456
ON THE DETERMINATION OF THE
[476
or, what is the same thing,
xi = cosec b
2// = V 3 cot b,
sin 6 — V3 cos b ’
or as these may also be written
sin b + V3 cos 5 ’
, _ V3 (cos 5 — V3 sin 5)
sin 6 + V3 cos b
, V3 (cos b + V3 sin 5) _
sin 6 — V 3 cos 5
aV = cosec Z> , yí = V3 cot 6 ,
= — cosec (& + 60°), y¿ — V3 cot (6 + 60°),
a?/ = — cosec (6 — 60°), y¿ = V3 cot (6 — 60°),
so that for each of these sets we have
(The curve is in fact a section of the hyperboloid of revolution, a; 2 + y 2 — = 1,
which passes through the three rays.)
101. As regards the equation of the orbit I will first consider the particular
cases 5 = 90°, b = 0°, which should agree with the orbits for c = 90° in the planograms
1 and 2 respectively.
For b = 90° we have x = x, y' — y and
and the orbit is at once found to be
r = J(l-Vl3)(«'-l),
the eccentricity (regarded as positive) being thus ^ (v^lS — 1), ='7685 as before. For
b = 0° there is a discontinuity, and I write successively b = +e, and b = — e. For b = + e
we have x' = — y, y' = x, and
Xi = oo , y{ — oo V3,