AN ESSAr OX CVCLOIDAL CURVES. 



561 



Peoposition viii. Problem. (Plate 8. 

 Fig. 63.) To find a geometrical equation for 

 tbe conchoidal epitrocboid. 



Let CP:=PK. On the centre C describe a circle equal to 

 GM, cutting SC in Z. Join MZF, then the arc DZ=:GM, 

 and MZ is parallel to CK, therefore EF is also equal to 

 DZ or GM, CF is parallel to KM, arid MF=CK : there- 

 fore this epitrochoid is the curve named by Delahire the 

 conchoid of acircular basis, as was first observed by Reau- 

 mur in 1708, and afterwards by Maclaurin, in 1720. Call 

 CK, a, DE, b, ZII, r, HM, y, ZM, s ; and let ZI be 



perpendicular to CK ; then FZzza — s, CI— - 



and 

 l< 



CIZ and ZHM being similar, CZ : Clr.ZM : ZH, or 



a— 5 



——:'.s:x; hence /i^as — ss, bx4-ssZlas, /I'a'+Shs'' 

 2 • . 



+j*z:dV, and by substituting for s', I'-l- it-r' — o'J^'■' + 

 6'I'^-2.T'J'-^- ibxy' -^ y' — a'y':ZO. 



Corollary 1. Join FN, and complete the parallelo- 

 gram MFNL; then since EFzzDZizEN, FN is perpendicu- 

 cular to EK, and ML to NL, and, NL being always equal 

 to FM or CK, L i» always in a circle described on the cen- 

 tre N, LM a tangent to that circle, and ZM a perpendicu- 

 lar to that tangent drawn from the point Z. 



Corollary 2. (Plate 8. Fig. 64.) Tlie unicuspi- 

 date epicycloid admits of a peculiar central equation, 

 with respect to the point S. Call SM, s, and let ST:=u be 



i 



For IP 



and the trian- 



perpendlcular to the tangent MT, then k= { — j 



being half of SM, or s, SP =: v^ (—) 



gles SIP andMTS being similar, SP5 : SMi/XIPj • ST.?, 



or — : i';;— : «', and 2at/^s'. 



Corollary 3. The unicuspidate epicycloid is one of 

 the caustics of a circle. For making the angle CllY 

 = MRC = JMKP=iSCP, the triangle CRY is isosceles, 

 and CY is constant; so that all rays in the direction of the 

 tangent MR will be reflected by the circle QR towards Y, 

 and consequently SM will be the caustic of a radiant point 

 atY. 



Proposition i.x. Peoble.m. (Plate 8. 

 Fig. 65.) To find a geomeliical equation for 

 the tricuspidate hypocycloid. 



Let PA and MF be perpendicular to CS. Join PMB, KM, 

 RMG, and PD. Then the angle APB is equal to the dif- 

 ference of APC and MPR, or to that of their complements 

 PRM, PCA : but PRM=i PKM=:|PCA, therefore AP 



VOL. II. 



= 1 PCA=ADP=:APS, and the triangles APS and APB 

 are similar and equal. Ixt SC=o, SF=.t, FM=:y, and 

 SB=r. Then SA ; SP::SP: SD, and SP^v' (ar). Draw 

 PE perpendicular to BP ; then BE—SDzz'ia, BC=:o— r, 

 £0=30— r, and by similar triangles, CP: CR'.IEC : CG 

 =1 EC=:a— ir; therefore GB=:ii- : but BE : BG :: BP .- 



BM or 2a:|r:: v" [ary.r —^Z («r)=:BM ; again, BP: BM :; 



BA: BF, or v'(ar) : -^*/(ar):;.I- -. IL, andSF=:x=r_ 



3a 2 6a 



rr 

 — , 6ax=:6ar~rr, andrZZ3a± ./ {gaa — 6ax). But 



r' r' 



MFq=BUq — BVg, 0Ty'=: , and 36a'- »/«=: ' 



<Ja 36aa •' 



iar'—r'. By adding to this the square of the former equa- 

 tion, and proceeding in the same manner to exterminate r, 

 we obtain an equalion of the value of x and y, which, 

 when the surds are brought to the same side, and the 

 square of tlie whole is taken, is at last reduced to x'~'^r.x' 

 +'ix'y' — \iaxy'+y' + iiay=:o, a regular equation of 

 the fourth order. 



Scholium. The equation of the corresponding liypotro- 

 choids may be investigated nearly in the same manner, by 

 dividing PR and PM in a given ralio, but the process will be 

 somewhat more tedious. 



Proposition x. Problem. (Plate 8. 

 Fig. 66.) To find a geometrical equation for 

 the bicuspidate epicycloid. 



Let CPzzPR. Join RMT, PM, PD ; draw CT perpen- 

 dicular U) RT, TE to CR, and EG, MB, RA, to SC. Then 

 the angle DRP=iMKP=SCP, and by equal triangles, RA 

 =CT, and RD=:CD, and by similar triangles RM : RP:: 

 RE : RT, and RP : RD::RT : RC : therefore RM : RU:: 

 RE : RC, and ME is parallel to SC, and EG=BM. Put 

 CI'=:a, BC=x, BM=:y, CM:=s, Cr=u ; then by prop. ;, 

 u'-i {s''—a'); or ^^'=^'-0'; but RC : CT::CT . CE 



:;CE : EG, or y, hence 2/=- 



16aY, -^u^zz ■ ' 



o<y'=(si—aa)'=:(j-x-t-i/y— as)'; whence by involution the 

 equation of the sixth order may be had at length. 



Corollary. Since CRM^SCR, a ray in the direction 

 of the tangent MR will be reflected, by a circle FR, always 

 parallel to SC ; therefore SM is the caustic of the circle 

 FR when the incident rays are parallel to CS. 



Proposition xi. Problem. (Plate 8. 

 Fig. 67.) To find a geometrical equation for 

 tbe quadricuspidate hypocycloid. 



Let CR=:PR, then the angle PRM = 1PKM=2PCS. 

 4C 



