Calculus: Analytic Geometry and Calculus, with Vectors

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10.2 Polar curves, tangents, and lengths 539

p > 0, so there can be no possible objection to use of Figure 10.21.
intrinsic equation (10.22) then gives


The

When cos ¢ < 1/e, we can solve for p to obtain the important "standard
form" equation


(10.23) p = ep
1-ecos4,-

To illustrate the fact that polar equations are usedas sources of infor-
mation, we proceed to study (10.23). When e < 1, the condition


cos j < 1/e is satisfied for each 0 and the polar graph of (10.23) is an
ellipse. In this case the point P with polar coordinates (p,4,) runs


repeatedly around the ellipse in the positive direction as 4, increases over
intervals of length 2,r. In case e = 1, the condition cos 0 < 1 means


that 4, is not an integer mutiple of 2a, and the polar graph of (10.23) isa
parabola. In this case P,,(p,4), the point having polar coordinates p
and 0, runs in the positive direction along arcs of the parabola as 0
increases over subintervals of the interval 0 < 0 < 2a. In case e > 1,
the restriction cos 0 < 1/e is more severe, and the polar graph of (10.23)
is one branch of a hyperbola. In this case, we confine 0 to the interval
¢o < 0 < 27r - 4o, where 4o is the first-quadrant angle for which
cos ¢o = 1/e. The point P,(p,4)) runs over arcs of the hyperbola as ¢


increases over subintervals of the interval 4,o < 0 < 27r - 4,o. Because
particles moving along one branch of a hyperbola do not suddenly jump
to another branch, and for other reasons, we are usually content to work
with only one branch of a hyperbola. We are, therefore, usually not
interested in the fact that if q5 is an angle for which I4I < 4)o so that
e cos 0 > 1, then the formula (10.23) determines a negative number p
and the point P,(p,o) lies on the other branch of the hyperbola. As
Figure 10.21 indicates, each conic K intersects the axis of K at a vertex
Y between the focus and directrix. Putting ¢ = xr in (10.23) shows that
the distance from F to V is ep/(1 + e). In case e = 1, this reduces, as
it should, to p/2. In case e 0 1, another vertex V' is obtained by setting
0 = 0. The polar coordinates of Y' are ep/(1 - e) and 0. In case
e < 1, our formulas give


(10.24) IPP'i=IYFI+IFP'I+e-


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The center of the conic lies midway between the vertices, and it is a
straightforward matter to continue this investigation to obtain additional
information.
In connection with a conic or other curve C having a manageable polar
equation, it is of interest to have information about the tangent to C

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