第5.5节 椭圆与双曲线

5.5 ELLIPSES AND HYPERBOLAS

In this section we shall study two important types of curves, the ellipses and hyperbolas. The intersection of a circular cone and a plane will always be either a parabola, an ellipse, a hyperbola, or one of three degenerate cases - one line, two lines, or a point. For this reason, parabolas, ellipses, and hyperbolas are called conic sections. We begin with the definition of an ellipse in the plane.

DEFINITION OF ELLIPSE

Given two points, F1 and F2, and a constant, L, the ellipse with foci F1 and F2 and length L is

the set of all points the sum of whose distances from F1 and F2 is equal to L.

If the two foci F1 and F2 are the same, the ellipse is just the circle with center at the focus and diameter L. Circles are discussed in Section 1.1.

We shall concentrate on the case where the foci F1 and F2 are different. The ellipse will be an oval curve shown in Figure 5.5.1. The orbit of a planet is an ellipse with the sun at one focus. The eye sees a tilted circle as an ellipse.

Figure 5.5.1

The line through the foci F1 and F2 is called the major axis of the ellipse. The point on the major axis halfway between the foci is called the center. The line through the center perpendicular to the major axis is called the minor axis.

An ellipse is symmetric about both its major and its minor axes. That is, for any point P on the ellipse, the mirror image of P on the other side of either axis is also on the ellipse. The equation of an ellipse has a simple form when the major and minor axes are chosen for the x-axis and y-axis.

THEOREM 1

For any positive a and b, the graph of the equation

_______________________

is an ellipse with its center at the origin. There are three cases:

(i) a=b. The ellipse is a circle of radius a.

(ii) a>b. This is a horizontal ellipse, whose major axis is the x-axis, and whose minor axis is the y-axis.

The length is 2a. The foci are at (-c, 0) and (c, 0), where c is found by

c² = a² - b².

(iii) a<b. This is a vertical ellipse whose major axis is the y-axis and whose minor axis is the x-axis.

The length is 2b. The foci are at (0, -c) and (0, c), where c is found by

c² = b² - a².

Figure 5.5.2

This theorem is illustrated by Figure 5.5.2. Here is the proof in case (ii), a > b. A point P (x, y) is on the ellipse with foci (-c, 0 ), (c, 0) and length 2a if and only if the sum of the distances from P to the foci is 2a. That is,

____________________________.

Rewrite this as

___________________________

Square both sides:

________________________________

Simplify:

_________________________________

Square both sides again:

a² (x² + 2cx + c² + y²) = a4 + 2a²cx + c²x².

Collect the x² and y² terms and simplify.

x² (a² - c² )+ y²( a² ) = a4 - a²c² = a²(a² - c²).

Using the equation b² = a² - c², write this as

x²b² + y²a² = a²b².

Finally, divide by a²b² to obtain the required equation

_______________________.

Setting x=0 we see that the ellipse meets the y-axis at the two points y= + b. Also, it meets the x-axis at

x= + a. Since all terms are ≥ 0, at every point on the ellipse we have

______________________

and ________________________

Using these facts we can easily sketch the ellipse. It is an oval curve inscribed in the rectangle bounded by the lines x= +a, y= + b.

Figure 5.5.3 shows a horizontal ellipse (where a > b) and a vertical ellipse ( where a < b).

Figure 5.5.3

EXAMPLE 1 Sketch the curve ___________________.

The curve is an ellipse that cuts the x-axis at + 3 and the y-axis at +1. To sketch the curve, we first draw the rectangle x=+3, y=+1 with dotted lines and then inscribe the ellipse in the rectangle. The ellipse, shown in Figure 5.5.4, is horizontal.

Figure 5.5.4

EXAMPLE 2 Sketch the curve 4x²+y²=9 and find the foci.

The equation may be rewritten as

__________________________

The graph (Figure 5.5.5) is a vertical ellipse cutting the x-axis at ______ and the y=axis at +3.

Figure 5.5.5

By Theorem 1, the foci are on the y-axis at (0, + c). We compute c form the equation

c² = b² - a².

a and b are the x and y intercepts of the ellipse, a= ____, b=3. Thus

___________________

____________________

The foci are at (0, ____ ).

We turn next to the hyperbola. A hyperbola, like an ellipse, has two foci. However, the distances between the foci and a point on the hyperbola must have a constant difference instead of a constant sum.

DEFINITION OF HYPERBOLA

Given two distinct points, F1 and F2, and a constant, l, the hyperbola with foci F1 and F2 and

difference l is the set of all points the difference of whose distances from F1 and F2 is equal to l.

