424 
ON THE DETERMINATION OE THE 
[476 
57. We may consider various particular forms of the hyperbola y' 2 — (x' — If tan 2 8 — to 2 . 
1°. If tanS = 0, the hyperbola is the pair of parallel lines y' 2 = m, 2 . 
This can only happen if h = 0, f cos 6 + g sin 6 = 0. The first equation gives af + bg = 0, 
, , 7 f b , ,, — a sin b + b cos 6 0 , . , . 
whence tan b — = - ; we have thus to = ^ = - which is consistent with 
g a if 0 
to finite. The equations show that the ray is parallel to the line of nodes. 
2°. If tan 8 = oo, the hyperbola is (x — If = 0, viz., the line x' = l twice : the 
condition is — f sin b + g cos b = 0; viz., the ray (not in general cutting the line of 
nodes) is at right angles to the line of nodes. 
3°. If to = 0, the hyperbola is the pair of intersecting lines y' 2 = (x — If tan 2 8. The 
condition is — a sin b + b cos 6 = 0, signifying that the ray cuts the line of nodes. 
4°. We may have simultaneously tanS = oo, to = 0. The hyperbola (as in 2°) is 
(x' — If = 0. The conditions are — f sin 6 + g cos 6 = 0, — a sin 6 + b cos 6 = 0, whence 
tan 6 = j = -, and therefore also ag — bf = 0; these signify that the ray cuts at right 
angles the line of nodes. 
o 
The line x' = l passes through the point ^, that is, we ought to have 
h 2 Z 2 = a 2 + b 2 . The value of l is in the first instance given in the form 
where 
l = ^ {(ah — cf) cos 6 + (bh — eg) sin 6}, 
O 2 = h 2 + (f cos 6 + g sin bf = h 2 + f 2 + g 2 — (— f sin 6 + g cos bf = f 2 + g 2 + h 2 . 
But observe that the equations 
ag — bf = 0, 
bg + af = — ch, 
give 
and thence 
consequently 
f: 
— ch 
— ch 
—a, gf = 1— b, 
a 2 + b 2 ° a 2 + b 2 
*->■***» 
, f , a 2 + b 2 + c 2 a 
a 2 + b 2 h ’ 
,, . , a 2 + b 2 + c 2 b 
bh — eg = bh —-—t—— = r LI; 
a 2 + b 2 h ’ 
I 2 
l = (a cos 6 + b sin 6) il 2 = ^ (a cos 6 + b sin 6) = ^ Va 2 4- b 2 , 
which is right.