in the higher Geometry. 387 
Now, it was before demonstrated, that the parameter of LM 
is equal to AP x MP -j- q . AG 
( m + Thig is therefore 
*■ * 1 * 3 m -J- 2 n 
^!—X.JZrf x ACD + q. AN x AG 
equal to — ===== , multiplying both by 
m + n , , 
■^$Tn * * ~ p + p 
+p, we have ^ x ACD + q . AN x AG 
= AP x (MP x (p + + 7 ■ AG). 
From these equals take q . AG x AN, and there remains 
ffffe x FF> X ACD equal to AP x PM x +/>) 
+ 5 . AG x AP — AN ; or, dividing by q — p ; x ACD 
= AP x -ZL±i_i_. 2 _ x PM + - 3 — x AG x AP - AN. Now, 
~A .! L x AP x PM is equal to the area APM; therefore, the area 
APM together with x AP. PM, and x AG x AP - AN, 
or APM with — p — x AP. PM Z_ x AG x AN — AP, 
or APM 4 - -Z-r x AP . PM q —x reel. PT is equal to 
1 q —p q —p ^ M+N 
xACD. Now,IC'is an hyperbola of the order/ -fg; therefore, its 
area is - - j — x rect. GH . MH. But q is greater than p\ therefore, 
jZT q is negative, and is the area MHKC' ; and the 
area NTKC' is equal to x GT x TN ; therefore, MNTH is 
equal to (MHKC 7 — NTKC'), or to^~x GH.MH — GT.TN? 
From these equals take the common rectangle AT, and there 
3 D 2 
