DR. BOOTH ON THE GEOMETRICAL PROPERTIES OF ELLIPTIC INTEGRALS. 381 
the line GOm. Through D draw the line DmG^, Join OG„ it will cut the sphere in 
rtp the vertex of the principal arc Z«,. Let OZ—k, then ZG=/l:tance, and as CZF 
is a right-angled triangle. CO=ZD=^^^ 7 |====:|, k and B being the semiaxes of the 
maximum cylinder. As all the bases of the cylinders are similar. 
F “ — * • 
Now as ZOG and ZDG' are similar triangles, ZG : ZO : : ZD : ZG', or 
k k 
k tana : k : : - : ZG', or ZG' = -^t — . But ZG'=k tana,, hence tana tanai=l, or the 
arcs a and a^ are connected by the equation established in (317.). 
When we require to know which of these successive curves on this sphere is the 
spherical parabola, the same construction will enable us to determine it. Draw 
ZT, a tangent to the circle on OD, take ZT'=ZT"=ZT, and join T' and T" with O 
cutting the sphere in c and c'. Zc=Zc' is the principal semitransverse arc of the 
spherical parabola, for ZT^=^" tan^a=OZ.ZD = y, or tan^a=j. 
TT 
As ZT' > ZO, cZd > or the principal arc of a spherical parabola is always greater 
than a right angle. Since in the spherical parabola 7 -l- 2 g= 2 , the angle COT'=2s, or 
COT is equal to the distance between the foci of the curve. 
LVI. If we revert to the general formula (307.) and take ff as the quadrant of a 
spherical parabola, the integrations with respect to 6 must take place between 6=0, 
and 6,=tan~'^-^^, for e^=i, in (300.) gives tan0=^^. Hence 
1 f 
Since 
dip 
"Tt 
[I 
v/jJ 
]- 
sin6 cosd 
(1— i), when tan0=^Y^ . 
I 
df 
”7T 
1 —j'^siu^O 
w TT 
Putting the sum of these integrals = a, we shall have <’'=2~ 
But (68.) gives for the quadrant of the spherical parabola 
/T = 
/ r 
( 1+0 
W- 
djix. 
42 
( 1+0 
sm> 
Comparing these expressions for the same arc ?, 
djtx. 
42 
■f A, 
(319.) 
(1+0^ 
fjb being taken between the limits |«-=0, and |a^=tan“‘^-^^. 
It is easy to show that the integrals of the first order in Art. LIII. may be represented 
