Calculus: Analytic Geometry and Calculus, with Vectors

372

and dividing by the right side gives the standard form

(6.36) a2 + b2= 1,

where b2 = a2(1 - e2) or

(6.37) at = a2 - b2.

Cones and conics

It is of interest to see that memorization of only a few details enables us

to find the graph, foci, and directrices of the graph of (6.36) when a and b

are given constants, say a = 5 and b = 2. Putting y = 0 shows that

points (a,0) and (-a,0) lie on the graph. The line segment joining these

points is the major axis of the graph. Putting x = 0 shows that the

points (0,b) and (0,-b) lie on the graph. The line joining these points

is the minor axis of the graph. The graph is an ellipse through these four

points. The foci always lie on the major axis. With or without the aid

of the string property of the ellipse,

Y we can remember that the foci lie on

(o,b) the major axis and on the circle of

Figure 6.38

radius a having its center at an end of

the minor axis. Then Figure 6.38,

which is a simpler version of Figure

6.34, shows that the distance from the

center of the ellipse to the foci can be

calculated by the Pythagorastheorem.

The distance is a2 - b2, and if we

will remember that this is at, then we

can calculate e. Finally, we can calculate the distance from the center to

the directrices if we remember that this distance is ale. The numbers

at and ale are the key numbers.

Problems 6.39

(^1) For each of the following pairs of values of a and b, sketch the ellipse
having the equation
x2
}
y2
as b,-= 1,
find the eccentricity, find the foci (give coordinates), and find the directrices
(give equations). Try to cultivate the ability to use the Pythagoras theorem and
key numbers without use of books or notes. Check the numerical results by use
of the fact that the distance p from a focus to its directrix must satisfy theequa-
tion e2p2 = b5(1 - e2).
(a) a=5,b=2 (b) a=3,b=1 (c) a=5,b=4