Calculus: Analytic Geometry and Calculus, with Vectors

378 Cones and conics

Hence (why?) P must lie on the line F1H2. Similarly (why?) P must lie on the
line F22H2. This proves that the lines F1H2 and F2H1 intersect at P. Therefore
(why?) the angles F1PG1 and F2PG2 are equal.
22 This problem involves a nested family of curves containing many ellipses
and one circle. Let a be a given positive number. For each positive number b,
the graph of the equation
2 2
a2+ b2

is an ellipse (or a circle) which intersects the x axis at the points (-a,0) and

(a,0). It is easy to generate interest in these graphs by sketching some of them.

When 0 < xi < a, the line having the equation x = x1 intersects each of these

graphs at two points. Prove that the tangents to these graphs at these points

all intersect at a point on the x axis. Solution: Using the result of Problem 8

(or working out the result again) shows that each tangent intersects the x axis

at the point (a2/x1, 0) which does not depend upon b.

23 For each 9 for which 0 < 8 < r/2 and 6 T r/4, the graph of the equation

x2 Y2

sine 0 cost 0 - 1

is an ellipse. Sketch several of these graphs.

24 The members of a family of confocal ellipses have foci at the points

(-1,0) and (1,0). Sketch good approximations to six of them. Suggestion:

Do not work too long on an easy problem. Select a point (0,b) and make Figure

6.38 tell you what the a in (a,0) must be.

25 Supposing that an ellipse E is given, tell how our little sister can use her

new drawing equipment to locate the center, the axes, and the foci of E. Hints:

The mid-points of parallel chords of E lie on a line through the center C of E.

It is easy to choose r such that the circle of radius r having center at C intersects

the ellipse in four points.

26 A rod of length L has a red end, a blue end, and a pink dot P at distance

q from the red end. Suppose that 0 < q < L. Show that if the red end is on

the x axis and the blue end is on the y axis, then (except when q = 4) P must lie

on an ellipse. 11ns.: The equation of the ellipse is

2 y2

(^27) Let F be a point which is inside a circle C but is not the center of C. A
Figure 6.397 little preliminary sketching shows that the set S of points
equidistant from C and F looks much like an ellipse
having a focus at F. What are the facts? Solution: As
in Figure 6.397, let F, be the center of the circle and let
F2 be the point F. The condition that P be equidistant
from C and F2 can be put in the form

(1) r - ZPI = API

where r is the radius of the circle. Before undertaking