Astronomy.

^o W order to determine the Area A, we must4 . ^ ern ‘-conjugate HK, thus: Because AB —C, therefore FC - x — a\ and by similarj, * es SDC and FEC we have S C : S D :: FC :

G that , b X — ab „ ,

4C is, a : b x a :- — F E j and

th,

lerefore F G = (FE — GE)Again, S C

^ & :• Tr ^ — /~— TT x — a

' * C ; C E : or a : a—x — s :-

a

0 -« V~

Hence DE or SG=CE+CD

X

- b z -f- sf ft

a

s/a'-h 1 . But

iqb

■ therefore F S _ \/ SG' F G

b z x z

>; .-j-—------

Y -— 4 a b 1 x -f- 4 cs- b % a* x

—^ a z

V ct G

~4 a b 1 x 4 ct l^T. And therefore S H = f S F

V' ■——___

A. jj' Moreover, by the Nature of an Ellipsis, SK ^and therefore \/ SFC*— SH~HKz=:^v* 2 . -f- 4 ab'- x — 4 a 1 b %

th,

4 st 2

y/ a x ■

a »

l ^f orc I K _ 2 H K ~ s/Ts- Consequent-

ly^ __

^ a = A P T B, the Semi-Area of the

Jf; Let Q = Diameter of the Circle ALB,

the f its Periphery; then since ' L H X P ~{ QJ? isQ, of the Circle, we shall have Qj : * Q_P (:: 1V GQR) :: AQRB-.ALB:: APTB-.AIB ::

bp x

4 a q

f: > " xb ,

? That is, T : 4 qp —\/ —«a :

= But2ATB = AJKB = A

F 4

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