209] A DEMONSTRATION OF SIR W. R. HAMILTON’S THEOREM &C. 263 
of the described arc and the sum of the two radius vectors. And this remark 
suggests a mode of investigation of the theorem. Consider the intercepted arc of one 
of the hodographs : the tangents to the hodograph at the extremities of this arc are 
radii of the orthotomic circle ; i.e. the corresponding arc of the orbit is the arc cut off 
by the polar (in respect to the directrix circle by which the hodograph is determined) of 
the centre of the orthotomic circle; the portion of this polar intercepted by the orbit 
is the elliptic chord, and this elliptic chord and the sum of the radius vectors at the 
two extremities of the elliptic chord determine the time of description of the arc ; 
and the values of these quantities, viz. the elliptic chord and the sum of the radius 
vectors, must be the same in each orbit. 
orbit, 
The analytical investigation is not difficult. I take as the equation of the first 
a( 1 — e 2 ) 
1 + e cos (6 — ct) ’ 
then the polar of the orbit with respect to a directrix circle r = c is 
c 2 e 
r 2 7t 5\ ^ cos (6-vr) —r- = 0, 
a( l-e 2 ) v ' a 2 (l — e 2 ) 
and putting c 2 = h V/s Va (1 — g-) (where Jc is a constant quantity, i.e. it is the same in 
each orbit), the equation becomes 
bJc Jc ,/i \ Jc? . 
r 2 —.. . .=r r cos (u — w) = 0. 
va (1 — e 2 ) a 
But since a is supposed to be the same in each orbit, we may for greater simplicity 
write Jc 3 = m 2 a; it will be convenient also to put e = sin k ; we have then 
r = 
a cos 2 k 
1 + sin K COS (6 — zr) 
for the equation of the orbit, r 2 = ma cos k for the equation of the directrix circle, and 
r 2 — r .m tan k. cos (6 — vr) — m 2 = 0 
for the equation of the hodograph. 
We have in like manner 
a cos 2 k 
1 + sin k! cos (6 — -sr') 
for the equation of the second orbit, r 2 = ma cos k for the equation of the corresponding 
directrix circle, and 
r 2 — r. m tan k' cos (0 — -nr') — m 2 = 0 
for that of the hodograph.