Calculus: Analytic Geometry and Calculus, with Vectors

385 Cones and conic,

Aristotle told Alexander the Great that there is no royal road to geometry,

but it is quite easy to see what we have done and it is even quite easy to see some

of the implications of what we have done. We have proved the theorem given

above in italics. We have described a simple ruler-and-compass procedure for

construction of points P on the hyperbola. We can use the construction to

produce 4 or even 40 points on the hyperbola, but we cannot thus produce all

of the points on the hyperbola and "construct the hyperbola." Our result

enables us to write the vector r from the origin 0 to a point P on the hyperbola

in the form

r = a sec: q5i+btan 0j,

but i is an "eccentric angle" which is not the angle between r and i. Persons

having inherent interest in geometry may, as an extramural excursion, consider

the special case in which the ray OT is the part of the asymptote in the first

quadrant. Considerable geometry is associated with the fact that, in this case,

the point Q is the focus F. To emphasize the fact that consideration of these

things need not be tedious, we observe that if tan (A = b/a, then cos _

a/ a2 -+b2, so lOQI = a2 + b2 and hence Q must be F.

18 A glance at Figure 6.494 suggests that the directrices may be the lines

determined by the points at which the asymptotes intersect the circle containing

the vertices. Prove that it is so.

19 Find the distance from a focus to a directrix of a hyperbola having the

standard equation x2/a2 - y2/b2 = 1. Ans.: b2/ a2 -{- b2.

20 Let a hyperbola have the standard equation x2/a2 - y2/b2 = 1. Let F

be the focus and let D be the directrix in the right half-plane. Let P1(x,,yl)

be a point on the hyperbola in the right half-plane. Show that

ON = ex, - a.

Let A' be the point in which the directrix D is intersected by the line through

P, parallel to an asymptote. Show that

I2PI = ex, - a.

21 The functions in the right members of the formulas

e° - e-e et+e e

sinh t = 2 cosh t = 2

are used often enough to justify introductions of special names and symbols for

them. They are called hyperbolic functions, and the h in sinh i and cosh t tells

us that the functions are hyperbolic sines and hyperbolic cosines. Show that

if x = a cosh t and y = b sinh t, then

x2y2

a=-b= =1.

This shows that if a particle P moves in the xy plane in such a way that, for each

t, the vector r running from the origin to P is

r = (a cosh kt)i + (b sinh kt)j,