xli 
y being an arbitrary constant, and a’ a variable vector ; and 
since it must evidently give simpler and more symmetric 
results to suppose the vector y co-axal with ¢, than to make 
the contrary supposition, since we shall thus place the ori- 
gin of the new vectors a‘ upon the axis of revolution of the 
surface, let 
ey — ye = 0, ory = ge, (25) 
g being an arbitrary scalar, to be disposed of according to con- 
venience. Equations (24) and (25), combined with (6) and 
(17), will give, for every point of space, 
—a®=—(a—yler+97e+g(ae+ ea); (26) 
and therefore, for every point of the locus (15), 
—a? =r — gr + ge? + gp. (27) 
The second member of this last equation may be made an 
exact square, by assuming 
ge + 2gp = 9g’, thatis, g = mT = es (28) 
the scalar quotient 
foe =a the transformation p = a(1 — e*), © (29) 
being thus suggested to our attention ; and with this value of 
g we shall have, by (27), 
= a* = (2a — r)}, (30) 
2a = ¥(— a) + ¥(— a”); (31). 
so that either the sum or the difference of the distances of any 
point of the locus (15) from the-two foci of which the vectors 
are respectively 0 and 2ae, is equal to the constant 2a. It is. 
not difficult to prove that the upper or the lower sign is to be 
taken, in the formula (31), according as e? is < or >. 1. For 
the case ec? = 1, the recent transformation fails. 
- Again, to find whether the locus has a centre, we may 
that is, 
make 
a=ytdag'e + 8, (32) 
