QUADRATURE o the CONIC SECTIONS, &c. 305 
(1 ee ee t Ls 
| ae —(Etant rite ptand 5 t tan's ae © tan ds. 
pM Sa aa : y 
t +08 m3 
36. Ler us now fuppofe the hyperbolic fector ACB to be di- 
vided into 2” equal parts, by lines drawn from the centre to the 
points I, 2, 3, 4,-..-7 in the curve, and tangents to be drawn 
at the extremities of the hyperbolic arch AB, and at the alter- 
nate intermediate points of divifion 2, 4, 6, &c. fo as to form 
the polygon AFF’ F’ F’ BC. Then, by a known property of 
the hyperbola, the triangles ACF, FC 2, 2 CF’, FC4,... F” CB 
are all equal, and as their number is 2”, the whole polygon 
bounded by the tangents, and by the ftraight lines AC, CB will 
be equal to the triangle ACF taken 2"times. But the area of 
this triangle is > AC xAF= - tan = (becaufe AF = tan ap 
: & 
therefore 2” tan— expreffes twice the area of the polygon 
AFF’ F’ F’ BG. Let Q denote this area, then, fubftituting = Q. 
for 2” tan = and multiplying all the terms of. the feries by 2, 
we have 
TO IT Ir. T I T 
ae tan —- 5s = tan = “Ss - tan- §s — tan — Sie. 
aint ne ( 2 x 7 8 mi 8 «16 
Q 2: I Ss 
+70 5) 
Now 
