1’ORM OF TANGENTIAL EQUATION. 
CHAPTER II. 
Section I. — Transformation of Tangential into Cartesian Equations. 
26. The tangential equations v=f(<p) of a curve being given, to find its Cartesian 
equations. 
Consider two consecutive positions of the line x-\-y cot <p— *=0, such as LP, L'P 
intersecting in P ; then P is the point of contact 
of LP with its envelope, and the diameter of 
the circle described about the infinitesimal tri- 
angle LL'P is evidently equal ^=~=f'(<p); 
.-. LQ =/ (<p). Hence PE =f (<p) sin 2 <p and 
OE=OL + LE=/(p+/'(<p) sin (f> cos <p ; 
.•. the Cartesian coordinates of the point P are 
x = f{ty)’ J t~f (<p) sin <p cos <p . . (46) 
y— —/'(?>) sin 2 (p .... (47) 
These values can be got also by the analytical 
method of finding envelopes. For differentiating 
the equation x-\-y cot <p—f<p = 0 with respect to <p, we get y— —f'<p sin 2 p ; and substi- 
tuting this in the equation x-\-y cot <p —f<p, we get x=f{<p)-\-f(<$) sin <p cos <p. 
27. From the results of the last article we get 
the subtangent=/’ , (^>) sin <p cos <p (48) 
subnormal =f'(<p) sin 2 <p tan <p (49) 
28. If the movable line x-\-y cot <p — v=0 be a double tangent, it is evident that for 
the same values of v and <p we must have two different values for LP, one value corre- 
sponding to each point of contact. Hence, since LP =f (<p) sin <p, we must have two 
different values for f(<p ). This will happen when f'(<p ) is given in the form of a 
fraction whose numerator and denominator each vanish. Thus, suppose the equation of 
the curve to be given in the form 
F(v, <p)=0 ; 
then we have 
dv_ df ' dF ' 
d<p dip ' dv ’ 
Fig. 4. 
and therefore the conditions for a double tangent are 
d F 
dip 
= 0 , 
(50) 
which correspond with the conditions in Cartesian coordinates for a double point. 
3 h 2 