In this definition, l must be a positive number less than the distance between the foci. A hyperbola will have two separate branches, each shaped like a rounded V. On one branch the points are closer to F1 than F2; and on the other branch they are closer to F2 than F1. Figure 5.5.6 shows a typical hyperbola. The path of a comet on an orbit that will escape the solar system is a hyperbola with the sun at one focus. The shadow of a cylindrical lampshade on a wall is a hyperbola (the section of the light cone cut by the wall).

The line through the foci is the transverse axis of the hyperbola, and the point on the axis midway between the foci is the center. The hyperbola crosses the transverse axis at two points called the vertices. The line through the center perpendicular to the transverse axis is the conjugate axis. The hyperbola never crosses its conjugate axis. A hyperbola is symmetric about both axes. A simple equation is obtained when the transverse and conjugate axes are chosen for the coordinate axes.

Figure 5.5.6

THEOREM 2

For any positive a and b, the graph of the equation

_______________________

is a hyperbola with its center at the origin. Its transverse axis is the y-axis.

and its conjugate axis is the x-axis. The vertices are at (0, + b), and the foci are at (0, + c), where c is

found by

a² + b² = c².

The graph of the equation

__________________

is a hyperbola with similar properties with the roles of x, a and y, b reversed. The proof of Theorem 2 uses a computation like the proof of Theorem 1 on ellipses and is omitted.

Using derivatives and limits, we can get additional information that is helpful in sketching the graph of a hyperbola. By solving the equation

___________________________

for y as a function of x, we see that the upper and lower branches have the equations

upper branch: __________________

lower branch: ____________________

We concentrate on the upper branch. Its first two derivatives, after some algebraic simplification, come out to be

_____________________________

Thus the first derivative is zero only at x =0 (the vertex), and the second derivative is always positive. We have the following table of values for the upper branch.

All the limit computations are easy except for dy/dx, which we work out for x →∞.

Let H be positive infinite.

We carry out a similar computation for the limit as x→ -∞.

The table shows that the upper branch is almost a straight line with slope -b/a for large negative x and almost a straight line with slope b/a for large positive x. In fact, we shall show now that the lines

y=bx/a, y= -bx/a

are asymptotes of the hyperbola. That is, as x approaches ∞ or -∞, the distance between the line and the hyperbola approaches zero. We show that the upper branch approaches the line y=bx/a as x→∞; that is,

______________________________

Let H be positive infinite. Then

This is infinitesimal, so the limit is zero. Here are the steps for graphing a hyperbola y²/b² - x²/a²=1.

GRAPHING A HYPERBOLA ____________________

Step 1 Compute the values of a and b from the equation. Draw the rectangle with sides x= + a, y= +b.

Step 2 Draw the diagonals of the rectangle. They will be the asymptotes.

Step 3 Mark the vertices of the hyperbola at the points (0, + b).

Step 4 Draw the upper and lower branches of the hyperbola. The upper branch has a minimum

at the vertex (0, b), is concave upward, and approaches the diagonal asymptotes from

above. The lower branch is a mirror image. See Figure 5.5.7.

A hyperbola of the form

______________________

is graphed in a similar manner, but with the roles of x and y reversed. There is a left branch and a right branch, which are vertical at the vertices ( + a, 0).

EXAMPLE 3 Sketch the hyperbola 4y² - x² = 1 and find its foci.

First compute a and b.

4y² = y²/b², b=____

x² = x²/a², a= 1.

The rectangle has sides x = +1, y= ___, and the vertices are at (0,___ ).

The hyperbola is sketched using Steps 1-4 in Figure 5.5.8. The foci are at (0, + c) where

______________________

_____________________

Using the method of this section, we can sketch the graph of any equation of the form

Ax² + Cy² + F = 0.

In the ordinary case where A, C, and F are all different from zero, rewrite the equation as

A1 x² + C1 y² = 1,

Where A1 = -A/F, C1 = -C/F. There are four cases depending on the signs of A1 and C1, which are listed in Table 5.5.1.

If one or two of A, C, and F are zero, the graph will be degenerate (two lines, one line, a point, or empty).

PROBLEMS FOR SECTION 5.5

In Problems 1-12, find the foci and sketch the given ellipse or hyperbola.

1 x² + 4y² = 1 2 _________________

3 _________ 4 __________________

5 9x² + 4y² = 16 6 x² + 9y² = 4

7 y² - 4x² = 1 8 y² - x² = 4

9 9y² - x² = 4 10 4y² - 4x² = 1

11 x² - y² = 1 12 _______________

13 Prove that the hyperbola x²/a² - y²/b² =1 has the two asymptotes y=bx/a and y= -bx/a.