The Variation in Attraction Due to the Attt-acting Bodies. 203 



Draw the tangent lines P c, P a and P 1. Through the points of tan- 

 gency c, a, 1 draw line c g so that the part 1 g is perpendicular to P C. 

 Then by known demonstration of conio sections the line c g is straight, and 

 g a is to g c as B to A, also g a is to g 1 as A. to B. tSince triangles 1 P g, a 

 P g, and c P g have each a right angle at g and same base P g, the tan- 

 gents of the angles 1 P g, a P g, and c P g are directly as g 1, g a, and g c. 

 Since the right angled triangles C 1 g, C a g, and C c g have the same per- 

 pendicular C g, the tangents of the angles C 1 g, C a g, and C c g are in- 

 versely as g 1, g a, and g c. But angle a P g equals C a g. therefore angle 

 1 P g equals Cog, and angle c P g equals C 1 g. Also angle c P a equals 

 a C 1, and angle a P 1 equals c C a, and line C 1 produced is perpendicular 

 to tangent line P c. 



Produce major axis A A to P', making distance C P' to C P as A to B. 

 Draw P'm tangent to circle having radius Cm equal to C A, and draw P' B, 

 tangent to ellipse having axes A A and B B. Through points of tangency 

 m and B, draw m g'. na g' is perpendicular to P' C and parallel to P C. By 

 constructions angle m P' C equals a P C or C a g. Because then m g' is to 

 B^ g' inversely as A to B, angle C B, g' equals c P g. Therefore B, C or B, B, 

 is ijarallel to tangent line P c, and B, B, is a conjugate diameter to C C,. 

 The angle c C B, is greater or less than a right angle by the angle c C 1 or a-. 



Let angle C a g be represented by 3^, C c g by 5— z, and C 1 g by 3^ + y. 

 Hereafter in this investigation Sr, or the angle in any ellipse in the situa- 

 tion of 3^, is called the Elliptic Angle, (3— z) is called the Alpha Elliptic, and 

 i^+j), the Beta Elliptic Angle. 



It is now evident that: 

 Cg = B sin 3. 



Cg' = A sin 3. 

 ag = B cos 2r. 



A 



eg =: -g B COS 3 ^ A cos 5. 



mg' = A COS 3. 



B g' = A COS 5 = B cos 5. 



In right angled triangle C c g; 



C c = A =^/ A'^ cos- 3 +B= sin- 3 = i/B^ + CA'^-B^) cos^ 3= 



V'A-^-CA'^-B'-Osin-'^. 

 -^.=B |/1 + E, cos- 3- = A^/l-E^ sin- 3. 



Likewise from right angled triangle C B, g', 



B, =Bv/1 + E;- sin'^ 3 = A v^l-E5cos-^3. 



