Methodus inveniendi lineas curvas

154 DE USU METHODI $y$, & formulae $\int dx \sqrt{(1+pp)}$ seu $dB = nv. dx ( - \frac{1}{x})$ $d. \frac{p}{\sqrt{(1+pp)}} = -nv. d. \frac{p}{\sqrt{(1+pp)}}$. Quibus valoribus in aequatione $BdA = AdB$ substitutis, prodibit pro curva quaesita sequens aequatio: $Bdx = -Ad. \frac{p}{\sqrt{(1+pp)}}$. Ponatur $\frac{A}{B} = e$, ita ut, pro abscissa $x = a$, area curvae fiat aequalis producto ex arcu in hanc constantem $e$. Erit ergo $dx = -ed. \frac{p}{\sqrt{(1+pp)}}$, & integrando $x = b - \frac{cp}{\sqrt{(1+pp)}}$, seu $cp = (b-x)\sqrt{(1+pp)}$, hincque $p = \frac{b-x}{\sqrt{(c^2-(b-x)^2)}} = \frac{dy}{dx}$. Erit ergo $y = \dots$ $\int \frac{(b-x)dx}{\sqrt{(c^2-(b-x)^2)}} = f \pm \sqrt{(c^2-(b-x)^2)}$ seu $(y-f)^2 + (b-x)^2 = [TRUNCATED]