466 
ON THE DETERMINATION OF THE [476 
or nearly x = = '08; a second approximation gives « = ‘0802; or we have A 2 = "9198, 
A = -9592, whence H = 43° 49'. Substituting in the equation 
4 o 
^ (9 + 4A 2 ) - (3 - 2A) R, + (3 + 2A) R 3 = 0, 
v3 
this will be satisfied by 6 = — 1, viz. the parabola belongs (as it obviously should do) 
to a point of AB' within the triangle BB'B". 
To obtain the other positive root we may write the equation in the form 
*o , 141-75 182-25 
x 63 + ~v sr- 
the approximate value A 2 = 63, gives more nearly A 2 = 65 and then 
141-75 128-24 
A 2 = 63 + 
nr> =65-177, 
65 4225 
whence A 2 =8"073 or H= 82° 56'. Substituting in the equation 
40 
V3 
(9 + 4A 2 ) - (3 - 2A) B 2 + (3 + 2A) R 3 = 0, 
we have 6 = +1, viz. this parabola belongs to a point of B'C' within the triangle 
BB'B". 
The two values of e for 6 = +1 and 6 — — 1, are each infinite for A = 0, and 
they become equal for A = oo (viz. when the orbit-pole is on the ecliptic), but 
not in any other case; in fact they can only do so for 9 + A 2 = 0, or else for 
(3 - 2A) B 2 = (3 + 2A) R 3 , that is, A (288 + 128 A 2 ) = 0, viz., A (9 + 4A 2 ) = 0. 
113. In further explanation I give a diagram of the eccentricity. 
The base AB'G'B is here the broken line AB'CB' of figure 10: the ordinates 
along the base AC'(=90°) of the two continuous curves exhibit the values of e, as