F = 0.

By Rotation of Axes, we get a new equation of the simpler form

A1X² + C1Y² + D1X + E1Y + F1 = 0.

If either A1=0 or C1 =0, the curve is a parabola that can be sketched by the method of Section 5.4. If A1 and C1 are both nonzero, Translation of Axes gives us a new equation of the simpler form

A2U² + B2V ²+F2 = 0.

The graph of this equation is an ellipse or hyperbola, which can be sketched by the method of section 5.5. The degenerate cases - two lines, one line, a point, or an empty graph - may also occur.

PROBLEMS FOR SECTION 5.7

In Problems 1-10, rotate the axes to transform the given equation into a new equation with no XY - term. Find the angle of rotation and the new equation.

1 xy+4 = 0 2 x² + xy + y² = 2

3 x² - 4xy + y² = 1 4 x² + 3xy + y² = 4

5 ________________ 6 ___________________

7 x² + xy = 3 8 2x² - xy - y² = 1

9 ________________ 10 _____________________

11 Prove that if we begin with a second degree equation with no first degree terms,

Ax² + Bxy + Cy² + F = 0, and then rotate axes, the new equation will again have no first degree terms.

□12 Prove that if we begin with a second degree equation with no first degree terms,

Ax² + Bxy + Cy² + F = 0, and then rotate axes, the new equation will again have no first degree

terms.

□13 Prove that the sum A + C is not changed by rotation of axes. That is, if Equation (2) is obtained

from Equation (1) by rotation of axes, then A+C = A1 + C1.

□14 Prove that the discriminant of a second degree equation is not changed by rotation of axes. That is, if

Equation (2) is obtained from Equation (1) by rotation of axes, then B² - 4 AC = _____ - 4A1C1.