[b]Folium of Descartes — Algebraic Proof& Key Properties[/b][br]Definition. The Folium of Descartes is the plane algebraic curve defined implicitly by  x³ + y³ − 3 a x y = 0, where a > 0 is a parameter.[br][br][b]1. Symmetry and Double Point[/b][br]Symmetry about y = x follows by exchanging x and y in the equation, which leaves it invariant. Setting x = y in the equation gives 2x³ − 3 a x² = x² (2x − 3 a) = 0, so the curve meets y = x at the origin (a double point) and at (3a/2, 3a/2). The origin (0, 0) is a double point because the lowest-degree terms in x and y are of total degree 2 (the term −3 a x y), so two real branches intersect there.[br][br][b]2. Parametric Representation (Rational Parametrization)[/b][br]Let t = y/x (the slope through the origin). Then y = tx. Substituting into the implicit equation:[br]x³ + (t x)³ − 3 a x (t x) = x³ (1 +t³) − 3 a t x² = 0.[br][br]Assuming x ≠ 0 and 1 + t³ ≠ 0, solve for x:[br]x = (3 a t)/(1 + t³).[br][br]Then y = t x gives[br]y = (3 a t²)/(1 + t³).[br][br]Thus a rational parametrization is x(t) = 3 a t / (1 + t³),    y(t) = 3 a t² / (1 + t³),    t ∈ℝ \ {t : 1 + t³ = 0}.[br][br][b]3. Asymptote[/b][br]Using the parametrization, compute x + y:[br]x + y = (3 a t + 3 a t²) / (1 + t³) = 3 a t(1 + t) / [(1 + t)(1 − t + t²)] = 3 a t / (1 − t + t²).[br][br]As t → −1, the denominator 1 − t + t² → 3 while 1 + t → 0 in the original fraction, and the point (x, y) recedes to infinity along a line. A more direct method is to set u = 1/t and expand x and y for small u; one obtains x + y →−a. Hence the oblique asymptote is x + y = −a.[br][br][b]4. Tangent Slope by Implicit Differentiation[/b][br]Differentiate F(x, y) = x³ + y³ − 3 a x y = 0[br]implicitly:[br]3 x² + 3 y² (dy/dx) − 3 a [y + x[br](dy/dx)] = 0.[br][br]Collect dy/dx terms:[br](3 y² − 3 a x) dy/dx = 3 a y − 3 x².[br][br]Therefore dy/dx = (a y − x²) / (y² − a x).[br][br]The slope is undefined when y² − a x = 0 (vertical tangents), and zero when a y− x² = 0 (horizontal tangents).[br][br][b]5. Intersection with y = x[/b][br]Substitute y = x into x³ + y³ − 3 a x y = 0 to get 2x³ − 3 a x² = 0. Hence x = 0 (double point) or x = 3 a / 2, so the non-origin intersection is (3 a / 2, 3 a / 2).[br][br][b]6. Area of the Loop[/b][br]Using the polar form r(θ) = [3 a sin θ cos θ] / [sin³[br]θ + cos³ θ] (for θ not causing the denominator to vanish), or using the[br]rational parametrization with Green’s Theorem, one evaluates the area A of the[br]interior loop as A = (3/2) a². Thus the area enclosed by the loop equals 1.5 a².[br][br]Drag Test:[br]Drag the slider aaa and watch the loop grow/shrink, the asymptote shift, and the intersection point move.