ON THE SPHERE AND CYLINDER, II 
49 
luce K'M to 
How this ingenious analysis was suggested it is not possible 
to say. It is the equivalent of' reducing the four unknowns 
h, h', k, k' to two, by putting h = r + x, h! — r—x and k! = y, 
r.h', 
and then reducing the given relations to two equations in x, y, 
which are coordinates of a point in relation to Ox, Oy as axes, 
where 0 is the middle point of AA', and Ox lies along OA', 
while Oy is perpendicular to it. 
Our original relations (p. 47) give 
A' + A'M), 
7 , ah' r — x 7 ah r + x 7 m h + k 
y = k = = a 5 k = = a, > and — = .. • 
h r + x h r—x n h+k 
We have at once, from the first two equations, 
(1) 
7 r + x 9 
ky = a y = a, 
J r — x J 
I)]- ( 2 ) 
rs equal to a. 
in the figure. 
A' or h', and 
whence (r + x) y = a (r — x), 
and (;x + r) (y + a) = 2 ra, 
which is the rectangular hyperbola (4) above. 
7 7 (i + a:) ("l + ^ 
. m h + k \ r — x/ 
Agam - *“*'+*' = .- ;/, Try 
whence we obtain a cubic equation in x, 
(3) 
diameter, and 
{r + x) 2 ('r + a — x) = ^ {r — x) 2 {r + a + x), 
which gives 
m / fr + a + xx 2 / v 2 2 
— (r—xY\ ) = (r + aY — x 2 . 
n y ' \ r+x / 
• 
tj , y a . y + r — x r + a + x 
J4ut = , whence = > 
r — x r + x r — x r + x 
and the equation becomes 
mce 
2ra, (4) 
ola with KF, 
^{y + r—x) 2 — {r + a) 2 — x 2 , 
which is the ellipse (3) above. 
1523,2 
E