One cardioid (that with a=1/2) is easiest defined in polar coordiantes by r = 1+sin(Q). In cartesian coordiantes we would have (x2+y2-y)2 = (x2+y2)
The slope of the trajectory is .
For instance, for Q=0, the slope at the point (1,0) is 1.
The length of the trajectory is , something which is not so easy to integrate (or is it?).
The area enclosed by the trajectory turns out to be
, which is equal to
0.75*Q - cos(Q) -cos(Q)sin(Q)/4 = 1.5*P = 4.712