in the higher Geometry. 393 
fourth proportional to the arc OO, the rectangle AB x 2 . FE, and 
the line CL ; this cycloid shall cut the ellipse in M, so that, if 
MC be joined, the area ACM shall be to CMB : : M : N. 
Demonstration. Let AP =2 x, PM=y, AC=c, AB=tf, 
and 2 . EF = h ; then, by the nature of the cycloid GMR, 
— PM : OO :: 2 . FE x AB : rL,andQO=AO — AQ= (by const.) 
m ' -p ' n x AK AQ-* a ^ so » — AB — 2 . AC, since AC — LB. 
Therefore, _ PM : j^x AK—AQ : : AB * 2 . EF: AB-a . AC ; or 
—y : - ^ —x arc . 90 0 — arc .vers. sin. x : : a b : a — 2 c ; therefore, 
< — y ( a — 2 c) or-j-y (2 c — a) — abx 
and, by transposition, 
I M 
ivi ' +N x arc '9° — arc.v.s.a: 
a b . M 
abx arc . v. s. x -f- y (s.£ — a) = x arc . 90°. To these 
equals, add 2y (x — x) = 0, and multiply by — 1 ; then will 
M , 
M-f N X X arC *9° 9 
ab x arc. v.s.x -f- (20: — a)y — 2 y {x — c)- 
of which the fourth parts are also equal ; therefore, 
y , \ a b M , 
T C ) ~ X M-f N X alC * 9 ° • 
x ax — x z , and 
abx arc. v. s. x , (2 x — a) y 
_ - -f 
by 
Now, because AFB is an ellipse, y L - 
y z=J—s/ ax — x 2 ; therefore, —■ x a J c • v - s - x _j_ x Ls/ t 
y , x ab M „ -» *■ 1 • 1 , 1 
{ x — c ) = — x m ^ N " x arc . 90 . Multiply both numerator 
ax — x 
2 4 
and denominator of the first and last terms by a; then will 
b a z .2 x — a b / I v . . /, 
- x 7 X arc . v.s. x + x-Vax—x'—A (x—c) = A x 
a* m 
x m"+n x arc * 9°°- Now, the fluxion of an arc whose versed 
* is equal to 
2 1 z vax~x i ’ 
sine is x and radius is equal to 
which is also the 
fluxion of the arc whose sine is^^ — and radius unity ;* wherefore, 
* The semi-transverse is supposed unity, through this demonstration. 
MDCCXCVIII. 3 E 
