Calculus: Analytic Geometry and Calculus, with Vectors

376 Cones and conics

18 Study Figure 6.395 and discover the procedure by which the encircled

points are determined, and then use the procedure to obtain another encircled

Figure 6.395

point. Prove that the set of points obtained

by this procedure lie on an ellipse. Solution:

If the inner and outer circles have radiuses

(this time we use English; the Latin is radii)

a and b, and if a line is drawn through the

origin making the angle 0 with the positivex

axis, then the coordinates (x,y) of the resulting

encircled point are

(1) x=acos6, y=bsin0.

From these equations we obtain

(2)

X2^2

a--b:= cost 0 + sine 8 = 1,

so the point (x,y) lies on an ellipse.

19 Let 0 < b < a and let w > 0. Let a particle move in a plane in such a

way that, at each time t, the vector running from the origin of an x, y coordinate

system to it is

r = (a cos wt)i + (b sin wt)j.

Show that its path is an ellipse. Show that it is always accelerated toward the

origin and that the magnitude of the acceleration is

w2 -%Ifb-2+ (a2 - b2) cos2 wt.

Hint: To get started, let x = a cos wt, y = b sin wt, and observe the result of

some dividing and squaring and adding.

20 A point P moves around an ellipse having foci

(1) a2 --b2, 0), F2(,-,442- b2, 0)

in such a way that the vector r running from the origin to P at time t is

(2) r = (a cos t)i + (b sin t)j.

Show that the vector

(3) v = -(a sin t)i + (b cos t)j

is a forward tangent to the ellipse at P. Show that

(4) FkP=(acost+A a2-b2)i+bsintj,

where A = 1 when k = 1 and A = -1 when k = 2, and then show that

(5) 17kP12 = a2 + 2aX a2 - b2 cos t + X2(a2 - b2) cos2 t

and hence

(6) fFkPI = a + A a2 --b*- cos t.