FORM OF TANGENTIAL EQUATION. 
427 
Here we have 
m= 
= ^T + c 8in " I ( csin ^+^ ( 255 ) 
(2) Find the intrinsic equation of the parallel to the curve 
v=jA (Q)dQ, 
or 
v=~E(cQ). 
In this case t /’ , (5) = A(0) ; and we find, as before, 
s= SsmS_A{6) + 1 sin _,^ gin ^ 
(256) 
( 3) The Cartesian equation of the curve parallel to the parabola is, by the equation of 
rt. 122, the result of eliminating <p between the equations 
#=2 a tan cp+£ sin <p, 
y= — a tan 2 <p — k cos <p. 
This problem may be solved more easily by finding the envelope of the line 
x-\-y cot <p — a tan <p — k cosec <p, 
or, what is the same thing, the line 
x sin 2<p+(«+y)cos 2<p — 2 k cos q>-\-(y— a— 2#)=0. 
Writing this in the form 
A sin 2<p-f B cos 2<p-j- C cos <p-fD=0, 
we get, by a known method, the required envelope to be 
{432(A 2 +B 2 )D + 9(D-6B 2 )C -2D 3 } 2 =4{12(A 2 +B 2 )+3C 2 +D 2 } 3 . . (257) 
Cor. The characteristics of this curve are 
We shall find the reciprocal of this curve in a future article. 
(4) The envelope of a fixed tangent to the generating circle of a cycloid is a parallel 
to another cycloid whose generating circle has a diameter equal to half the diameter of 
the former. 
