Methodus inveniendi lineas curvas

IN RESOLVENDIS QUAESTIONIBUS. 167 P, &c. fient $=0$. Porro, ob $\Pi = \int dx \sqrt{(1+pp)}$, erit $[Z] = \sqrt{(1+pp)}$, & $d[Z] = \frac{pdp}{\sqrt{(1+pp)}}$, ex quo erit $[M] = 0$, $[N] = 0$, & $[P] = \frac{p}{\sqrt{(1+pp)}}$. Jam est $\int L dx = n \int e^{n \int dx \sqrt{(1+pp)}} dx$, cujus valor, posito $x=a$, erit $=nB$; hincque $V = n(B - \int e^{n \int dx \sqrt{(1+pp)}} dx)$. Per Regulam ergo datam fiet $dB = nr.dx(-\frac{d.[P]V}{dx}) = -nr.d.\frac{p(B - \int e^{n \int dx \sqrt{(1+pp)}} dx)}{\sqrt{(1+pp)}} = -nr.d.\frac{nBp}{\sqrt{(1+pp)}}$, ob $dB = BdA$. Integrando itaque erit $\frac{np(B - \int e^{n \int dx \sqrt{(1+pp)}} dx)}{\sqrt{(1+pp)}} = \frac{nBp}{\sqrt{(1+pp)}} - nb$; hincque $\frac{ [TRUNCATED]