138

the solution is impossible , and if ER touchthe curve ACB, t is then at its minimumwith respect to y, and the problem may inthat case be solved by art. 19. and 24. Butif ER cut the curve ACB, it must by art.11. cut it in two points as G and S ; and inthat case the problem admits of two solu-tions agreeable to the observation made inCase XII. of art. 9. For, through C and Gdraw CGM cutting KR in M, and throughM draw the ordinate ML; and constructthe catenary CPM, similar to CNG as inthe last case; then shall CPM be one ofthe catenaries required; and by proceedingwith the point S, as we have done with G,the other catenary having the proposed data,may be constructed. For since EQ =MR,and LE = MK, being opposite sides ofparallelograms, by adding equals to equals,we have QL = KR = t and LM = EK==y,

and these being proportional to the likequantities EF and FG in the catenary CGF,the problem is evident by art. 13.