Methodus inveniendi lineas curvas

142 DE USU METHODI $dy\sqrt{(yy+uu)}$. Porro cum sit $adx = ydt + \frac{1}{2}(udy+ydu)$ & $dt = \frac{2dy}{u}$, erit $audx = yydy + \frac{1}{2}uudy + \frac{1}{2}yudu$, & $du = \frac{2adx}{y} - \frac{2ydy}{u} - \frac{udy}{y}$. Jam ponatur $dy=pdx$, minimum esse debebit $\int \frac{pdx\sqrt{(yy+uu)}}{u}$, atque $u$ est quantitas cujus valor ex hac aequatione $du = dx(\frac{2a}{y} - \frac{2yp}{u} - \frac{up}{y})$ definiri debet. Pertinet itaque haec quaestio ad Casum quintum; cum quo si comparatio instituatur, fit $u=n$ & $Z = \frac{p\sqrt{(yy+n^2)}}{n}$, unde $L = \frac{-yyy}{n^2\sqrt{(yy+n^2)}}$; $M=0$, $N = \frac{yp}{n\sqrt{(yy+n^2)}}$, & $P = \frac{y\sqrt{(yy+nn)}}{n}$. Deinde cum fit $n = \int dx(\frac{2a}{y} - [TRUNCATED]