( 761 ) 
By the transformation 
esv+u, YU 
the pencil «— ym passes into the pencil 2 = m, the polar line 
into v= 0, and we find, as we ought to, the divided equation 
dv u du 
one ae 
Finally we find as integral curves the conics 
(eg)? + Cet)? =4, 
touching one another in the singular points «= +1,y== 1. The 
point of intersection of the lines «— y= m is the double point of 
the pair of lines (e—y)*? = 4. 
2 2v 
When applying the transformation « + y= —, a—y = — we find 
u u 
that #—y =m passes into v= $ mu, whilst the polar line is brought 
at infinity. We find then, as we ought to, a homogeneous equation, 
namely 
dv u 
du ov 
For a ray af the pencil 
y 4+ 1=m(e—1) 
we find 
nen wl) y — Y—2) 9+) Lg Se et) may OS 2m 7e) 
~ (@—1) (@ +2) —e(y+l)”  24+2—mz (m—1) «—2 , 
thus for the tangent 
[(m—1) e—2] Y = [(m—1) me + (1—2m—m’)] X + 2 (m1). 
Therefore this pencil is also critical. The poles lie on the line 
y—e=2. 
ExAMPLE IV. 
The equation 
et ae 
— @—ay—y+ 
has «= 0, y=1 as singular point. 
The pencil y—1= me is critical and has y= 0 as polar line. 
By the substitution y— 1 = wa, 
1 v 
ut v uty 
