1879.] proof of Lambert's theorem. 265 



P and p respectively ; then it may be shewn as before that the 

 chords are equal and subtend isochronous arcs, and that SQ + SQ' 

 is equal to sq + sq. 



With this latter construction it follows from Newton's Prin- 

 cipia (Lib. i. Sect. ii. prop. 6, theor. 5) as an immediate conse- 

 quence of the equality of the intercepts FE and pe that (the forces 

 at P and p being equal) the arcs QQ' and qq, supposed infim- 

 tesimal, are described in equal times : Lambert shews that these 

 arcs are still isochronous when the equal intercepts oxe finite. 



III. 



Another proof for the Ellipse by orthogonal projection is im- 

 plicitly contained in Sectio IV. §§ 180, 181, 186. 



a. Through a variable point E on SP draw a double ordinate 

 QQ' (bisected in 0) to the diameter through P, and let the ellipse 

 be projected on to its auxiliary circle, an-d let small letters be now 

 used to denote the points in the circle corresponding to the above. 

 Then, since 



PE : PK=PO : PC = po : pG, 

 (the point K being determined as in the first proof,) it follows 

 that the diametral sagitta po of the circular arc qq is equal to PE. 

 By making E coincide with 8 we deduce that the altitude of 

 the triangle Sqq is equal to SE. 



b. Hence, if we take two ellipses having equal auxiliary 

 circles, as in the construction II., and take SP and PE in the one 

 and equal corresponding lengths in the other, it follows (1) that 

 po is the same for both, and therefore that the isochronous arcs of 

 the ellipses correspond to equal arcs of their auxiliary circles ; and 

 (2) that the altitude of the triangle Sqq is the same for both. 



Hence the area of the sector Sqq is the same in both circles, 

 and the corresponding areas in the ellipses are therefore in the 

 subduplicate ratio of their latera recta*. 



* We can at once write clown an expression for the circular area Sqq' in terms 

 of the segments of the line PESK, viz. from the relation 



sector ySg2' = sector Cqq' - A [Cqq' - Sqq'). 

 Let the semi-axes of the ellipse be a and h, and let PE=x=po, and SP^r. Then 



sector Sqq'^a^ vers"! - - {a-r) V(2cfa; - «")• 

 therefore equal to 

 a" vers~i - - (a - r) ^(2«a; -x-)\ . 



The elliptic sector SQQ' is therefore equal to 

 b 



