第5.4节 抛物线

5.4 PARABOLAS

In this section we shall study the graph of the equation

y = ax² + bx +c,

which is a U-shaped curve called a vertical parabola. We begin with the general definition of a parabola in the plane.

Recall that the distance between a point P and a line L is the length of the perpendicular line from P to L, As in Figure 5.4.1. If we are given a line L and a point F not on L, the set of all points equidistant from L and F will form a U- shaped curve that passes midway between L and F. This curve is a parabola, shown in Figure 5.4.2.

DEFINITION OF PARABOLA

Given a line L and a point F not on the line, the set of all points equidistant from L and F is called the parabola with directrix L and focus F.

The line through the focus perpendicular to the directrix is called the axis of the parabola. The point where the parabola crosses the axis is called the vertex. These are illustrated in Figure 5.4.3.

As we can see from the figure, the parabola is symmetric about its axis. That is, if we fold the page along the axis, the parabola will fold upon itself. The vertex is just the point halfway between the focus an directrix. It is the point on the parabola which is closest to the directrix and focus.

When a ball is thrown into the air, its path is the parabola shown in Figure 5.4.4, which the highest point at the vertex.

Telescope mirrors and radar antennae are in the shape of parabolas. This is done because all light rays coming from the direction of the axis will be reflected to a single point, the focus (see Figure 5.4.5). For the same reason, reflectors for search lights and automobile headlights are shaped like parabolas, with the light at the focus.

Figure 5.4.3

A parabola with a vertical axis (and horizontal directrix) is called a vertical parabola. The vertex of a vertical parabola is either the highest or lowest point, because it is the point closest to the directrix.

EXAMPLE 1 Find an equation for the vertical parabola with directrix y= -1 and focus F(0,1)

(Figure 5.4.6).

Figure 5.4.6

Given a point P (x, y), the perpendicular from P to the directrix is a vertical line of length _____. Thus

distance from P to directrix = ___________________.

Also, distance from P to focus = ___________________.

The point P lies on the parabola exactly when these distances are equal,

_______________________________

The equation of a parabola is particularly simple if the coordinate axes are chosen so that the vertex is at the origin and the focus is on the y-axis. The parabola will then be vertical and have an equation of the form y=ax².

THEOREM 1

The graph of the equation

y = ax²

(where a ≠ 0) is the parabola with focus F (0,1/4a) and directrix y= -1/ (4a). Its vertex is (0,0), and its axis is the y-axis.

PROOF Let us find the equation of the parabola with focus F(0, d) and directrix y= -d,

shown in Figure 5.4.7.

Our plan is to show that the equation is y=ax² where d=1/(4a). Given a point P(x, y), the perpendicular from P to the directrix is a vertical line of length _______ . Thus

distance from P to directrix = ____________________.

Also, distance from P to focus = _____________________.

The point P lies on the parabola exactly when these distances are equal,

______________________________.

Figure 5.4.7

Simplifying we get

Putting a= 1/4d, we have d=1/4a where y=ax² is the equation of the parabola.

Note that if a is negative, the focus will be below the x-axis and the directrix above the x-axis.

EXAMPLE 2 Find the focus and directrix of the parabola

In Theorem 1, a= -1/2 and d=1/4a = _______. The focus is F (0, _______), and the

directrix is y = _____.

The next theorem shows that the graph of y= ax² + bx + c is exactly like the graph of y=ax², except that its vertex is at the point (x0, y0) where the curve has slope zero. The focus and directrix are still at a distance 1/(4a) above and below the vertex.

THEOREM 2

The graph of the equation

y = ax² +bx + c

( where a ≠ 0 ) is a vertical parabola. Its vertex is at the point. (x0, y0) where the curve has slope zero, the focus is F((x0, y0 + 1/4a) and the directrix is y=y0 - 1/4a.

PROOF We first compute x0 . The curve y = ax² +bx + c has slope dy/dx= 2ax + b.

The slope is zero when 2ax + b = 0, x= -b/ 2a. Thus

x0 = - b/2a.

Let p be the parabola with focus F(x0, y0 + 1/4a) and directrix y=y0 - 1/4a.

Put X=x - x0 and Y = y -y0. In terms of X and Y, the focus and directrix are at

(X, Y) = (0, 1/4a), Y = -1/4a.

By Theorem 1, p has the equation

Y= a X ²,

or y - y0 = a(x - x0)²,

y = ax² - 2ax0x + (_________+ y0).

Substituting -b/2a for x0, we have

y =ax² + bx + (b²/4a+ y0).

This shows that the parabola p and the curve y=ax² + bx + c differ at most by a

constant. Moreover, the point (x0, y0) lies on the curve. (x0, y0) is also the vertex of

the parabola p, where (X, Y) = (0, 0). Therefore the curve and the parabola are the

same.

EXAMPLE 3 Find the vertex, focus and directrix of the parabola

y =2x² - 5x + 4

First find the point x0 where the slope is 0.

____________________

Then 4x0 - 5 = 0,

_____________.

Substitute to find y0.

_________________

The vertex is _______________.

We have a =2, so 1/4a =_______ . By Theorem 2, the focus is

_____________________

The directrix is

_______________________.

The vertex, axis, focus, and directrix can be used to sketch quickly the graph of a vertical parabola.

GRAPHING A PARABOLA y = ax² + bx + c

Step 1 Make a table of values of x, y, dy/dx, and d²y/dx² at x→ -∞, x= -b/2a

(the vertex), and x→∞.

Step 2 Compute the axis, vertex, focus, and directrix, and draw them.

Step 3 Draw the two squares with sides along the axis and directrix and a corner at the focus.

The two new corners level with the focus, P and Q, are on the parabola because they are

equidistant from the focus and the directrix.

Step 4 Draw the diagonals of the squares through P and Q. These are the tangent lines to the

parabola at P and Q. ( The proof of this fact is left as a problem.)

Step 5 Draw the parabola through the vertex, P, and Q, using the table and tangent lines. The

parabola should be symmetrical about the axis x= -b/2a. See Figure 5.4.8(a).

A horizontal parabola x= ay² + by + c can be graphed by the same method with the roles of x and y interchanged, as in Figure 5.4.8(b).

Figure 5.4.8

EXAMPLE 2 ( Continued) Sketch the parabola y= _________.

The first two derivatives are

_______________________.