When is a quadratic Bézier curve nearest the origin?

Question

Consider a planet moving along a quadratic Bézier curve through points A B C, with $t$ = time: $\qquad \operatorname{curve}( t, A, B, C ) \equiv t' (t' A + t (2B - A)) \ + \ t (t' (2B - C) + t C ) $, where $ t' = 1 - t $
When is the planet nearest the sun ? That is, is there a way of finding
$\qquad \operatorname{argmin}_t \lVert \operatorname{curve}( t ) \rVert^2 $
better than running a 1d minimizer on this 4 th degree polynomial ?
("Better" means simple, foolproof code, preferably self-contained. I don't need much accuracy, but do need $t_{min}$, not the nearest point — when, not where.)

Answer 1

At the point where the distance is minimal, changes of $t$ in either direction increase the distance. This means that the tangent direction has to be orthogonal to the radius. If you have

\begin{align*} x(t) &= at^2 + bt + c & \dot x(t) &= 2at + b \\ y(t) &= dt^2 + et + f & \dot y(t) &= 2dt + e \end{align*}

then the vectors are orthogonal if their dot product is zero, i.e. if

$$ (at^2 + bt + c)(2at + b) + (dt^2 + et + f)(2dt + e) = 0 $$

That's just a cubic equation in $t$. I must confess that I'm slightly confiused by the degree, because I'd expect the points of minimal and maximal distance as two solutions, but what's the third?

Well, perhaps I'm thinking too much about real planets. Note that planets in general don't really move along quadratic Bézier curves, and that even if they approximately do, the parameter of the curve might not correspond to time in the physical sense. Turning a position back into a time for an exact conic trajectory is much harder.