DR. BOOTH ON THE GEOMETRICAL PROPERTIES OF ELLIPTIC INTEGRALS. 403 
Again, (49.) shows, when we measure the arc from the minor principal arc, that 
cot0=^ I ; or cot 0=^^ tan -4;'. Now in order that we may compare these arcs 
together, we must have 6=K. Hence 
„ , , tan^/3 1 
(b.) 
When we substitute for <p any particular value, (a.) and (b.) will give the coi respond- 
sina 
ing values of tan-v// and tan ■4/' ; but when we make tan^9=^T^=j, the values of %4' and 
become equal, or the compared arcs together constitute the quadrant. 
LXXIV. To determine the inclination, to the horizontal plane, of the tangent 
drawn to any point of the spherical ellipse. The spherical ellipse being taken as the 
curve of intersection of a cylinder by a sphere as in (X.), through a side Rr of the 
cylinder let a plane be drawn, it will cut the sphere in a small circle, which will touch 
the spherical ellipse in the point r, and will cut the base of the hemisphere in the right 
line RP, which touches the base of the cylinder at the 
point R. Let O be the centre of the sphere and Z the 
centre of the spherical hyperconic. Through the line OZ 
let a plane be drawn at right angles to the plane of the 
small circle RrrP, it will cut the sphere in the arc of a 
great circle Zt at right angles to the arc rr ; and as the 
three planes, namely, the horizontal plane, the plane of 
the small circle, and the plane of the great circle ZOP-r, 
are mutually at right angles, the right lines in which 
they intersect PR, P^r, PO are mutually at right angles, 
therefore P is the foot of the perpendicular drawn from the centre O of the base of 
the cylinder, to the tangent RP which touches the curve. P is also the centre of 
the small circle ArT, since AB is a cord of the sphere. Hence At is a quadrant, 
and therefore, vt or v is the inclination of the element of the spherical ellipse at 
r to the base of the hemisphere. Now ZO is the radius of the sphere, and Pr that 
of the small circle. RPO is a right angle, and therefore OR^=OP^+PR^. Hence 
Rr*=Or^— OR^ Now for the moment putting A and B for the semiaxes of the base 
of the cylinder, OP^=A^ cosX-l-B^sinX, and 
Fig. 26. 
z 
sin^X, cos% 
cos®A;+ sin^X, ' 
Whence OR'^= 
cos®X^+ B"* sin^X^ 
A^ cos^Xj + B^ sin^X^’ 
(a.) 
and therefore Rr^=Or^ 
Let Or=l, A=sina, B=:sin/3, . . (b.) 
