Posts tagged 'cubic'

F.1 Albert’s Parabola

Given a parabola y = x^2 and a point P = (P_x, P_y) on it, the tangent line T and normal line N at P satisfy y = 2 P_x x - P_x^2 and y = -x/(2 P_x) + P_x^2 + 1/2; requiring N to pass through (M, 0) yields the depressed cubic P_x^3 + (1/2) P_x - M/2 = 0.

F.30 Isosceles Triangles

An isosceles triangle of given semiperimeter s and area K has half-base x = c/2 governed by the cubic 2 s x^3 - s^2 x^2 + K^2 = 0, which under x = y + s/6 reduces to Cardano’s depressed cubic y^3 - (s^2/12) y - (s^3/54 - K^2/(2s)) = 0.