Conic Sections. 259 
of the conic section ARH relative to the axis of abscisse AM, be 
represented by (t), (uv), and QK,HK, the rectangular coordinates 
of the same point H relative to the axis of abstisse: QK and origin 
Q, by (t’) (w’). Draw RF parallel to AM, and put 2 VRF=s, 
ZKCM=¢, AL=d, AG=k, RQ=7, QG=i, RV=x, and VT, 
_the common section of the planes HTZ and RVWT, equal to (y). 
Since ARH is a conic section, the relation between (¢) and (uw) may 
be exhibited by the equation w? = (pt+-ct? ;) or a?u? —pb?t—— 
cb*t? =0; which evidently characterizes a right line when p=0 and 
c=1; acircle when a=b=p and c= —1; an ellipse when (a) and 
(5) are unequal, a=p and c=—1; an hyperbola when a=p and 
e=1; anda parabola when a=p and c=0. Draw KE parallel to 
HM and QE,KI perpendicular to KE, meeting KE and HM re- 
spectively at E and I. The relation between (¢’) (u’) may be de- 
termined, agreeably with the usual method of transforming coordi- 
nates, by putting t=AM=AG+QE —IK=k-+ cos. 9.t/—sin. 9.u’, 
and w=HM=QG+KE-+HI=é-+sin. 9.t/+cos. 9.u’, and substi- 
tuting these last values of (¢), (w) in the above equation of ARH. 
From this substitution we have 
a?42 —b? (pk-+-ck? (. 
+ [2a?4 sin. 9 —b3 (p+2ck) cos. ot! ot 
+ [2a?0cos. 94-5? (p+2ck) sin. glu’ | _ He ey, ST 
“hits 1-0) sin. fs eer " | +-mi'? 
where h, 6, e, f, m, n, represent the corresponding saetaats in the 
left rhetebers Resolving the abridged equation (1) in reference to 
ci ((ekf\>_ Maem)! But 
2n 2n n 
(u’) we obtain u/=— 
(drawing RP parallel to CK meeting HK at P) 2 VRP=a-—g, 
and consequently t/=QK=RP=cos: (« —¢).7; therefore, supply- 
ing the place of (¢’) by cos. (w—-¢).x in the equations above, 
se ehe ((Eem oon) 
results u’= — 
h+6 cos. So) ae cos. (w =x) ay, The plane HTZ, 
being Sinclar to CK, is perpendicular to. the plane ARHM, 
and is also a circular section of the solid. The planes HTZ and * 
