第5.6节 二阶曲线

5.6 SECOND DEGREE CURVES

A second degree equation is an equation of the form

(1) Ax² + B x y +C y² + D x + E y + F = 0.

The graph of such an equation will be a conic section: a parabola, ellipse, hyperbola, or one of several

degenerate cases. In section 5.4 we saw that the graph of a second degree equation of one of the forms

(2) Ax² + D x + F = 0.

or

(3) Cy² + D x + Ey+F = 0.

is a parabola or degenerate. In section 5.5 we saw that the graph of a second degree equation of the form.

(4) Ax² + Cy² + F = 0.

is an ellipse, a hyperbola, or degenerate.

In this and the next section we shall see how to describe and sketch the graph of any second degree equation. We will begin with the Discriminant Test, which shows at once whether a nondegenerate curve is a parabola, ellipse, or hyperbola. The next topic in this section will be translation of axes, which can change any second degree equation with no xy-term,

(5) Ax² + Cy² + Dx + Ey + F = 0,

into an equation of one of the simple forms (2),(3), or (4).

In the following section we will study rotation of axes, which can change any second degree equation into an equation of the form (5) with no xy-term. We will then be able to deal with any second degree equation by using first rotation and then translation of axes.

Here is the Discriminant Test.

DEFINITION

The quantity B² - 4AC is called the discriminant of the equation

Ax² + Bxy + Cy² + Dx + Ey + F = 0.

DISCRIMINANT TEST

If we ignore the degenerate cases, the graph of a second degree equation is:

A parabola if the discriminant is zero.

An ellipse if the discriminant is negative.

A hyperbola if the discriminant is positive.

For example, the equation

x y - 1 = 0

Has positive discriminant 1² - 4·0 = 1, and its graph is a hyperbola. The equation

2x² + xy + y² -1 = 0

has negative discriminant 1² - 4·2·1 = -7, and its graph is an ellipse.

The degenerate graphs that can arise are: two straight lines, one straight line, one point, and the empty graph. The Discriminant Test alone does not tell whether or not the graph is degenerate. However, a degenerate case can usually be recognized when one tries to sketch the graph. For the remainder of this section we shall ignore the degenerate cases.

We now turn to the method of Translation of Axes. This method is useful for graphing a second degree equation with no xy - term.

Ax² + Cy² + Dx + Ey + F = 0.

If A or C is zero, the graph will be a horizontal or vertical parabola, which can be graphed by the method of Section 5.4. If both A and C are nonzero, the graph turns out to be an ellipse or hyperbola with horizontal and vertical axes X and Y, as in Figure 5.6.1. In the method of Translation of Axes, we take X and Y as a new pair of coordinate axes and get a new equation for the curve in the simple form

AX² + CY² + F1 = 0.

This curve can be sketched as in Section 5.5. The name “Translation of Axes” means that the original coordinate axes x and y are replaced by new coordinate axes X and Y, which are parallel to the original axes.

The new axes are found using a procedure from algebra called “completing the squares”. This procedure changes an expression like Ax² + Dx into a perfect square plus a constant.

FORMULA FOR COMPLETING THE SQUARES

Let A be different from zero. Then

Ax² + Dx = AX² + K,

Where __________________________

For example,

4x² - 3x = 4X² - 9/16

Where X = x- ____________.

We shall illustrate the method of Translation of Axes with an example and then describe the method in general.

EXAMPLE 1 Sketch the curve 4x² - y² - 16x - 2y + 11 = 0

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

B² - 4AC = 0² - 4·4·(-1) = 16.

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

Step 2 Simplify by completing the squares. This is done by putting.

______________________

and writing the original equation in terms of X and Y.

The X and Y terms cancel, and

4X² + 16 - 32 -Y² - 1 +2 +11 = 0,

4X² - Y²-4 =0.

Step 3 Draw dotted lines for the X and Y axes, and sketch the curve as in Section 5.5. This is a hyperbola in the (X,Y) -plane. The X-axis is the line Y=0, or y = -1. The Y-axis is the line X = 0, or x = 2. The graph is shown in Figure 5.6.2.

Figure 5.6.2

METHOD OF TRANSLATION OF AXES

When to Use To graph an equation of the form Ax² + Cy²+Dx+ Ey+ F = 0 where A and C are both nonzero.

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

Step 2 Completing the Squares: Put

______________________________

And rewrite the original equation in terms of X and Y. The new equation will have the simple form

Ax² + Cy² + F1 = 0

Where F1 is a new constant.

Step 3 Draw dotted lines for the X and Y axes and sketch the curve as in Section 5.5.

PROBLEMS FOR SECTION 5.6

In Problems 1-6, given that the graph is nondegenerate, use the Discriminant Test to determine whether the graph is a parabola, ellipse, or hyperbola.

1 x²+2xy-3y²+5x+6y-100 = 0

2 4x² - 8xy + 6y²+10x - 2y-20 = 0

3 4x² + 4xy + y²+7x + 8y = 0

4 9x² + 6xy + y²+6x - 22 = 0

5 x² + 5xy + 10y² - 16 = 0

6 4xy + 5x - 10y + 1 = 0

In Problems 7-18, use the method of Translation of Axes to sketch the curve.

7 x² + y² -4x + 3 = 0 8 x² + y² +2x -6y+6 = 0

9 x² - y² + 4x - 2y + 2 = 0 10 -x² + y² + 8x - 6y - 16 = 0

11 x² + 4y² - 4x + 24y + 36 = 0 12 4x² - 9y² + 8x + 18y - 41 = 0

13 9x² - 4y² - 36x - 24y - 36 = 0 14 -x² + 4y² + 16y + 12 = 0

15 -x² + 3y² + 8x + 30y + 56 = 0 16 5x² + 2y² + 10x + 12y + 28 = 0

17 16x² + 9y² -320x -108y + 1780 = 0

18 25x² + 4y² +250x -40y + 625 = 0