Parabola from Vertex and Two Points on the Arc

Notes: [list] [*]Finding the roots of a cubic equation: {link} [*][b]Orientation:[/b] Ialways let [math]{\small {\bf x}(t)} [/math] open upward (in the direction of [math]{\small {\bf \hat v}} [/math]). From a given value of tan θ , I take 0 < θ < π . The angles are signed. I have used complex rotation, not because I like being abstract, but because (for example) complex multiplication by [math]{\small {\bf z}_α} = (x({\bf a}) +i y({\bf a}))/a [/math] corresponds always to "rotation by angle α". More on Signed Angles:[url]http://www.geogebratube.org/material/show/id/99837[/url] Complex Rotation: [url]http://www.geogebratube.org/material/show/id/115348[/url] [*]Equation (3) usually has ONE real solution, and the parabola is uniquely determined by the constraints. When there are instead THREE real solutions, I choose the value of θ which is closest to the previous value. This can always be done so that the axes rotate smoothly (no leaps). As a result, if A and C begin on opposite arms of the parabola, they will stay on opposite arms. Put another way, v will always point inside the triangle ΔVAC). [/list]