476] ORBIT OF A PLANET FROM THREE OBSERVATIONS, 
or observing that 
423 
tan 8 = 
il 
we have 
f sin b — g cos b ’ 
(b sin b + a cos b — hi) tan 8 = — — (f cos b + g sin b) (— a sin b + b cos b); 
d L 
and hence the result of the substitution is at once found to be 
(— a sin b + b cos b) 2 — ^ (— a sin b 4- b cos b) 2 (g sin b + f cos b) 2 
= m 2 h 2 , = 
h 2 (— a sin b + b cos b) 2 
il 2 
viz., the factor (— a sin b + b cos b) 2 divides out, and the equation then becomes 
1 h 2 
1 “ jp (g sin b + f cos b) 2 = ^, 
il 2 ' 
that is 
il 2 = h 2 + (g sin b + f cos b) 2 , 
which is in fact the value of il 2 . 
56. I seek for the direction of the hyperbola at the point (r, — ^ in question. 
We have 
dx : dy = (b cos b — a sin b) sin b + cos b tan 2 S (b sin b + a cos b — hZ) 
: — (b cos b — a sin b) cos b + sin b tan 2 8 (b sin b + a cos b — hi), 
and from the above values of (b sin b + a cos b — hi) and tan 8, we have 
_ . 7 , , .. g sin b + f cos b . . 7 . 7 N 
tan 2 o (b sm b + a cos b — hi) = j , (-asmo+b cos b); 
v i sm b — g cos b 
whence 
dx : dy= (b cos b — a sin b) sin b (f sin b — g cos b) + (g sin b + f cos b) cos b (— a sin b + b cos b) 
: — (b cos b — a sin b) cos b (f sin b — g cos b) + (g sin b + f cos b) sin b (— a sin b + b cos 6), 
which, multiplying out and reducing by means of the relation af+bg + ch = 0, becomes 
dx : dy = (— a sin b + b cos b) (sin 2 b + cos 2 b) f : (— a sin b + b cos b) (sin 2 b + cos 2 b) g; 
that is 
dx : dy = i : g, or = |, 
which shows that the hyperbola, at the point ^^ where it meets the ray, touches 
the projection 
x — A_y — B 
f ~~ 
of the ray on the plane of xy, which contains the hyperbola.