Methodus inveniendi lineas curvas

210 DE METHODO cus $s$ evanescit, posito $p = \frac{A}{B}$; ex quo valor ipsius $x$ ita de- terminari debet, ut posito $p = \frac{A}{B}$ fiat $= 0$. Quod si ponatur $\mathfrak{N} = 1$; habebitur ex priore constructio- ne, statim $dx = \frac{(B+s)ds}{\sqrt{(A^2+(B+s)^2)}}$; ideoque $x = \sqrt{(A^2}$ $+ B^2 + 2Bs + ss)} - \sqrt{(A^2+B^2)}$, seu posito $B=b$, & $\sqrt{(A^2+B^2)} = c$, erit $x+c = \sqrt{(c^2+2bs+ss)}$. Ex posteriore autem construendi modo, oritur $x = -$ $\int \frac{dp}{p\sqrt{(1+pp)}} = \frac{A\sqrt{(1+pp)}}{p} + b$, seu $(x-b)p =$ $c\sqrt{(1+pp)}$; hincque $p = \frac{c}{\sqrt{((x-b)^2-c^2)}} = \frac{dy}{dx}$. Quare cum fit $y = \int \frac{cdx}{\sqrt{((x-b)^2-c^2)}}$; [TRUNCATED]