CONIC SECTIONS. 
LEMMA, I. 
Fig. 15. If there be any number of right 
lines, as D E, P Q, and F G, all parallel to 
one another, and all terminating in the 
same two right lines D F and EG; then a 
right line, as B C, which bisects two of 
the parallels will bisect all the rest. 
Draw D H L and E K R parallel to B C : 
because D B = B E and F C = C G, there- 
fore F L = R G. It is plain that F L : 
P H : : R G : K Q ; and therefore P H == 
K Q, consequently P O = O Q. 
LEMMA. II. 
Fig. 16. If a right line A B, or a right 
line produced, be so divided in C and D, 
that ACxCB=ADxDB: then 
AC = BD and AD = CB. 
Bisect A B in O. Then the difference of 
A and A C X C B is equal to C (5 
and 6 . 2 . E); and the difference of A 
and A D X D B is equal to D : there- 
fore C O- = D O^, whence the lemma is 
manifest. 
PROP. XI. 
Fig. ir, 18, 19, 20. If a right line, as 
B C, bisect two parallel right lines, D E 
and F G, terminated both ways by a conic 
section, or opposite sections: the same 
right line B C will bisect every other right 
line, as P Q, terminated by the section, or 
opposite sections, and parallel to the two 
fonner right lines. 
Join D F and E G: then these lines are 
either parallel to one another, or, being 
produced, if necessary, they will meet, 
I. When D F and E G are parallel : (fig. 
ly.) let P Q meet these lines in M and 
]M ; then DMxMF:PMxMQ :: 
EN X NG : PN X NG (Pr. 10): but 
it is plain that DM x MF=:ENx 
N G ; therefore PMx MQ=PN x 
N G. Therefore P M = N Q (Lem. 2.) j 
and it is obvious that the right line B C, 
which bisects D E and F G, likewise.bisects 
P Q (Lem. 1.) 
II. Let F D and E G meet in a point 
H: (fig. 18, 19, 20.) assume any point, O, 
in the plane of the conic section, and through 
it draw T K, R S, and L I, terminated by 
the conic section, and respectively parallel 
to E G, D F, and D E or F G ; let P Q 
meet D F and E G in M and N. It is ma- 
nifest that D F and E G are similarly divid- 
ed in M and N, and also in the point of 
concourse H. Therefore 
DMxMF:ENxNG::FHxHD: 
GH X HE. 
Because T K is parallel to E G, and R .5 td 
D F : therefore 
FH X HD : GHxHE :: RO x OS; 
T O X O K. 
Consequently DM X M F : E N x N G :: 
R O X O S : T O x O K. Hence, and by 
Prop. 10, we have the following propor- 
tions : 
PM X MQ ; D Mx M F LOxOI ; 
ROxOS. 
DMxMF:EN xNG::ROxOS: 
T O X O K. 
ENxNG:PNxNQ:;TOxOK: 
LOxOI, 
Therefore, ex aequo, P M x M Q : P N X 
NQ::LOxOI:LOxOI. 
Consequently PMxMQ = PNxNQ; 
and P M =3 Q N (Lem. 2). Therefore tlie 
right line B C, which bisects D E and F G, 
will likewise bisect P Q (Lem. 1). 
Def. 9. A right line which bisects two 
parallel right lines, both terininafed by a 
conic section, or opposite sections, is called 
a diameter of the section, or opposite sec- 
tions. This definition relates merely to the 
position of the iliaineters, and not to their 
magnitude. 
Def. 10. The centre of an ellipse, or 
opposite hyperbolas, is d point in which is 
bisected every right line drawn through it, 
and terminated both w'ays by the ellipse, or 
opposite hyperbolas. 
PROP. XII. 
Fig. 21, 22. To find the centre of an 
ellipse, or opposite hyperbolas, given by 
position. 
Draw two parallel right lines, as D E and 
F G, terminated botli ways in the ellipse, 
or one hyperbola, or one of them in one hy- 
perbola, and the other in the opposite hy- 
perbola : draw' the right line B C to bisect 
both the parallels D E and F G : then it is 
plain that B C will in all cases meet both 
the opposite hyperbolas ; for it will bisect 
all the right lines that can be drawn in 
both, parallel to D E and FG (11.) : let it 
meet the ellipse and opposite hyperbolas 
in B and C, and bisect B C in .4, then is A 
the centre required. 
Let H be a point in the ellipse, or one of 
the hyperbolas, and draw H L M parallel 
to D E or F G ; take A N = A L, and draw 
P K through N parallel to D E or F G. 
— Then H M and K P terminated by 
the ellipse, or opposite hyperbolas, are bi- 
sected in L and N : and, because B L x 
LC=BNxNC, therefore H L x L M, 
or H L' = K N X N P, oi- K N' (Pr. 10 ): 
