422 
ON THE DETERMINATION OF THE 
[476 
and with these values 
x = l + m cot S tan (c — A), 
y = m sec (c — A), 
and thence 
y' 2 — (x — l) 2 tan 2 S = m 2 , 
viz., this is the hyperbola obtained by rotating the orbit-plane about the line of nodes, 
longitude b — 90°. 
54. Imagine the orbit-plane (having upon it the hyperbola) brought by such 
rotation into the plane z = 0, or plane of the ecliptic, so that the hyperbola will be 
a curve in this plane, the inclination to Sx, or longitude of the axis Sx', being of 
course = b — 90°. Transforming the equation to axes Sx, Sy, we must write in the 
equation 
x' = x sin b — y cos b, 
y' = x cos b + y sin b, 
and the equation thus becomes 
(x cos b + y sin b) 2 — (x sin b —y cos b — Vf tan 2 & = m 2 . 
55. It will be recollected that the equations of the ray were 
writing herein 0 = 0 we find 
x — A __y — B _z — C 
f -~Y ~ h ’ 
x = A — Î G, = 
y = B - - G, = 
* g 
b 
h’ 
a 
h’ 
and it is clear that this point f ^^J should lie on the hyperbola 
Substituting for (x, y) the values in question, we have first 
b sin b + a cos b — hi «■ 
= ^ {(h 2 + (f cos b + g sin b) 2 ) (b sin b + a cos b) — h (ah — cf) (cos b + (bh — eg) sin b)} 
= ^ {(f cos b + g sin b) 2 (b sin b + a cos b) + (f cos b + g sin b) ch} 
'12' 
= — (f cos b + g sin b) {(f cos b + g sin b) (b sin b + a cos b) + ch (cos 2 b 4- sin 2 b)} 
= — (f cos b + g sin b) (— a sin b + b cos b) (f sin b — g cos b) ;