128 PHILOSOPHICAL SOCIETY OF WASHINGTON. 
represents a surface of the second degree, it must be a hyperboloid 
of one sheet, for this and its varieties are the only ruled surfaces of 
that order. In the general form (9) it has a center in finite space. 
It is then the elliptic hyperboloid ; but if a = p (ora = qor p = q), 
it has its center at an infinite distance, and it is a parabolic hyperbo- 
loid. In this case the base surface a satis 
Ss ae = (11) 
which is a surface of ei of the Sec degree. 
If a = p= q, then (9) becomes a plane and the base surface a 
sphere. (9) is evidently satisfied by the center (a, 0, 0), therefore 
the intersecting surface always passes through the center of the 
base surface. 
I consider now the ellipsoid : 
0= 
2 
x? 2 
0O=C+R+5-1 (12) 
We have then the intersecting surface of the diorthode: 
# y 
0 = %%— 4% + Ay — ydz) + (a, 2% — %, % + Az — 2A) pr 
@ 
+ (Ys %, — I, % + Yaw — wAy) = (13) 
Let (0, y,, 2) be the eae wnire ane chord CL, 2) pierces the yz - plane 
(aw, 0, 2,) “ “ ae. * 
(z,, ws 0) éc “cc “ce cc “cc xy ‘a “ce 
then we can easily verify the relations: 
ts A WRIT WL eRe aan hig AO ba 
Sipe AX I ear Ax (14) 
Jee hp Ge re 
Ae Ay Pig aye Ay (14,) 
g, = HAS . y, = AAs (14,) 
and if we assume: 
a i 
a, =1— ;@42=1-— 
> (15,) 
B b ) 
ee hy yal (15,) 
e 
Bb 
(15,) 
