536 Polar, cylindrical, and spherical coordinates
This gives
(2)and squaring gives
(3)(2a - x)y = x x(2a - x),
The graph of this equation, including points for which x = a, x = 0, and y < 0,
is the cissoid. We now come to the polar coordinate problem. With the aid of
the fact that Q2 has polar coordinates (pi,g5) for which pi = 2a cos 0, try to use
Figure 10.192 to derive the polar equation of the cissoid. If unsuccessful, use
(2) and formulas which give x and y in terms of p and 0. Ans.:
(4) p = 2a sin 0 tan ¢.
Remark: Diodes employed the cissoid to "duplicate a cube," the problem being
to start with some line segment of length x (the length of an edge of a particular
cube) and construct a segment of length - x (the length of the edge of a cube
having double the volume of the original one). The line L through the points
(a,2a) and (2a,0) has the equation (2a - x) = y/2 and intersects the cissoid at
a point (x,y) for which y' = 2x' and hence y = x. What Diodes really
wanted to do was duplicate a cube by ruler-and-compass construction. This
has been proved to be impossible. It is possible to construct the line L with a
ruler and compass, and it is possible to construct points on the cissoid one by
one with a ruler and compass. The reason why Diodes failed to accomplish
his purpose should be explained. Life is too short to enable us to produce ruler-
and-compass constructions of all of the points on the cissoid, and there is no way
to prescribe rules for a ruler-and-compass construction of the particular point
where L intersects the cissoid.
17 The conchoids of Nicomedes provide a method (but not a ruler-and-com-
pass method, because no such method exists) for trisecting angles. Let p and q
be given positive numbers. Let 0 (the pole of the conchoid) be the origin and
let L (the directrix of the conchoid) be the line having the rectangular equation
x = p and the polar equation p cos 0 = p or p = p sec 0 as in Figure 10.193.
The conchoid consists of two parts or branches. When -7r/2 < 0 < it/2, the
line OM of the figure meets the line L at the point M and meets the far branch
(the branch most remote from the pole) at a point P whose distance from M is
q and whose distance from the origin is p sec q5 + q. The polar equation of the
far branch is therefore(1) p=psec 4,+q.
The same line OM meets the near branch (the branch nearer to the pole) at a
point P' whose distance from M is q and whose polar equation is(2) p=psecg5-q-
The equation
(3) p=psec4'±q
or the equation(4) (p - p sec 4')2 = q2