146 
OX THE AREA OE THE CONIC SECTION REPRESENTED BY THE 
[192 
we may eliminate 77, /¿, and the result is 
KL 
a + 
Pr 2 * 
h , 
9 » 
1 , 
6 + 
h , 9 
KM 
Pr 2 ’ 
/ > 1 
, KN _ 
/ > C+ P^> 1 
= 0, 
which may also be written 
(ST, 53', 6', S', $'$i,1, 1) 2 = 0, 
where (2T, ...) are what (.21, ...) become when a, b, c are changed into 
we in fact have 
KL , KM KN 
a Pr 2 ’ ^ Pr 2 ’ C Pr 2 ’ 
r “ a+ ^ (6Jr+CJf)+ ^ JfJ ’ 
8'=8 ~p V ,Lf. 
and (observing the value of P) the result consequently is 
P + ^ i Kb + e-2f)L + (e + a-2g)M + (a + b-2h)N] + ~ i (MN + NL + LM) = 0, 
which may also be written 
P 3 r 4 + PKr 2 {(b + c — 2f) L + (c + a — 2g) M + (a + b — 2h) iV} + 4A 2 K 2 = 0. 
Hence if r l} r 2 are the two semiaxes, we have 
4A 2 iT 2 
TiT 2 
ps ' ’ 
and the area is irr^ which is equal to 
which agrees with Mr Ferrers’ result. 
2irKA 
\/(P s ) ’ 
The formula r 2 = L% 2 + Mrf + Ng 2 which is assumed in the preceding investigation 
may be proved as follows: