M. B. PORTER—SPHERICS. 
47 
4. The polar reciprocal of the spherical ellipse is a curve enveloped 
by the base of a triangle of constant area whose vertical angle is fixed. 
In the case of the ellipse, therefore, the sum of the base angles is con¬ 
stant, and in the case of the hyperbola the difference. The lines polar 
to F and F 7 are called the cyclic lines for reasons that will appear later. 
THE TANGENT. 
tfig 
5. Theorem. A tangent to a spherical ellipse bisects the angle be¬ 
tween the focal radii drawn to that point. Let P be the given point, 
P 7 a point very near the first, and let 
F 7 s and Ft be laid off on FP and F 7 P 
equal to FP 7 and F 7 P 7 . In the right- 
angled triangles PP't and PP 7 s we will 
have PP 7 common and Pt equal to Ps. 
As the point P 7 approaches P the arcs 
P 7 t and P 7 s will more nearly coincide 
with the perpendiculars and PP 7 continually approach the bisector of 
angle tPF 7 . Conversely, it may be proved that if PT bisect the angle 
tPF' it can have only one point in common with the curve. 
(a) . In the reciprocal figure we have: the tangent intercepted be¬ 
tween the cyclic lines is bisected by the points of tangency. 
(b) . From I., above, it follows that confocal conics intersect at 
right angles. 
6. Theorem. If two tangent lines be drawn to the reciprocal curve 
they intersect the cyclic lines in four concyclic points , and the center 
of the circle is the pole of the line through the points of tangency. 
I shall prove the converse. In figure 3, afb—gfd. 
If C is the center f-\-d=t-\-a , and if s is added to 
both sides, we get a-j-b=g-j-d. C is a quadrant from 
<[> and where # bisects dg and C is on the perpen¬ 
dicular to the supplement of gd at its middle point, 
(c). Passing back to the spherical ellipse. The 
Fi gi 3 , four focal vectors drawn to any two points on an 
ellipse touch a circle whose center is the intersection of the tangents 
drawn at these points. 
7. Many focal properties are readily deducible from [2 and 5]. In 
figure 4, produce TF so that TL=F'T. The triangle PFT=triangle 
PTL and PL—PF'. 
Producing FT' so that T'S=T'F'; trian¬ 
gles PT'S and PT'F 7 are equal and PS=PF 7 
=PL, so that triangles PSF and PFL are 
equal. Thus it follows that angle LFS is 
Fig . 4 . bisected by PF. 
