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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
2013-08-17 00:26:08
