[b]Algebraic Proof of the Cissoid of Diocles[/b][br]Goal. Show that the cissoid of Diocles has implicit Cartesian equation (x² + y²) x = 2 a y² given its geometric construction. Throughout, a > 0.[br][br][b]1. Classical set‑up (translation/rotation to standard position)[/b][br]Place the origin at O = (0, 0). Let C be the circle centered at (a, 0) with radius a, so its equation is x² + y² = 2 a x. Let L be the vertical line x = 2a. For a ray from O with polar angle θ, define the two intersections along this ray: the intersection with L lies at polar radius r_L = 2 a sec(θ), and the intersection with C lies at polar radius r_C = 2 acos(θ). By the cissoid construction, the cissoid point on this ray is at polar radius[br] r = r_L − r_C = 2 a (sec(θ) − cos(θ)).[br][br][b]2. Polar simplification[/b][br]Use the identity sec(θ) − cos(θ) = (1 − cos²(θ)) /(cos(θ)) = sin²(θ) / cos(θ). Hence the polar radius can be written as: r = 2 a (sec(θ) − cos(θ)) = 2a · sin²(θ) / cos(θ) = 2a · sin(θ) · tan(θ).[br][br][b]3. Parametric form via t = tan(θ)[/b][br]Let t = tan(θ). Then sin(θ) = t / √(1 + t²) and cos(θ) = 1 / √(1 + t²). In Cartesian coordinates (x, y) = (r cos(θ), r sin(θ)), we therefore get:[br]x = r cos(θ) = (2a · sin²(θ) / cos(θ)) · cos(θ) = 2a · sin²(θ) = 2a · (t² / (1 + t²))[br]y = r sin(θ) = (2a · sin²(θ) / cos(θ)) · sin(θ) = 2a · sin³(θ) / cos(θ) = 2a · (t³ / (1 + t²))[br][br][b]Thus a rational parametrization is obtained:[/b][br]x(t) = (2 a t²) / (1 + t²),[br]y(t) = (2 a t³) / (1 + t²).[br][br][b]4. Elimination of the parameter[/b][br]From the parametrization we immediately have y = t x (since y/x = t). Using x = (2at²)/(1 + t²), multiply by (1 + t²): x (1 + t²) = 2 at².[br]Substitute t = y/x so that t² = y² / x². Then: x (1 + y² /x²) = 2 a (y² / x²).[br][br]Multiply both sides by x²: x³ + x y² 2+ ay². Factor x from the left-hand side: (x² + y²) x = 2 ay².[br]This is exactly the desired implicit Cartesian equation of the cissoid of Diocles.[br][br][b]5. Consistency checks and key features (algebraic)[/b][br]• Cusp at the origin: Setting x = 0 in (x² + y²) x = 2a, y² gives 0 = 2a, y² ⇒y = 0; thus the curve meets the origin only at a single point (a cusp).[br]• Vertical asymptote x = 2a: From x(t) = (2 at²)/(1 +t²), as |t| → ∞ we have x(t) → 2a while |y(t)| → ∞[br](since y(t) ~ 2at). Hence the curve has the vertical asymptote x = 2a.[br]• Agreement with the geometric construction: Along each ray (fixed θ), the cissoid point is placed at radius[br]r = 2a(sec(θ) − cos(θ)), which matches the difference of the radii to L and C. Converting this to (x, y) yields the same locus as the Cartesian equation.[br][br]6. Summary[br]Starting from the classical definition (difference of distances to the vertical line x = 2a and the circle [br]x² + y² = 2ax along each ray from the origin), we obtained the polar equation r = 2a(sec(θ) − cos(θ)),[br]simplified it to r = 2a · sin²(θ) / cos(θ), converted to the rational parametrization x = 2at² / (1 + t²), [br]y = 2 at³ / (1 + t²), and then eliminated the parameter to derive the Cartesian implicit equation[br] (x² + y²) x = 2 ay².[br][br][b]Drag Test[/b][br][list=1][*]Moving [b]M2[/b] along the circle dynamically moves [b]OM2[/b].[br][/*][*][b]Point P[/b] automatically traces the Cissoid.[br][/*][*]Adjusting [b]slider a[/b] changes the circle and Cissoid proportional[br][/*][/list]